Class 11 Engineering Maths - Statistics

Find the median of the following data: $2, 7, 3, 15, 12, 17$ and $5$ .

Median is the middle value of a given set of number arranged in the ascending order.

Given set of numbers are

$2,7,3,15,12,17$ and

$\ 5.$

Arranging the numbers in ascending order we get,

$2,3,5,7,12,15,17$ .

The middle number is

$7$ . Hence the median is

$7$ .

Find the mode of $14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18$ .

Ascending order:

$14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28$

$14$ has highest frequency

$: 4$ times

Mode

$= 14$

Calculate the coefficient of range for the following data:
Heights in cm. $120-124$ $125-129$ $130-134$ $135-139$ $140-144$
No. of students $6$ $9$ $10$ $12$ $8$

First let us find the limits of heights (limits of class intervals) and their mid points.

Height	Height limits	Midpoint of heights	No. of students
$120-124$	$119.5 - 124.5$	$122$	$6$
$125 - 129$	$124.5 - 129.5$	$127$	$9$
$130 - 134$	$129.5 - 134.5$	$132$	$10$
$135 - 139$	$134.5 - 139.5$	$137$	$12$
$140 - 144$	$139.5 - 144.5$	$142$	$8$

Upper limit of the highest height,

$H = 144.5$ , lower limit of the lowest height,

$L = 119.5$ .
Coefficient of range

$= \dfrac {H - L}{H + L} = \dfrac {144.5 - 119.5}{144.5 + 119.5} = \dfrac {25}{264} = 0.095$ .

Calculate the coefficient of range for the following data.
No. of wards $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
No. of houses $32$ $57$ $28$ $96$ $138$ $90$ $66$ $58$

Here

$H = 8, L = 1$ . Hence,
Coefficient of range

$= \dfrac {H - L}{H + L} = \dfrac {8 - 1}{8 + 1} = \dfrac {7}{9} = 0.78$ (approximately).

Calculate the mean deviation from the mean for the scores given below:
$15, 11, 13, 20, 26, 18, 21$

Arrange the scores in an order:

$11, 13, 15, 18, 20, 21, 26$ .
You know how to calculate the mean for the set of scores. Just take the sum of the scores and divide it by the number of scores to get mean.
Mean

$\overline {X} = \dfrac {\sum X}{N}$ .
We have

$\sum X = 11 + 13 + 15 + 18 + 20 + 21 + 26 = 124$ and

$N = 7$ .
Therefore

$Mean\ X = \dfrac {\sum X}{N} = \dfrac {124}{7}\approx 17.7$ .

The formula for coefficient of variation (C.V.) is given by _______________.

The formula for coefficient of variation

$(C.V)$ is given by

$C.V.=\dfrac {\sigma}{\mu}\times 100$ , where

$\mu$ is mean for the population and

$\sigma$ is standard deviation.

The weight(in kg) of $13$ students in a class are $42.5, 47.5, 48.6, 50.5, 49, 46.2, 49.8, 45.8, 43.2, 48, 44.7, 46.9, 42.4$ . Find the range and coefficient of range.

Let us arrange the given data in the ascending order.

$42.4, 42.5, 43.2, 44.7, 45.8, 46.2, 46.9, 47.5, 48, 48.6, 49, 49.8, 50.5$
From the given data, the largest value

$L=50.5$ and the smallest value

$S=42.4$
Range

$=L-S$

$=50.5-42.4=8.1$
The coefficient of range

$=\displaystyle\frac{L-S}{L+S}=\frac{50.5-42.4}{50.5+42.4}=\frac{8.1}{92.9}$

$=0.087$

The smallest value of a collection of data is $12$ and the range is $59$ . Find the largest value of the collection of data.

$\Rightarrow$ Here,

$S=12$ and

$R=59$

$\Rightarrow$

$R=L-S$

$59=L-12$

$L=59+12$

$\therefore$

$L=71$

$\therefore$ The largest value of the collection of data is

$71$ .

Write two merits of mode.

(1) It is a simple and popular average.
(ii) It is the possible value of the distribution.

The largest value in a collection of data is $7.44$ . If the range is $2.26$ , then find the smallest value in the collection.

Range

$=$ largest value

$-$ smallest value

$\rightarrow 7.44-$ smallest value

$=2.26$

$\therefore$ The smallest value

$=7.44-2.26=5.18$

Find the mode of the following data. 12, 24, 36, 46, 25, 38, 72, 36, 25, 38, 12, 24, 46,25, 12, 24, 46, 25, 72, 12, 24, 36,25, 38 and 36.

Arranging given numbers in ascending order,

$12,12,12,12,24,24,24,24,25,25,25,25,25,36,36,36,36,38,38,38,46,46,46,72,72.$

$x$	$f$
$12$	$4$
$24$	$4$
$25$	$5$
$36$	$3$
$38$	$3$
$46$	$3$
$72$	$2$

Arranging given numbers in ascending order,

$12,12,12,12,24,24,24,24,25,25,25,25,25,36,36,36,36,38,38,38,46,46,46,72,72.$

$\Rightarrow$ From above table we can see frequency of

$25$ is maximum.

$\therefore$

$Mode=25$

Write the relation between standard deviation of a set of scores and its variance.

Standard deviation

$(\sigma)$ is the square of Variance

$(\sigma^{2})$

The standard deviation is the square root of variance.

If variance is

$9$ then the standard deviation is

$3$

What is the range of the data : $48, 65, 27, 23, 44, 41, 25, 70, 49$ ?

Solution -

Range of data

$\rightarrow 48,65,27,23,44,41,25,70,49$

Range = Highest value in data - Lowest value in data

$= 70-23$

$= 47$

1100659_1187759_ans_a66ba96c4cf64dd8a44316c205918f44.jpg

Find the mean $S.D$ of $1,2,3,4,5,6$ .

We have,

$1,2,3,4,5,6$

$x$	${{x}^{2}}$
1	1
2	4
3	9
4	16
5	25
6	36
$\sum{x}=21$	$\sum{{{x}^{2}}}=91$

We know that

$\sigma =\sqrt{\dfrac{\sum{{{x}^{2}}}}{n}-{{\left( \dfrac{\sum{x}}{n} \right)}^{2}}}$

$\sigma =\sqrt{\dfrac{91}{6}-{{\left( \dfrac{21}{6} \right)}^{2}}}$

$\sigma =\sqrt{\dfrac{91}{6}-{{\left( \dfrac{7}{2} \right)}^{2}}}$

$\sigma =\sqrt{\dfrac{91}{6}-\dfrac{49}{4}}$

$\sigma =\sqrt{\dfrac{364-294}{24}}$

$\sigma =\sqrt{\dfrac{70}{24}}$

$\sigma =\sqrt{\dfrac{35}{12}}$

$\sigma =\sqrt{2.916}\approx 1.70$

Hence, the $S.D$ is $1.70$ .

Find the sum of deviation of all observations of the data 5, 8, 10, 15, 22 from their mean.

5, 8, 1, 15, 22

${\text{Moon = }}\dfrac{{{\text{sum of observation}}}}{{{\text{No}}{\text{. of obesuation}}}}$

$= \dfrac{{5 + 8 + 10 + 15 + 22}}{5}$

$= \dfrac{{60}}{5} = 12$

Sum of deviations

$= |5 - 12| + |8 - 12| + |10 - 12| + |15 - 12| + |22 - 12|$

$= |7| + | - 4| + | - 2| + |3| + |10|$

$= 7 + 4 + 2 + 3 + 10$

$= 26$

Find the mode of this data
$2,5,4,3,5,2,5,2,4,5$

Let us put the data in a tabular form-

Digits	Tally bar	Frequency
$2$	$\|\|\|$	$3$
$3$	$\|$	$1$
$4$	$\|\|$	$2$
$5$	$\|\|\|\|$	$4$
	Total	$10$

$5$ has occurred the most number of times, thus the mode of the given data is

$5$ .

Hence, the mode of the given data is

$5$ .

The rainfall ( in mm) in a city on $7$ days of a certain week was recorded as follows:

Days

Mon

Tue

Wed

Thurs

Fri

Sat

Sun

Rainfall

( in mm)

0.0

12.2

2.1

0.0

20.5

5.5

1.0

i) Find the range of the rainfall in the above data.

Since, Range = Largest variate - Lowest variate

$= 20.5 mm - 0.0 mm = 20.5 mm$

The runs scored in a cricket match by $11$ players is as follows:
$6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15$
Find the mean, mode and median of this data. Are the three same?

Ascending order the data is,

$6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120$

Sum of runs

$= 6+8+10+10+15+15+15+50+80+100+120 = 429$

Number of matches

$= 11$

$Mean = \dfrac{429}{11} = 39$

$Median =$ Middle value of the data

$= 15$

$Mode =$ Most occuring observation of the data

$= 15$ (3 times)

Three are not same

Calculate the variance and standard deviation of the following continuous frequency distribution
Class interval $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$
Frequency $3$ $7$ $12$ $15$ $8$ $3$ $2$

Class Interval Mid Value

$(x)$ frequency

$(f)$

$f{x}$

$f{x^2}$

$30-40$

$35$

$3$

$105$

$3675$

$40-50$

$45$

$7$

$315$

$14175$

$50-60$

$55$

$12$

$660$

$36300$

$60-70$

$65$

$15$

$957$

$63375$

$70-80$

$75$

$8$

$600$

$45000$

$80-90$

$85$

$3$

$255$

$21675$

$90-100$

$95$

$2$

$190$

$18050$

$n=\sum f=50$

$\bar x=\dfrac{1}{n}\displaystyle\sum {f x_{i}}=\dfrac{3082}{50}=61.64$

Variance

$(\sigma ^{2})=\dfrac{1}{n}\displaystyle\sum f{x^2} -(\bar x)^2=\dfrac{202250}{50}-(61.64)^2=245.5104$

Standard deviation

$(\sigma)=\sqrt{245.5104}=15.67$

Find the standard deviation of $9,16,23,30,37,44,51$ .

$Mean(\overline{x})=\dfrac{9+16+23+30+37+44+51}{7}=\dfrac{210}{7}=30$

$\Rightarrow$ Standard deviation

$=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{7}}$

$=\sqrt{\dfrac{(9-30)^2+(16-30)^2+(23-30)^2+(30-30)^2+(37-30)^2+(44-30)^2+(51-30)^2}{7}}$

$=\sqrt{\dfrac{(-21)^2+(-14)^2+(-7)^2+(0)^2+(7)^2+(14)^2+(21)^2}{7}}$

$=\sqrt{\dfrac{441+196+49+49+196+441}{7}}$

$=\sqrt{\dfrac{1372}{7}}$

$=\sqrt{196}$

$=14$

$\therefore$ Standard of deviation

$=14$

Find the median of:
$26, 33, 41, 18, 30, 22, 36, 45$ and $24$

Data :

$26, 33, 41, 18, 30, 22, 36, 45, 24$

Data in increasing order :

$18, 22, 24, 26, 30, 33, 36, 41, 45$

No. of terms

$:9$

Median

$=\left(\dfrac{n+1}{2}\right)^{th}$ term

Median

$=5^{th}$ term

$\therefore$ Median

$=30$

The S. D of $15$ items is $6$ and if each item is decreased by $1$ , then standard deviation will be

$\Rightarrow$ If each item of a data is increased or decreased by the same amount.

$\Rightarrow$ The SD of the data remains unchanged.

$\therefore$ S.D of new set will be

$6$

Find the median of:
$3.2, 4.8, 5.6, 7.3, 8.9$ and $9.1$

Data :

$3.2, 4.8, 5.6, 7.3, 8.9, 9.1$

No. of terms

$=6$

Median

$=\dfrac{(n/2)^{th}\ term +(n+1/2)^{th}\ term }{2}$

Median

$=\dfrac{3^{rd}\ term +4^{th}\ term}{2}$

Median

$=\dfrac{5.6+7.3}{2}$

Median

$=\dfrac{12.9}{2}=6.45$

$\therefore$ Median

$=6.45$

Let $a, b, c, d$ and $e$ be the observations with mean ${ m }$ and standard deviation $s$ . The standard deviation of the observations $a$ $+ { k } , b+k ,c+k , d+k$ and $e+k$ is

The given observations are

$a,b,c,d,e$

The give mean is

$m$ .

$\therefore$

$m=\dfrac{a+b+c+d+e}{5}$

Let new mean be

$n$

The new observations are

$a+k,b+k,c+k,d+k,e+k$

$\Rightarrow$

$n=\dfrac{a+k+b+k+c+k+d+k+e+k}{5}$

$=\dfrac{a+b+c+d+e+5k}{5}$

$=\dfrac{a+b+c+d+e}{5}+\dfrac{5k}{5}$

$\therefore$

$n=m+k$

The standard deviation

$(s)=\sqrt{\dfrac{(a-m)^2+(b-m)^2+(c-m)^2+(d-m)^2+(e-m)^2}{5}}$

Find the median of:
$80, 48, 66, 61, 75, 52, 45$ and $70$

1100934_1192594_ans_dfba3fe4bcfc4d0886c498cf3129de56.jpg

Find the coefficient of range for the following data.
Size $10-15$ $15-20$ $20-25$ $25-30$
Frequency $2$ $4$ $6$ $8$

Largest value

$\left( L \right) = 30$

Smalles value

$\left( S \right) = 10$

Therefore,

Coefficient of range

$= \cfrac{L - S}{L + S} \\ = \cfrac{30 - 10}{30 + 10} \\ = \cfrac{20}{40} = \cfrac{1}{2} = 0.5$

Following tare the lives in hours of $15$ pieces of the components of aircraft engine $.$ Find the median $:$ $715,724,725,710,729,745,694,699,696,712,734,728,716,705,719$

Lives in hours of is pieces are

$=$

$714,724,725,710,729,745,694,699,696,712,734,728,716,705,719$

Arrange the above data in ascending order

$=$

$694,696,699,705,710,712,715,716,719,721,725,728,729,734,745$

$N=15(odd)$

Medium

$= {\left( {\frac{{N + 1}}{2}} \right)^{th}}\,terms$

$= {\left( {\frac{{15 + 1}}{2}} \right)^{th}}\,terms = {8^{th}}\,terms = 716$

Find the median of $8, 15, 7, 19, 16, 12$ and $10$

Arranging the given data in ascending order-

$7, 8, 10, 12, 15, 16, 19$

$\because n = 7 =$ Odd

Therefore,

Median

$= {\left( \cfrac{n+1}{2} \right)}^{th} \text{ term}$

$= {\left( \cfrac{7 + 1}{2} \right)}^{th} \text{ term}$

$= {4}^{th} \text{ term} = 12$

What is mode ?

The observation, which occurs most frequently among the given observations, i.e., the value of the observation having maximum frequency is called mode.

If the median of $\cfrac{a}{3},\cfrac{a}{2},\cfrac{a}{4},\cfrac{{2a}}{5},\cfrac{a}{6}$ is $12$ then find the value of $a\left( {a > 0} \right)$ .

We have,

$\cfrac{a}{3},\cfrac{a}{2},\cfrac{a}{4},\cfrac{{2a}}{5},\cfrac{a}{6}$

On taking LCM of denominators,

$\cfrac{a}{3}=\dfrac{20a}{60}$

$\cfrac{a}{2}=\dfrac{30a}{60}$

$\cfrac{a}{4}=\dfrac{15a}{60}$

$\cfrac{2a}{5}=\dfrac{24a}{60}$

$\cfrac{a}{6}=\dfrac{10a}{60}$

Now, arranging in asscending order

$\cfrac{a}{6},\cfrac{a}{4},\cfrac{a}{3},\cfrac{{2a}}{5},\cfrac{a}{2}$

So, the median term is

$3^{rd}$ , we get

$\dfrac{a}{3}$

Therefore,

$\dfrac{a}{3}=12$

$a=36$

Hence, this is the answer.

The median of the given data 6,8,9,15,x,29,25,21,22 is 18, find value of x ?

Given data is

$6,8,9,15,x,29,25,21,22$

The ascending order of data excluding

$x$ is

$6,8,9,15,21,22,25,29$

For median to be

$18$ ,the middle element should be

$18$ (as the data is odd)

So the only possibility is

$x=18$

Calculate variance and standard deviation of the following data :
10, 12, 8, 14, 16

$AM, \overline x =\dfrac{10+12+8+14+16}{5}=\dfrac{60}{5}=12$

Variance =

$\dfrac{(10-12)^{2}+(12-12)^{2}+(8-12)^{2}+(14-12)^{2}+(16-12)^{2}}{5}$

$=\dfrac{40}{5}=8$

$SD(\sigma) =\sqrt{variance} =\sqrt{8}$

Find the median of the following :
33,31,48,45,41,92,78,51,61

The numbers are

$33, 31, 48, 45, 41, 92, 78, 51, 61$ .

Arranging in ascending order , we get,

$31, 33, 41, 45, 48, 51, 61, 78, 92$

Therefore, median = central term =

$48$

A die is tossed thrice. Getting an even number is considered as success. What is the variance of the binomial distribution ?

Here,

$n=3$

p = Probability of getting an even number in a single throw of a die

$p=\dfrac{3}{6}=\dfrac{1}{2}$

And,

$q=1-p=1-\dfrac{1}{2}=\dfrac{1}{2}$

Therefore,

Variance =

$npq = 3\times \dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{3}{4}$

Find the mean deviation from the mean of the following data, using the step deviation method:
Marks No. of students
0-10 6
10-20 5
20-30 8
30-40 15
40-50 7
50-60 6
60-70 3

Marks	${x}_{i}$	${f}_{i}$	${d}_{i} = \cfrac{{x}_{i} -A}{h}$	${f}_{i} {d}_{i}$
$0-10$	$5$	$6$	$-3$	$-18$
$10 - 20$	$15$	$5$	$-2$	$-10$
$20 - 30$	$25$	$8$	$-1$	$-8$
$30 - 40$	$35 = A$	$15$	$0$	$0$
$40 - 50$	$45$	$7$	$1$	$7$
$50 - 60$	$55$	$6$	$2$	$12$
$60 - 70$	$65$	$3$	$3$	$9$
		$\sum{{f}_{i}} = 50$		$\sum{{f}_{i} {d}_{i}} = -8$

Mean

$\left( \bar{x} \right) = A + \cfrac{\sum{{f}_{i} {d}_{i}}}{\sum{f}_{i}} \times h$

$\Rightarrow \bar{x} = 35 + \cfrac{-8}{50} \times 10 = 33.4$

Now,

$\because$ Mean deviation

$= \cfrac{\sum{{f}_{i} \left| {x}_{i} - \bar{x} \right|}}{\sum{{f}_{i}}}$

${x}_{i}$	$\left\| {x}_{i} - \bar{x} \right\|$	${f}_{i} \left\| {x}_{i} - \bar{x} \right\|$
$5$	$28.4$	$170.4$
$15$	$18.4$	$92$
$25$	$8.4$	$67.2$
$35$	$1.6$	$24$
$45$	$11.6$	$81.2$
$55$	$21.6$	$129.6$
$65$	$31.6$	$94.8$
		$\sum{{f}_{i} \left\| {x}_{i} - \bar{x} \right\|} = 659.2$

Therefore,

Mean deviation

$= \cfrac{659.2}{50} = 13.184$

Hence the mean deviation of the given data is

$13.184$ .

From the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range
Variate
15
18
20
22
25
27
30
Frequency
4
6
8
9
7
8
6

We write the variates in cumulative frequency table.

Variate	Frequency (f)	Cumulative frequency
15	4	4
18	6	10
20	8	18
22	9	27
25	7	34
27	8	42
30	6	48

(i) Here the number of observations, $n = 48$ which is even.

So $median = \frac{1}{2}( n/2^{th} term + ((n/2)+1)^{th} term)$

$= \dfrac{1}{2}(48/2^{th} term + ((48/2)+1)^{th} term)$

$= \dfrac{1}{2}(24^{th} term + (24+1)^{th} term)$

$= \dfrac{1}{2} (24^{th} term + 25^{th} term)$

$= \dfrac{1}{2} (22+22)$ [Since all observations from $19^{th}$ to $27^{th}$ are $22$ ]

$= \dfrac{1}{2} \times 44$

$= 22$

Hence the median is 22.

(ii) Lower quartile, $Q_1 = (n/4)^{th} term$

$= (48)/4$

$= 12^{th} term$

$= 20$

Hence the lower quartile is $20$ .

(iii)Upper quartile, $Q_3 = (3n/4)^{th} term$

$= (3\times48/4)^{th} term$

$= (3\times12)^{th} term$

$= 36^{th} term$

$= 27$

Hence the upper quartile is $27$ .

(iv)Interquartile range $= Q_3-Q_1$

$= 27-20$

$= 7$

Hence the Interquartile range is $7$ .

A data has $13$ observations arranged in descending order. Which observation represents the median of data?

Total no. of observations

$=13$ , which is odd

$\therefore$ The median will be

$\left(\dfrac {n+1}{2}\right)^{th}$ term

$=\left(\dfrac {13+1}{2}\right)^{th}$ term

$=\left(\dfrac {14}{2}\right)^{th}$ term

$=7^{th}$ term
i.e.,

$7^{th}$ term will be the median.

The distribution below gives weights of $30$ students of a class. Find the median weight of the students.

Calculation of median

Weight (in kg)	Number of students $(f_{i})$	Cumulative frequency $(cf)$
$40-45$	$2$	$2$
$45-50$	$3$	$5$
$50-55$	$8$	$13$
$55-60$	$6$	$19$
$60-65$	$6$	$25$
$65-70$	$3$	$28$
$70-75$	$2$	$30$
Total	$\displaystyle\sum f_{i}=30$

We have

$\displaystyle\sum f_{i}=n=30\Rightarrow \dfrac{n}{2}=15$

The length of $40$ leaves of a plant are measured correctly of the nearest millimeter, and the data obtained is represented in the following table:
Find the median length of the leaves.

Here, the classes are not in inclusive form. So, we first convert them in inclusive form by subtracting

h2h2

The average weight of $40$ Students is $32$ . If the Heaviest and Lightest are excluded the average weight reduces by $1$ . If only the Heaviest is excluded then the average is $31$ . Then what is the weight of the Lightest?

Average weight of

$40$ students

$= 32$ .

Total weight

$= 40 \times 32= 1280$ .

Average weight of

$38$ students except the heaviest and the lightest ones

$= 32 - 1 = 31$ .

Total weight

$= 38 \times 31 = 1178$ .

Sum of weights of the heaviest and the lightest

$= 1280 - 1178 = 102$ .

Average weight of

$39$ students except the heaviest one

$= 31$ .

Total weight

$= 39 \times 31 = 1209$ .

Weight of the heaviest

$= 1280 - 1209 = 71$

Weight of the lightest

$= 102 - 71 = 31$

The exam scores of all 500 students were recorded and it was determined that these scores were normally distributed. If Jane's score is 0.8 standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?

$s$ be the standard deviation and find the

$z$ score.

$z =\dfrac{x-m}{s}= \dfrac{0.8s+m-m}{s} = 0.8$

The percentage of student who scored above Jane is (from table of normal distribution)

$=1 - 0.7881 = 0.2119 = 21.19$ percentage

The number of student who scored above Jane is (from table of normal distribution)

$=21.19% 0f 500 =\dfrac{21.19}{100} \times 500=21.19\times 5=106$

25, 16, 26, 16, 32, 31, 19, 28 and 35

Firstly arrange the numbers in ascending order

16, 16, 19, 25, 26, 28, 31, 32, 35

Now since

n = 9(odd)

Therefore Median

$= \left ( \frac{n + 1}{2} \right )^{th}$

$= \left ( \frac{9 + 1}{2} \right )^{th}$

$= 5^{th}$

Thus the median is 26

Find the mode of the following distribution:
Income (in rupees) 500-600 400-500 300-400 200-300 100-200
Number of persons 17 20 30 18 15

Here, the maximum no. of persons i.e. 300 have income in the interval 300-400.
So, the modal class is 300-400.
lower limit (I) of modal class=300,
class size (h)= 100
frequency

$({ f }_{ 1 })$ of the modal class=30
frequency

$({ f }_{ 0 })$ of class preceding the modal class=18
frequency

$({ f }_{ 2 })$ of class preceding the modal class=20
Now, let us substitute these values in the formula

Mode=

$I+\left( \frac { { f }_{ 1 }-{ f }_{ 0 } }{ { 2f }_{ 1 }-{ f }_{ 0 }-{ f }_{ 2 } } \right) \times h$

$=300+\left( \cfrac { 30-18 }{ 2\times 30-20-18 } \right) \times 100$

$=300+\dfrac { 12 }{ 60-38 } \times 100$

$=300+54.54$

$=Rs. 354.54$

Find the mode of the following data:
Marks below 10 20 30 40 50
Number of students 3 8 17 20 22

Find the mode of the following distribution:
Marks more than 0 10 20 30 40 50
Number of students 50 46 40 20 10 3

Find the mode of the following distribution:
Marks 4 5 6 7 8 9 10
Frequency 3 8 10 12 16 12 10

Here,
the marks 8 has the maximum frequency '16'.

$\therefore$ Mode =8

Find the mode of the following distribution:
Age in (years) 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 8 14 25 28 25 10 5

Here, the maximum class frequency is 28, and the class corresponding to this frequency is 30-40.
So, the modal class is 30-40
Now, modal class = 30-40,
lower limit (I) of the modal class= 60, class size(h)=10
frequency

$({ f }_{ 1 })$ of the modal class=28
frequency

$({ f }_{ 0 })$ of class preceding the modal class=25
frequency

$({ f }_{ 2 })$ of class preceding the modal class=25
Now, let us substitute these values in the formula

$Mode = 1+\left( \frac { { f }_{ 1 }-{ f }_{ 0 } }{ 2{ f }_{ 1 }-{ f }_{ 0 }-{ f }_{ 2 } } \right) \times h$

$=30+\left( \frac { 28-25 }{ 2\times28-25-25 } \right) \times 10$

$=30+\frac { 3 }{ 6 } \times 10$

$=30+5$

$=35$

Find the mode of the following data:
Class interval 11 12 13 14 15 16 17 18
Frequency 2 9 3 4 8 7 8 3

Here, the class interval 12 has the maximum frequency '9'

$\therefore$ Mode =12

Find the median of
$3.2,4.8,5.6,5.6,7.3,8.9$ and $91$

Given
Numbers are

$3.2,4.8,5.6,5.6,7.3,8.9$ and

$91$
Total numbers

$=7$
Total number of numbers is odd
Hence, the median will be the middle number
Here, middle number is

$5.6$
Therefore, the median is

$5.6$

The marks (out of $100$ ) obtained by $7$ students in Mathematics examination are given below. Find the mode for these marks.
$99, 100, 95, 100, 100, 60, 90$

The marks (out of

100100) obtained by

77 students in Mathematics' examination are given below:

99,100,95,100,100,60,9099,100,95,100,100,60,90
Since

100100 repeats for maximum times.
So, the mode for these marks is

100100.

2.5 2.5 2.1 2.5 2.7 2.8 2.5

Here 2.5 has maximum frequency (4 times) so mode of data will be 2.5.

2 4 6 2 6 6 7 8

Here 6 has maximum frequency (3 times) so mode of data will be 6.

Find the mode of the following data.
Marks 35 36 37 38 39 40
No of Student 09 07 09 04 04 02

We have,

Marks	35	36	37	38	39	40
No of Student	09	07	09	04	04	02

Since the marks that occur the most are 35 and 37.
So, the mode of the given data is 35 and 37.

Find mode of the following frequency distribution :
x 3 4 5 6 7 8
f 2 4 6 3 2 1

5 has maximum frequency (6) so mode = 5

A data has 19 observations arranged in descending order which observation represents the median data?

Total number of observations = 19 (odd).

$\therefore$ Median will be

$= \left ( \dfrac{n + 1}{2} \right )^{th} \;term = \dfrac{(19 + 1)^{th}}{2}$

$= \dfrac{20}{2} = 10^{th} \;term.$
10th term will be the median.

Find mode of the following frequency distribution :
x 1.1 1.2 1.3 1.4 1.5 1.6
f 20 50 80 60 15 8

1.3 has maximum frequency (80) so mode = 1.3

2 5 7 5 3 1 5 8 7 5

Here 5 has maximum frequency (4 times) so mode of data will be 5.

The weights of $10$ students (in kg) are given below :
$40, 35, 42, 43, 37, 35, 37, 37, 42, 37$
Find the mode of the data.

In a data, the observation that occurs maximum number of times is called mode.

The weight of

$10$ students is as given below:

$40,35,42,43,37,35,37,37,42,37$

Fr7From the data it can be observed that the observation

$37$ occurs maximum number of times. Hence, mode of the data is

$37$ .

The number of members in 30 families of a village are according to following table.
Find their mode.
No. of members 2 3 4 5 6 7 8
No. of families 1 2 4 6 10 3 5

Here no. of members 6 has maximum frequency 10 so mode = 6

Marks obtained by students are given below, find the mode of marks.
Marks 10 20 30 40 50 60 70 80
No. of Students 2 8 16 26 20 16 7 4

From table it is clear that students who got 40 marks have maximum frequency (26) so mode = 40

Age in years of 20 students of a class are as follows:
15 16 13 14 14
13 15 14 13 13
14 12 15 14 16
13 14 14 13 15
Find mode by expressing them in frequency distribution:

Age	Frequency
12	1
13	6
14	7
15	4
16	2

From data it is clear that students of age 14 years have maximum frequency, so mode = 14

Find the median of :
$66, 98, 54, 92, 87, 63, 72$

Arrange the given data in increasing

$54, 63, 66, 72, 87, 92, 98$

Median

$=$ Middle element of the data arranged in increasing order.

The middle element of the arranged data is

$72$ .

Find mode of the following frequency distribution
Marks No. of Students
0 - 10 5
10 - 20 12
20 - 30 14
30 - 40 10
40 - 50 8
50 - 60 6

Here, Maximum frequency is 16 and its corresponding class is 20 - 30
Thus, modal class = 20 - 30
Now, I = 20,

$f_1$ = 14,

$f_0$ = 12,

$f_2$ = 10 and h = 10

$\because$ Mode

$= l + \left ( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right ) \times h$

$= 20 + \left ( \frac{14 - 12}{(2 \times 14) - 12 - 10} \right ) \times 10$

$= 20 + \left ( \frac{2}{28 - 22} \right ) \times 10$

$= 20 + \frac{2 \times 10}{6} = 20 + 3.33 = 23.33$
Thus, mode = 23.33 marks (approx)

Find the mode of $0, 5, 2, 7, 2, 1, 1, 3, 2, 4, 5, 7, 5, 1$ and $2$

$\textbf{Step-1: Arrange given numbers is ascending order}$

$\text{Given numbers : 0,5,2,7,2,1,1,3,2,4,5,7,5,1 and 2}$

$\text{Ascending order: 0,1,1,1,,2,2,2,2,3,4,5,5,5,7,7}$

$\textbf{Step-2: Find mode by checking most occurring number}$

$\text{We conclude that 2 occurs the maximum number of times.}$

$\text{And mode is the value that occurs the maximum number of times in a given set of elements.}$

$\textbf{Hence, the mode is 2}$

Find the mean and variance for the data $6, 7, 10 , 12, 13, 4, 8, 12$

Given date is,

$6, 7, 10, 12, 13, 4, 8, 12$
Mean

$\displaystyle \bar{x}=\frac{\sum_{i=1}^{8}x_{i}}{n}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9$
Now,

$\displaystyle x_{i}$	$\displaystyle \left (x_{i}-\bar{x} \right )$	$\displaystyle \left (x_{i}-\bar{x} \right )^{2}$
6	-3	9
7	-2	4
10	-1	1
12	3	9
13	4	16
4	-5	25
8	-1	1
12	3	9
		74

$\therefore$ Variance

$\displaystyle \left ( \sigma ^{2} \right )=\frac{1}{n}\sum_{i=l}^{8}(x_{i}-\bar{x})^{2}=\frac{1}{8}\times 74=9.25$

Find the mean and variance for the data
$\displaystyle X_{i}$ 6 10 14 18 24 28 30
$\displaystyle f_{i}$
2 4 7 12 8 4 3

The data is obtained in tabular form as follows

$\displaystyle X_{i}$	$\displaystyle f_{i}$	$\displaystyle f_{i} X_{i}$	$\displaystyle X_{i}-\bar{X}$	$\displaystyle \left (X_{i}-\bar{X} \right )^{2}$	$\displaystyle f_{i} \left (X_{i}-\bar{X} \right )^{2}$
6	2	12	-13	169	338
10	4	40	-9	81	324
14	7	98	-5	25	175
18	12	216	-1	1	12
24	8	192	5	25	200
28	4	112	9	81	324
30	3	90	11	121	363
	40	760			1736

Here,

$N=40$ ,

$\displaystyle \sum_{i=1}^{7}f_{i}x_{i}=760$

$\displaystyle \therefore \bar{x}=\frac{\sum_{i=1}^7f_{i}x_{i}}{N}=\frac{760}{40}=19$
and Variance

$\displaystyle \left (\sigma ^{2} \right )=\frac{1}{N}\sum_{i=1}^{7}f_{i}(x_{i}-\bar{X})^{2}=\frac{1}{40}\times 1736=43.4$

Find the mean deviation about the mean for the data
$4, 7, 8, 9, 10, 12, 13, 17$

The given data is

$4, 7, 8, 9, 10, 12, 13, 17$
Mean of the data

$\displaystyle \bar{x}= \frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}=10$

The deviations of the respective observations from the mean

$\displaystyle \bar{x} i.e. x_{i}-\bar{x},$ are

$-6, -3, -2, -1, 0, 2, 3, 7$ .

The absolute values of the deviations i.e.

$\displaystyle \mid x_{i}-\bar{x} \mid$ , are

$6, 3, 2, 1, 0, 2, 3, 7$ .

The required mean deviation about the mean is

$\displaystyle M.D.=\frac{\sum_{i=1}^{8}\mid x_{i}-\bar{x} \mid }{8}$

$\displaystyle M.D.=\frac{6+3+2+1+0+2+3+7}{8}=\frac{24}{8}=3$

Find the mean and standard deviation using short-cut method
$\displaystyle X_{i}$ 60 61 62 63 64 65 66 67 68
$\displaystyle f_{i}$ 2 1 12 29 25 12 10 4 5

$x_i$	$f_i$	$u_i=\cfrac{x_i-A}{h}$	$u_i^2$	$f_iu_i$	$f_iu_i^2$
60	2	-4	16	-8	32
61	1	-3	9	-3	9
62	12	-2	4	-24	48
63	29	-1	1	-29	29
64	25	0	0	0	0
65	12	1	1	12	12
66	10	2	4	20	40
67	4	3	9	12	36
68	5	4	16	20	80
Total	100			0	286

Here,

$N=100, h=1$

Let assumed mean

$A=64$

Mean

$\displaystyle \bar{x}=A+\frac{\sum_{i=1}^{9}f_{i}u_{i}}{N}\times h$

$=64+\cfrac{0}{100}\times 1=64+0=64$

Variance

$\displaystyle \sigma ^{2}=\frac{h^{2}}{N^{2}}\left [ N\sum_{i=l}^{9}f_{i}u_{i}^{2}-\left ( \sum_{i=l}^{9}f_{i}u_{i} \right )^{2} \right ]$

$\displaystyle =\frac{1}{100^{2}}\left [ 100\times 286-0 \right ]$

$=2.86$

$\displaystyle \therefore$ standard deviation

$\displaystyle \left (\sigma \right )=\sqrt{2.86}=1.69$

Find the mean and variance for the following frequency distrubution
Classes 0-10 10-20 20-30 30-40 40-50
Frequencies 5 8 15 16 6

Class interval	$f_i$	$x_i$	$u_i=\cfrac{x_i-A}{h}$	$u_i^2$	$f_iu_i$	$f_iu_i^2$
0-10	5	5	-2	4	-10	20
10-20	8	15	-1	1	-8	8
20-30	15	25	0	0	0	0
30-40	16	35	1	1	16	16
40-50	6	45	2	4	12	24
Total	50				10	68

We have

$N=\sum f_i=50$

Width of class

$h= 10$

Assumed mean

$A=25 (say)$

Mean,

$\displaystyle \bar{x}=A+\frac{\sum_{i=1}^{5}f_{i}u_{i}}{N}\times h$

$=25+\cfrac{10}{50}\times 10=25+2=27$

$\Rightarrow \bar{x}=27$

Variance

$\sigma ^{ 2 }=h^{ 2 }\left[ \cfrac { \sum _{ i=1 }^{ 5 } f_{ i }u_{ i }^{ 2 } }{ N } -\left( \cfrac { \sum _{ i=1 }^{ 5 } f_{ i }u_{ i } }{ N } \right) ^{ 2 } \right]$

$\sigma ^{ 2 }=10^{ 2 }\left[ \cfrac { 68 }{ 50 } -\left( \cfrac { 10 }{ 50 } \right) ^{ 2 } \right]$

$=10^{ 2 }\cfrac { \left[ 50\times 68-(10)^{ 2 } \right] }{ \left( 50 \right) ^{ 2 } }$

$=\displaystyle \frac{1}{25}\left [ 3400-100 \right ]$

$=\cfrac{3300}{25}$

$=132$

Find the mean and variance for the following frequency distribution
Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210
Frequencies 2 3 5 10 3 5 2

Class	$f_i$	$x_i$	$u_i=\cfrac{x_i-A}{h}$	$u_i^2$	$f_iu_i$	$f_iu_i^2$
0-30	2	15	-3	9	-6	18
30-60	3	45	-2	4	-6	12
60-90	5	75	-1	1	-5	5
90-120	10	105	0	0	0	0
120-150	3	135	1	1	3	3
150-180	5	165	2	4	10	20
180-210	2	195	3	9	6	18
Total	30				2	76

Here,

$N=30, h=30$

Mean,

$\displaystyle \bar{x}=A+\frac{\sum_{i=1}^{7}f_{i}u_{i}}{N}\times h$

$=105+\cfrac{2}{30}\times 30$

$=105+2=107$

Variance

$\displaystyle \left (\sigma ^{2} \right )=\cfrac{h^{2}}{N^{2}}\left [ N\sum_{i=l}^{7}f_{i}u_{i}^{2}-\left ( \sum_{i=l}^{7}f_{i}u_{i} \right )^{2} \right ]$
=

$\displaystyle \frac{\left ( 30 \right )^{2}}{\left ( 30 \right )^{2}}\left [ 30\times 76-(2)^{2} \right ]$

$=2280-4$

$=2276$

Find the mean deviation about the mean for the data
$38, 70, 48, 40, 42, 55, 63, 46, 54, 44$

The given data is

$38, 70, 48, 40, 42, 55, 63, 46, 54, 44$

Mean

$\displaystyle \bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}$

$=\dfrac{500}{10}=50$

The deviations of the respective observation from the mean

$\displaystyle \bar{x},$ i.e.

$x_{i}=\bar{x}$ are

$-12, 20, -2, -10, -8, 5, 13, -4, 4, -6$ .

The absolute values of the deviations i.e.

$\displaystyle \mid x_{1}-\bar{x} \mid$ are

$12, 20, 2, 10, 8, 5, 13, 4, 4, 6$

The required mean deviation about the mean is

$\displaystyle M.D.=\dfrac{\sum_{i=1}^{10}\mid x_{i}-\bar{x} \mid}{10}$

$= \displaystyle \frac{12+10+2+10+8+5+13+4+4+6}{10}$

$= \displaystyle \frac{84}{10}$

$=8.4$

Find the mean and variance for the data
$\displaystyle X_{i}$ 92 93 97 98 102 104 109
$\displaystyle f_{i}$
3 2 3 2 6 3 3

The data is obtained in tabular form as follows

$\displaystyle X_{i}$	$\displaystyle f_{i}$	$\displaystyle f_{i}-\bar{X}$	$\displaystyle X_{i}-\bar{X}$	$\displaystyle \left ( X_{i}-\bar{X} \right )^{2}$	$\displaystyle f_{i}\left ( X_{i}-\bar{X} \right )^{2}$
92	3	276	-8	64	192
93	2	186	-7	49	98
97	3	291	-3	9	27
98	2	196	-2	4	8
102	6	612	2	4	24
104	3	312	4	16	48
109	3	327	9	81	243
109	3	327	9	81	243

2200

640

Here ,

$N=22$ ,

$\displaystyle \sum_{i=1}^{7}f_{i}X_{i}=2200$

$\displaystyle \therefore \bar{X}=\frac{1}{N}\sum_{i=1}^{7}f_{i}x_{i}=\frac{1}{22}\times 2200=100$
and Variance

$\displaystyle \left (\sigma ^{2} \right )=\frac{1}{N}\sum_{i=1}^{7}f_{i}\left ( X_{i}-\bar{X} \right )^{2}=\frac{1}{22}\times 640=29.09$

Find the mean variance and standard deviation using short-cut method
Height
in cms No. of children
70-75 3
75-80 4
80-85 7
85-90 7
90-95 15
95-100 9
100-105 6
105-110 6
110-115 3

Class interval	$f_i$	$x_i$	$u_i=\cfrac{x_i-A}{h}$	$u_i^2$	$f_iu_i$	$f_iu_i^2$
70-75	3	72.5	-4	16	-12	48
75-80	4	77.5	-3	9	-12	36
80-85	7	82.5	-2	4	-14	28
85-90	7	87.5	-1	1	-7	7
90-95	15	92.5	0	0	0	0
95-100	9	97.5	1	1	9	9
100-105	6	102.5	2	4	12	24
105-110	6	107.5	3	9	18	54
110-115	3	112.5	4	16	12	48
Total	60				6	254

We have

$N=\sum f_i=60$

Width of class

$h= 5$

Assumed mean

$A=92.5 (say)$

Mean,

$\displaystyle \bar{x}=A+\frac{\sum_{i=1}^{9}f_{i}u_{i}}{N}\times h$

$=92.5+\cfrac{6}{60}\times 5=92.5+0.5=93$

$\Rightarrow \bar{x}=93$

Variance

$\sigma ^{ 2 }=h^{ 2 }\left[ \cfrac { \sum _{ i=1 }^{ 9 } f_{ i }u_{ i }^{ 2 } }{ N } -\left( \cfrac { \sum _{ i=1 }^{ 9 } f_{ i }u_{ i } }{ N } \right) ^{ 2 } \right]$

$\sigma ^{ 2 }=5^{ 2 }\left[ \cfrac { 254 }{ 60 } -\left( \cfrac { 6 }{ 60 } \right) ^{ 2 } \right]$

$=5^{ 2 }\cfrac { \left[ 60\times 254-6^{ 2 } \right] }{ \left( 60 \right) ^{ 2 } }$

$=\displaystyle \frac{25}{3600}\left [ 15240-36\right ]$

$=\cfrac { 25 }{ 3600 } \times 15204$

$\sigma^2=105.58$

Standard Deviation

$=\sqrt{\text{Variance}}$

$\sigma =\sqrt{105.58}$

$\sigma =10.27$

Find the mean deviation about the median for the data $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$ .

Arranging the data in ascending order

$10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$
Here the numbers of observations

$n=12$ (even)

$\displaystyle M= \frac{\left ( \frac{12}{2} \right )^{th}observation\, +\left ( \frac{12}{2}+1 \right )^{th}observation}{2}$

$\displaystyle M= \frac{6^{th}observation\, +7^{th}observation}{2}$

$\displaystyle M= \frac{13+14}{2}=\frac{27}{2}=13.5$

The deviations of the respective observation from the median, i.e.

$\displaystyle x_{i}-M$ are

$-3.5, -2.5, -2.5, -1.5, -0.5, -0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5$

The absolute values of the deviations

$\displaystyle \mid x_{i}-M \mid$ are

$3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5$

The required mean deviation about the median is

$\displaystyle M.D.(M) =\frac{\sum_{i=1}^{12}\left | x_{1}-M \right |}{12}$

$\displaystyle =\frac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5}{12}$

$\displaystyle =\frac{28} {12} =2.33$

The sum and sum of squares corresponding to length $X$ (in cm ) and weight $y$
(in gm) of $50$ plant products are given below:
$\displaystyle \sum_{i=l}^{50}X_{i}=212, \sum_{i=l}^{50} X_{i}^{2}=902.8, \sum_{i=l}^{50}y_{i}=261,\sum_{i=l}^{50}y_{i}^{2}=1457.6$
Which is more varying the length or weight ?

$\displaystyle \sum_{i=1}^{50}X_{i}=212, \sum_{i=1}^{50} X_{i}^{2}=902.8,$
Here,

$N = 50$

Mean

$\displaystyle \bar{X_i}=\frac{\sum_{i=1}^{50}}{N}=\frac{212}{50}=4.24$

Variance

$\displaystyle \left (\sigma _{1}^{2} \right )=\frac{1}{N}\sum_{i=1}^{50}\left ( X_{i}-\bar{X} \right )^{2}$

$\displaystyle =\frac{1}{50}\sum_{i=1}^{50}(X_{i}-4.24)^{2}$

$\displaystyle =\frac{1}{50}\sum_{i=1}^{50}\left [ X_{i}^{2}-8.48X_{i}+17.97 \right ]$

$\displaystyle =\frac{1}{50}\left [ \sum_{i=1}^{50}X_{i}^{2}-8.48\sum_{i=1}^{50}X_{i}+17.97\times 50 \right ]$

$\displaystyle =\frac{1}{50}\left [ 902.8-8.48\times(212)+898.5 \right ]$

$\displaystyle =\frac{1}{50}\left [ 1801.3-1797.76 \right ]$

$\displaystyle =\frac{1}{50}\times 3.54$

$= 0.07$

Also, given

$\displaystyle \sum_{i=1}^{50}y_{i}=261,\sum_{i=1}^{50}y_{i}^{2}=1457.6$

Mean

$\displaystyle \bar{y}=\frac{1}{N}\sum_{i=1}^{50}y_{i}=\frac{1}{50}\times 261=5.22$

Variance

$\displaystyle \left ( \sigma _{2}^{2} \right )=\frac{1}{N}\sum_{i=1}^{50}\left ( y_{i} -\bar{y}\right )^{2}$

$=\displaystyle \frac{1}{50}\sum_{i=1}^{50}(y_{i}-5.22)^{2}$

$\displaystyle =\frac{1}{50}\sum_{i=1}^{50}\left [ y_{1}^{2}-10.44y_{i}+27.24 \right ]$

$\displaystyle =\frac{1}{50}\left [\sum_{i=1}^{50} y_{i}^{2}-10.44y_{i}+27.24\times 50 \right ]$

$\displaystyle =\frac{1}{50}\left [ 1457.6-10.44\times(261)+1362 \right ]$

$\displaystyle =\frac{1}{50}\left [ 2819.6-2724.84 \right ]$

$\displaystyle =\frac{1}{50}\times 94.76$

$= 1.89$

Since, variance of weights (

$y_i$ ) is greater than the variance of lengths (

$X_i$ )

i.e

$\sigma_2 >\sigma_1$

So, weight vary more than the lengths .

The mean and variance of eight observation are $9$ and $9.25,$ respectively. If six of the observation are $6, 7, 10, 12, 12$ and $13$ , find the remaining two observations.

Let the remaining two observations be

$x$ and

$y$
Therefore, the observations are

$6, 7, 10, 12, 12, 13, x,y$
Thus mean

$\displaystyle \bar{X}=\frac{6+7+10+12+12+13+x+y}{8}=9$

$\displaystyle \Rightarrow 60 + x + y = 12$

$\displaystyle \Rightarrow x + y = 12$ ..........(i)
Variance

$= 9.25 = \displaystyle \frac{1}{N}\sum_{i=l}^{8}(x_{i}-\bar{x})^{2}$

$9.25= \displaystyle \frac{1}{8}\left [ (-3)^{2}+(-2)^{2}+(1)^{2}+(3)^{2}+(3)^{2}+(4)^{2}+x^{2}+y^{2}-2\times9(x+y)+2\times(9)^{2} \right ]$

$9.25 = \displaystyle \frac{1}{8}\left [ 9+4+1+9+9+16+x^{2}+y^{2}-18(12)+162 \right ]$ .........[Using (i) ]

$\displaystyle 9.25=\frac{1}{8}[48+x^{2}+y^{2}-216+162]$

$\displaystyle 9.25=\frac{1}{8}[x^{2}+y^{2}-6]$

$\displaystyle \Rightarrow x^{2}+y^{2}=80$ ......(2)
From (1) we obtain

$\displaystyle x^{2}+y^{2}+2xy=144$ ......(3)
From (2) and (3), we obtain

$2xy = 64$ .......(4)
Subtracting (4) from (2), we obtain

$\displaystyle x^{2}+y^{2}-2xy=80-64=16$

$\displaystyle \Rightarrow x-y=\pm 4$ ..........(5)
Therefore from (1) and (5), we obtain

$x = 8$ and

$y = 4$ when

$x - y = 4$

$x = 4$ and

$y = 8$ when

$x - y = -4$
Thus the remaining observation are

$4$ and

$8$

The mean and standard deviation of marks obtained by $50$ students of a class in three
subjects Mathematics, Physics and chemistry are given below :
Subject Mathematics Physics Chemistry
Mean 42 32 40.9
Standard deviation 12 15 20
Which of the three subjects shows the highest variability in marks and which shows the
lowest ?

Standard deviation of mathematics

$= 12$
Standard deviation of physics

$= 15$
Standard deviation of chemistry

$= 20$
The coefficient of variation (C.V) is given by

$\displaystyle \frac{Standard\, deviation }{Mean}\times 100$
Thus C.V (in mathematics )=

$\displaystyle \frac{12}{42}\times100 = 28.57$
C.V.(in physics ) =

$\displaystyle \frac{15}{32}\times100 = 46.87$
C.V.(in Chemistry ) =

$\displaystyle \frac{20}{40.9}\times100 = 48.89$
The subject with greater C.V is more variable than others
Therefore the highest variability in marks is in chemistry and the lowest variability in
marks is in mathematics

The mean and standard deviation of a group of $100$ observations were found to be $20$ and $3$ respectively. Later on it was found that three observations were incorrect which were recorded as $21, 21$ and $18$ . Find the mean and standard deviation if the incorrect observation are omitted

Given, number of observation

$n = 100$
Incorrect mean

$\displaystyle (\bar{x})= 20$

We know that

$\bar{x}=\cfrac{\sum x_i}{n}$

$\displaystyle \Rightarrow 20=\frac{1}{100}\sum_{i=1}^{100}x_{i}$

$\displaystyle \Rightarrow \sum_{i=l}^{100}x_{i}= 20\times 100 = 2000$
So, the incorrect sum of observations

$= 2000$ .

Given incorrect observations are

$21, 21, 18$ and these has to be omitted.
So, correct sum of observations

$= 2000 - 21 - 21 - 18 = 2000 - 60 = 1940$

$\displaystyle$ Correct mean

$=\cfrac{Correct\, sum}{100-3}=\cfrac{1940}{97} = 20$

Given, incorrect standard deviation

$\sigma =3$

$\displaystyle \sigma =\sqrt { \frac { \sum _{ i=1 }^{ n } x_{ i }^2 }{ n } -\left( \frac { \sum _{ i=1 }^{ n } x_{ i } }{ n } \right) ^{ 2 } }$

$\displaystyle \sigma ^{ 2 }=\frac { \sum _{ i=1 }^{ n } x_{ i }^2 }{ n } -(\bar{x}) ^{ 2 }$

$\displaystyle \Rightarrow 3^{ 2 }=\frac { \sum x_{ i }^{ 2 } }{ 100 } -(20)^{ 2 }$

$\displaystyle \Rightarrow Incorrect \sum x_{i}^{2}=100(9+400) = 40900$

Correct

$\displaystyle \sum_{i=l}^{n}x_{i}^{2}=Incorrect\sum_{i=l}^{n}x_{i}^{2}-(21)^{2}-(21)^{2}-(18)^{2}$

$= 40900 - 441 - 441 - 324$

$= 39694$

$\displaystyle \therefore$ Correct standard deviation =

$\displaystyle \sqrt{\frac{correct \sum x_{i}^{2}}{n}-(Correct\, mean)^{2}}$

$= \displaystyle \sqrt{\frac{39694}{97}-(20)^{2}}$

$= \displaystyle \sqrt{409.216-400}$

$= \displaystyle \sqrt{9.216}$

$= 3.036$

From the data given below state which group is more variable $A$ or $B$
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Group A 9 17 32 33 40 10 9
Group B 10 20 30 25 43 15 7

Class	$f_i$	$x_i$	$u_i=\cfrac{x_i-A}{h}$	$u_i^2$	$f_iu_i$	$f_iu_i^2$
10-20	9	15	-3	9	-27	81
20-30	17	25	-2	4	-34	68
30-40	32	35	-1	1	-32	32
40-50	33	45	0	0	0	0
50-60	40	55	1	1	40	40
60-70	10	65	2	4	20	40
70-80	9	75	3	9	27	81
Total	150				-6	342

Here

$h = 10, N = 150$

Let assumed mean

$A = 45$
Mean

$\displaystyle \bar{x}=A+\frac{\sum_{i=1}^{7}f_iu_{i}}{N}\times h$

$=45+\cfrac{(-6)\times10}{150}=45-0.4=44.6$

Variance

$\displaystyle \left (\sigma_{1} ^{2} \right )=\frac{h^{2}}{N^{2}}\left [ N\sum_{i=l}^{7}f_{i}u_{i}^{2}-\left ( \sum_{i=l}^{7}f_{i}u_{i} \right )^{2} \right ]$

$\displaystyle =\frac{100}{22500}\left [ 150\times 342-\left ( 6 \right )^{2} \right ]$

$= \displaystyle \frac{1}{225}(51264)$

$=227.84$

$\displaystyle \therefore$ Standard deviation

$(\sigma_{1} )=\sqrt {227.84}=15.09$

Group B:

Class	$f_i$	$x_i$	$u_i=\cfrac{x_i-A}{h}$	$u_i^2$	$f_iu_i$	$f_iu_i^2$
10-20	10	15	-3	9	-30	90
20-30	20	25	-2	4	-40	80
30-40	30	35	-1	1	-30	30
40-50	25	45	0	0	0	0
50-60	43	55	1	1	43	43
60-70	15	65	2	4	30	60
70-80	7	75	3	9	21	63
Total	150				-6	366

Here

$h = 10, N = 150$

Let assumed mean

$A = 45$
Mean

$\displaystyle \bar{x}=A+\frac{\sum_{i=1}^{7}f_iu_{i}}{N}\times h$

$=45+\cfrac{(-6)\times10}{150}=45-0.4=44.6$

Variance

$\displaystyle \left (\sigma_{2} ^{2} \right )=\frac{h^{2}}{N^{2}}\left [ N\sum_{i=l}^{7}f_{i}u_{i}^{2}-\left ( \sum_{i=l}^{7}f_{i}u_{i} \right )^{2} \right ]$

$\displaystyle =\frac{100}{22500}\left [ 150\times 366-\left ( 6 \right )^{2} \right ]$

$= \displaystyle \frac{1}{225}(54864)$

$=243.84$

$\displaystyle \therefore$ Standard deviation

$(\sigma_{2} )=\sqrt {243.84}=15.61$

Since, the mean of both the groups is same the group with greater standard deviation will be more variable.
Thus group

$B$ has more variability in the marks.

The diameters of circles (in $mm$ ) drawn in a design are given below :
Diameters No. of circles
$33-36$ $15$
$37-40$ $17$
$41-44$ $21$
$45-48$ $22$
$49-52$ $25$

Since, the given frequency distribution is not continuous, so to make it continuous, we will subtract

$0.5$ from lower limit and add

$0.5$ to the upper limit of each class.

Class	$f_i$	$x_i$	$u_i=\cfrac{x_i-A}{h}$	$u_i^2$	$f_iu_i$	$f_iu_i^2$
32.5-36.5	15	34.5	-2	4	-30	60
36.5-40.5	17	38.5	-1	1	-17	17
40.5-44.5	21	42.5	0	0	0	0
44.5-48.5	22	46.5	1	1	22	22
48.5-52.5	25	50.5	2	4	50	100
Total	10 0				25	199

Here,

$N =100, h=4$
Let the assumed mean

$A = 42.5$
Mean

$\displaystyle \bar{x}=A+\frac{\sum_{i=1}^{5}f_{i}u_{i}}{N}\times h$

$=42.5+\cfrac{25}{100}\times 4=43.5$

Variance

$\displaystyle \left (\sigma ^{2} \right )=\frac{h^{2}}{N^{2}}\left [ N\sum_{i=1}^{5}f_{i}u_{i}^{2}-\left ( \sum_{i=1}^{5}f_{i}u_{i} \right )^{2} \right ]$
=

$\displaystyle \frac{16}{10000}\left [ 100\times 199-\left ( 25 \right )^{2} \right ]$

$= \displaystyle \frac{16}{10000}\left [ 19900-625 \right ]$

$= \displaystyle \frac{16}{10000}\times 19275$

$=30.84$

$\displaystyle \therefore$ Standard deviation

$(\sigma ) = 5.55$

Given that $\displaystyle \bar{X}$ is the mean and $\displaystyle \sigma ^{2}$ is the variance of $\displaystyle n$ observations $\displaystyle X_{1},X_{2}...X_{n }$ . Prove that the mean and variance of the observations $\displaystyle aX_{1},aX_{2},aX_{3}....aX_{n }$ are $\displaystyle \bar{ax}$ and $\displaystyle a^{2}\sigma ^{2}$ respectively $\displaystyle (a\neq 0)$ .

The given

$\displaystyle n$ observation are

$\displaystyle x_{1},x_{2}..x_{n}$
Mean =

$\displaystyle \bar{x}$
variance =

$\displaystyle \sigma ^{2}$

$\displaystyle \therefore\sigma ^{2}=\frac{1}{n}\sum_{i=l}^{n}y_{i}\left ( x_{i}-\bar{x} \right )^{2}$ ....(i)
If each observation is multiplied by a and the new observation are

$\displaystyle y_{i}$ then

$\displaystyle y_{i}=ax_i, i.e,x_{i}=\frac{1}{a}y_{i}$

$\displaystyle \therefore \bar{y}=\frac{1}{n}\sum_{i=l}^{n}y_{i}=\frac{1}{n}\sum_{i=l}^{n}ax_{i}=\frac{a}{n}\sum_{i=l}^{n}x_{i}=\bar{ax}, \left (\because \bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_{i} \right )$
Therefore mean of the observation,

$\displaystyle ax_{1},ax_{2}....ax_{n}$ is

$\displaystyle \bar{ax}$
Substituting the values of

$\displaystyle x_{i}$ and

$\displaystyle \bar{x}$ in (1) we obtain

$\displaystyle \sigma ^{2}=\frac{1}{2}\sum_{i=l}^{n}\left ( \frac{1}{a}y_{i}-\frac{1}{a}\bar{y} \right )^{2}$

$\displaystyle \Rightarrow a^{2}\sigma ^{2}=\frac{1}{n}\sum_{i=l}^{n}\left (y_{i}-\bar{y} \right )^{2}$
Thus the variance of the observation

$\displaystyle ax_{1},ax_{2}...ax_{n}$ is

$\displaystyle a^{2}\sigma ^{2}$

From the prices of shares $X$ and $Y$ below find out which is more stable in value :
X 35 54 52 53 56 58 52 50 51 49
Y 108 107 105 105 106 107 104 103 104 101

$X_i$	$\|X_i-\bar{X}\|$	$\|X_i-\bar{X}\|^2$
35	16	256
54	3	9
52	1	1
53	2	4
56	5	25
58	7	49
52	1	1
50	1	1
51	0	0
49	2	4
Total= 510

Here, the number of observations

$n=10$

$\displaystyle \therefore Mean,\bar{X}=\frac{1}{n}\sum_{i=1}^{10} X_{i}=\frac{1}{10}\times 510=51$

Variance

$\displaystyle \left (\sigma_{1} ^{2} \right )=\frac{1}{n}\sum_{i=1}^{10}(X_i-\bar{X})^{2}$

$=\cfrac{1}{10}\times 350=35$

$\displaystyle \therefore$ Standard deviation

$(\sigma_{1} ) =\sqrt{35}=5.91$

$Y_i$	$\|Y_i-\bar{Y}\|$	$\|Y_i-\bar{Y}\|^2$
108	3	9
107	2	4
105	0	0
105	0	0
106	1	1
107	2	4
104	1	1
103	2	4
104	1	1
101	4	16
Total= 1050		40

Here, the number of observations

$n=10$

$\displaystyle \therefore Mean,\bar{Y}=\frac{1}{n}\sum_{i=1}^{10} Y_{i}=\frac{1}{10}\times 1050=105$

Variance

$\displaystyle \left (\sigma_{2} ^{2} \right )=\frac{1}{n}\sum_{i=1}^{10}(Y_i-\bar{Y})^{2}$

$=\cfrac{1}{10}\times 40=4$

$\displaystyle \therefore$ Standard deviation

$(\sigma_{2} ) =\sqrt{4}=2$

Since, the variance of prices of share

$X$ is greater than that of prices of share

$Y$ .

So, prices of share

$X$ are more variable than prices of share

$Y$ .

So, the prices of shares

$Y$ are more stable than the prices of share

$X$ .

The mean and standard deviation of $20$ observation are found to be $10$ and $2$ respectively. On rechecking it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases :
(i) If wrong item is omitted
(ii) If it is replaced by $12$

(i) Given, number of observations

$n=20$
Incorrect mean

$= 10$
Incorrect standard deviation

$= 2$

$\displaystyle \bar{x}=\frac{1}{n}\sum_{i=1}^{20}x_{i}$

$\displaystyle 10=\frac{1}{20}\sum_{i=1}^{20}x_{i}$

$\displaystyle \Rightarrow \sum_{i=1}^{20}x_{i}=200$
So, the incorrect sum of observations

$= 200$
Correct sum of observation

$= 200 - 8 = 192$

$\displaystyle \Rightarrow$ Correct mean

$=\cfrac{Correct\, sum}{19}=\cfrac{192}{19}= 10.1$

Standard deviation

$\displaystyle \sigma =\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\frac{1}{n^{2}}\left ( \sum_{i=1}^{n}x_{i} \right )^{2}} = \sqrt{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\left ( \bar{x} \right )^{2}}$

$\displaystyle \Rightarrow 2=\sqrt{\frac{1}{20}Incorrrect \sum_{i=1}^{n}x_{i}^{2}-(10)^{2}}$

$\displaystyle \Rightarrow 4=\frac{1}{20}Incorrect\, \sum_{i=1}^{n}x_{i}^{2}-100$

$\displaystyle \Rightarrow Incorrect\sum_{i=1}^{n}x_{i}^{2}=2080$

$\displaystyle \therefore$ Correct

$\displaystyle \sum_{i=1}^{n}x_{i}^{2}=Incorrect \sum_{i=1}^{n}x_{i}^{2}-\left ( 8 \right )^{2}$

$= 2080 - 64$

$= 2016$

$\displaystyle \therefore$ Correct Standard deviation

$=\sqrt{\frac{Correct\sum x_{i}^{2}} {n}-(Correct\, Mean)^{2}}$
=

$\displaystyle \sqrt{\frac{2016}{19}-(10.1)^{2}}$

$\displaystyle \sqrt{106.1-102.01}$
=

$\displaystyle \sqrt{4.09}$
= 2.02

(ii) When

$8$ is replaced by

$12$
Incorrect sum of observation

$= 200$

$\displaystyle \therefore$ Correct sum of observations

$= 200 - 8 + 12 = 204$

$\displaystyle \therefore$ Correct mean =

$\displaystyle \frac{Correct sum}{20}=\frac{204}{20}= 10.2$

Standard deviation

$\displaystyle \sigma = \sqrt{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\frac{1}{n^{2}}\left ( \sum_{i=1}^{n}x_{i} \right )^{2}}=\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}- \left (\bar{x} \right )^{2} }$

$\displaystyle \Rightarrow$

$\displaystyle 2=\sqrt{\frac{1}{20}Incorrect \sum_{i=1}^{n}x_{i}^{2}-(10)^{2}}$

$\displaystyle \Rightarrow$

$\displaystyle 4 =\frac{1}{20}Incorrect \sum_{i=1}^{n}x_{i}^{2}-100$

$\displaystyle \Rightarrow Incorrect\sum_{i=1}^{n}x_{i}^{2}=2080$

$\displaystyle \therefore Correct\, \sum_{i=1}^{n}x_{i}^{2}=Incorrect \sum_{i=1}^{n}x_{i}^{2}-(8)^{2}+(12)^{2}$

$= 2080 - 64 + 144$

$= 2160$

$\displaystyle Correct\, standard\, deviation\, =\sqrt{\frac{Correct\sum x_{i}^{2}}{n}-(Correct\, mean)^{2}}$
=

$\displaystyle \sqrt{\frac{2160}{20}-(10.2)^{2}}$

$\displaystyle =\sqrt{108-104.04}$

$\displaystyle =\sqrt{3.96}$

$= 1.98$

The mean and standard deviation of six observation are $8$ and $4$ respectively. If each observation is multiplied by $3$ , find the new mean and new standard deviation of the resulting observations.

Let the observations be

$\displaystyle x_{1},x_{2},x_{3},x_{4},x_{5},$ and

$x_{6}$
It is given that mean is

$8$ and standard deviation is

$4$ .

$\displaystyle\Rightarrow$ Mean

$=\displaystyle \bar{x}=\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}=8$ ....(1)
If each observation is multiplied by

$3$ and the resulting observation are

$y_{i}$ , then

$\displaystyle y_{i}=3x_i, \text{for} \, i=1 \text{to} 6$

$\displaystyle \therefore$ New mean

$\displaystyle \bar{y}=\frac{y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}}{6}$

$\displaystyle =\frac{3(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6})}{6}$

$= 3 \times 8$ ........[Using (1) ]

$= 24$
Standard deviation

$\displaystyle \sigma =\sqrt{\frac{1}{n}\sum_{i=l}^{6}(x_{i}-\bar{x})^{2}}$

$\displaystyle \therefore \left ( 4 \right )^{2}=\frac{1}{6}\sum_{i=l}^{6}(x_{i}-\bar{x})^{2}$

$\displaystyle \sum_{i=l}^{6}\left ( x_{1}-\bar{x} \right )^{2}=96$ .......(2)
From (1) and (2) it can be observed that

$\displaystyle \bar{y}=3\bar{x}$

$\displaystyle \bar{x}=\frac{1}{3}\bar{y}$
Substituting the values of

$\displaystyle x_{i}$ and

$\displaystyle \bar{x}$ in (2) we obtain

$\displaystyle \sum_{i=l}^{6}\left ( \frac{1}{3}y_{i}-\frac{1}{3}\bar{y} \right )^{2}=96$

$\displaystyle \Rightarrow \sum_{i=l}^{6}\left ( y_{i}-\bar{y} \right )^{2} = 864$
Therefore, variance of new observation

$=$

$\displaystyle \left ( \frac{1}{6}\times 864 \right )= 144$
Hence, the standard deviation of new observations is

$\displaystyle \sqrt{144}= 12$

The following number of goals were scored by a team in a series of $10$ matches:
$2, 3, 4, 5, 0, 1, 3, 3, 4, 3$
Find the mean, median and mode of these scores.

Mean

$= \dfrac{\text{Sum of all observations}}{\text{Total number of observations}}$

$= \dfrac{2+3+4+5+0+1+3+3+4+3}{10}$

$= \dfrac{28}{10} = 2.8$ ;

Total observations

$= 10$ (even)

$\therefore$ Median

$= \dfrac{10}{2} =5$ th &

$\dfrac{10}{2} + 1 = 6$ th

Median score

$= \dfrac{5^{th} + 6^{th} \text{ observations}}{2} = \dfrac{3+3}{2} = \dfrac{6}{2} = 3$ ;

Maximum frequency

$=4$ of

$3$ ;

$\therefore$ Mode

$=3$

The following observations have been arranged in ascending order. If the median of the data is $63$ , find the value of $x$ .
$29, 32, 48, 50, x, x+2, 72, 78, 84, 95$

Total observations

$= 10$ (even)
Median

$= \dfrac{10}{2} = 5$ th &

$\dfrac{10}{2} + 1 = 6^{th}$ observations
Median

$= \dfrac{5^{th} + 6^{th} \text{ observations}}{2}$

$63 = \dfrac{x+x+2}{2}$

$\Rightarrow x+1 = 63$

$\Rightarrow x=62$

Write the marks wise frequencies in the following frequency distribution table.
Marks Up to $5$ Up to $6$ Up to $7$ $ Up to $8$ Up to $9$ Up to $10$
No of students $5$ $11$ $19$ $31$ $40$ $45$

Marks No of students Frequency (f)

5 5 5

6 11 6

7 19 8

8 31 12

9 40 9

10 45 5

Find the mean deviation about mean for the following data:
$14, 21, 28, 21, 18$

Mean Deviation

$= \sum \dfrac{ |\overline{x}-x_{1}|+ |\overline{x}- x_{2}|+...+ |\overline{x}-x_n|}{n}$

where,

$\overline x = \dfrac{x_1 +x_2 +x_2...+x_n}{n}$

on substituting values

$\overline{x} = \dfrac {14+21+28+21+18}{5} = \dfrac{102}{5} = 20.4$

Mean Deviation

$= \dfrac{|20.4-14| +|20.4-21| + |20.4-28| +|20.4-21| +|20.4-18|}{5}$

$= \dfrac{6.4 +0.6 + 7.6 +0.6 +2.4}{5} = \dfrac{17.6}{5} =3.52$

The relative humidity (in %) of a certain city for a September month of $30$ days was as follows:
$98.1$ $98.6$ $99.2$ $90.3$ $86.5$ $95.3$ $92.9$ $96.3$ $94.2$ $95.1$
$89.2$ $92.3$ $97.1$ $93.5$ $92.7$ $95.1$ $97.2$ $93.3$ $95.2$ $97.3$
$96.0$ $92.1$ $84.9$ $90.0$ $95.7$ $98.3$ $97.3$ $96.1$ $92.1$ $89$
What is the range of the data?

Relative humanity (in %) Tally No of days ( frequency)

$84-86$ I

$1$

$86-88$ I

$1$

$88-90$ II

$2$

$90-92$ II

$2$

$92-94$ IIIII II

$7$

$94-96$ IIIII I

$6$

$96-98$ IIIII II

$7$

$98-100$ IIII

$4$

Total

$30$

Minimum humanity =

$84.9$

Maximum humanity=

$99.2$

Then range of data =Maximum value - Minimum value=

$99.2-84.9=14.3$

The marks scored by $60$ students in a science test are given below.
Marks $(x)$ $10$ $20$ $30$ $40$ $50$ $60$
No. of students $(f)$ $8$ $12$ $20$ $10$ $7$ $3$
Calculate the variance and standard deviation

$\bar x=\cfrac{\sum Fx}{\sum F}=\cfrac{1850}{60}=30.83$

${\sigma}^{2}=\sqrt { \cfrac { \sum { F } { ({ x }_{ i }-\bar { x } ) }^{ 2 } }{ \sum { F-1 } } } =\cfrac{10858.396}{1849}=5.873$

$\therefore \sigma=2.423$

1017735_561645_ans_b5358b2de7a64256b19a255431001817.png

Calculate the range and coefficient of range from the following data:
Number of trees planted in $6$ months: $186, 234, 465, 361, 290, 142$

$X_{max}=168,\quad X_{min}=465$

So, Range

$= X_{max}-X_{min}=465-142=323$

Coefficient of range

$=\cfrac{X_{max}-X_{min}}{X_{max}+X_{min}}=\cfrac{465-132}{465+142}=\cfrac{323}{607}=0.53$

The number of books bought by $200$ students in a book exhibition is given below.
No. of books $0$ $1$ $2$ $3$ $4$
No. of students $35$ $64$ $68$ $18$ $15$
Find the variance and standard variation

$\bar x=\cfrac{\sum Fx}{\sum F}=\cfrac{314}{200}=1.57$

$\sigma =\sqrt { \cfrac { \sum { F } { ({ x }_{ i }-\bar { x } ) }^{ 2 } }{ \sum { F-1 } } } =\sqrt { \cfrac {244.988 }{ 199 } }=1.1095$

${\sigma}^{2}=1.2310$

1017870_561643_ans_6edaa70f4c4a40cea4e2e7717a5f72cb.png

Find the mean deviation about median for the following data.
$18, 23, 9, 11, 26, 4, 14, 21$

First arranging data in ascending order

$4, 9, 11, 14, 18, 21, 23, 26$

No. of terms

$n= 8$ ( even)

Since

$n$ is even,

median

$\displaystyle=\left( \frac {n }{ 2 }\right )$ th term,

$=( \dfrac {8 }{ 2 } )$ th term,

$=4$ th term,

$=14$

Now, We calculate mean deviation about median

median

$(M)=14$

mean deviation about median

$=\displaystyle\frac{\sum |x_i - M|}{8}$

$M.d.=\displaystyle\frac{(| 4-14 | + | 9-14 | + | 11-14 | + | 14-14 | + |18-14 | + | 21-14 | + | 23-14 | + |26-14|)}{8}$

$=\displaystyle\frac{(|-10| + |-5 | +|-3 |+|0|+|4| +|7|+|9|+|12|)}{8}$

$=\displaystyle\frac{10+5+3+4+7+9+12}{8}$

$=\displaystyle\frac{50}{8}$

$=6.25$

The daily wages of $40$ workers of a factory are given in the following table.
Wages in Rs $30-34$ $34-38$ $38-42$ $42-46$ $46-50$ $50-54$
No. of workers $4$ $7$ $9$ $11$ $6$ $3$
Calculate i) Mean (ii) Variance and (iii) Standard deviation of wages and interpret the findings

$i)$

$\bar x=\cfrac{\sum Fx}{\sum F}=41.7$

$ii)$

${\sigma}^{2}=\cfrac { \sum { F } { (xi-\bar { x } ) }^{ 2 } }{ \sum { F-1 } } =\cfrac{1244.4}{39}=31.908$

$iii)$

$\sigma =\sqrt{31.908}=5.649$

1017846_561646_ans_622c4eb4eb5a403e8acb47d026441670.png

Calculate the range and coefficient of range from the following data:
Marks scored by $12$ students in a test: $31, 18, 27, 19, 25, 28, 49, 14, 41, 22, 33, 13$

$X_{min}=13,\quad X_{max}=49$

So, Range

$= X_{max}-X_{min}=49-13=36$

Coefficient of range

$=\cfrac{X_{max}-X_{min}}{X_{max}+X_{min}}=\cfrac{49-13}{49+13}=\cfrac{36}{62}=0.58$

Find the mean deviation about mean for the following data.
$15, 18, 13, 16, 12, 24, 10, 20$

Mean Deviation

$= \sum \dfrac{ |\overline{x}-x_{1}|+ |\overline{x}- x_{2}|+...+ |\overline{x}-x_n|}{n}$

where,

$\overline x = \dfrac{x_1 +x_2 +x_2...+x_n}{n}$

on substituting values

$\overline{x} = \dfrac {15+18+13+16+12+24+10+20}{8} = \dfrac{128}{8} = 16$

Mean Deviation

$= \dfrac{|16-15 +|18-16| + |13-16|+|16-16| +|16-12|+|16-24| +|16-10| +|16-20|}{}$

$= \dfrac{1 +2 + 3 +0 +4+8+6+4}{8} = \dfrac{28}{8} =3.5$

If ' $x$ ' varies inversely as ' $y$ ' and $x=7$ when $y=9$ :
(a) Find constant of variation $(k)$ .
(b) Write equation of variation.
(c) Find ' $y$ ' when $x=9$ .

(a)

$x\propto \cfrac{1}{y}$

$\Rightarrow$

$x=\cfrac{k}{y}$

$\Rightarrow$

$7=\cfrac{k}{9}$

$\Rightarrow$

$k=7\times 9$

$\therefore$

$x=63$

$x=\cfrac{63}{y}$

$\Rightarrow$

$y=\cfrac{63}{x}$

$\Rightarrow$

$y=\cfrac{63}{9}$

$\therefore$

$y=7$

The coefficient of range of the data set $\{3,1,4,2,5,9,8\}$ can be represented as $\dfrac{p}{q}$ where $p$ and $q$ are co-primes. Then the value of $(q-p)^2$ is:

Given Dataset

$\{3, 1, 4, 2, 5, 9, 8\}$

Here

(H) high value

$= 9$

(L) Low value

$= 1$

Know

We Range

$= H - L$

$= 9 - 1$

$= 8$

and

Coefficient of the range

$= \dfrac{H - L}{H + L}$

$= \dfrac{9 - 1}{9 + 1}$

Common factor

$=1$

$= \dfrac{8}{10}$

$\therefore P = 4, q = 5$

$= \dfrac{4}{5}$ it is solution of

$\dfrac{p}{q}$

then p and q are coprimes

$\therefore$ value of

$(q - p)^2 = (5 - 4)^2 = 1^2 = 1$ .

Find the range and the coefficient of range of $43, 24, 38, 56, 22, 39, 45$

Here, larger item

$L=56$
Smaller item

$S=22$

$Range=L-S=56-22=34$

$\text{Coefficient of Range=}\dfrac{L-S}{L+S}=\dfrac{56-22}{56+22}=\dfrac{34}{78}=\dfrac{17}{39}$

Find the range and the coefficient of range of $43, 24, 38, 56, 22, 39, 45$

Given,

$43,24,38,56,29,45$

Thus,

$L = 56, S = 22$
Range :

$L - S = 56 - 22 = 34$
Coefficient of Range

$= \dfrac {L - S}{L+ S}$

$=\dfrac {56 - 22}{56 + 22}$

$= \dfrac {34}{78} = 0.436$

Find the mean deviation about the median for the following data.
$13, 17, 16, 11, 13, 10, 16, 11, 18, 12, 17$

Total number of observations =

$11$

When arranged in ascending order, as shown:

$10, 11, 11, 12, 13, 13, 16, 16, 17, 17, 18$ , the middle number is

$13$ and hence the median is

$13$ .

The mean deviation about the median is given by

$=\cfrac{(13-10) + 2(13-11) + (13-12) + 2(13 - 13) + 2(13 - 16) + 2(13 - 17)+ (13-18)}{11}$

$= \cfrac{3 + 2(2)+ 1 + 2(0)+ 2(-3) +2 (-4) + (-5)}{11}$

$= \cfrac{-11}{11}$

$= -1$

Find the standard deviation of the first $10$ natural numbers.

The standard deviation of the first

$10$ natural number

$=\sqrt { \dfrac { { n }^{ 2 }-1 }{ 12 } }$

$=\sqrt { \dfrac { { 10 }^{ 2 }-1 }{ 12 } }$

$=\sqrt { \dfrac { 100-1 }{ 12 } }$

$=\sqrt { \dfrac { 99 }{ 12 } }$

$=\sqrt { \dfrac { 33 }{ 4 } }$

$=\sqrt { 8.25 }$

$=2.87$

Thus, standard deviation of first

$10$ natural numbers is

$2.87$ .

Calculate the Standard Deviation and coefficient of variation for the given frequency table:
Class-interval Frequency
$1-5$ $1$
$6-10$ $2$
$11-15$ $3$
$16-20$ $4$

C.I	$f$	Mid point $x$	$fx$	$D=X-\overline { X }$	${D}^{2}$	$f{D}^{2}$
$1-5$	$1$	$3$	$3$	$-10$	$100$	$100$
$6-10$	$2$	$8$	$16$	$-5$	$25$	$50$
$11-15$	$3$	$13$	$39$	$0$	$0$	$0$
$16-20$	$4$	$18$	$72$	$+5$	$25$	$100$

$N=10$

$\sum { fx } =130$

$\sum { f{ D }^{ 2 } } =250$
Standard Deviation

$=\sigma =\sqrt { \cfrac { \sum { f{ D }^{ 2 } } }{ N } } =\sqrt { \cfrac { 250 }{ 10 } } =\sqrt { 25 }$
Therefore,

$\sigma =5$

Coefficient of variation, C.V

$=\cfrac { \sigma }{ \overline { x } } \times 100=\cfrac { 5 }{ 13 } \times 100=\cfrac { 500 }{ 13 } =38.46$

Find the variance for an ungrouped data $5, 12, 3, 18, 6, 8, 2, 10.$

Arranging the ungrouped data in ascending order, we can write

$2, 3, 5, 6, 8, 10, 12, 18$

The mean of the data is given by

$\cfrac{2 + 3 + 5 + 6 + 8 + 10 + 12 + 18}{8} = \cfrac{64}{8} = 8$

Variance is given by

$\cfrac{(x_i - x)^2}{n}$

i.e. Variance

$= \cfrac{(2 - 8)^2 + (3 - 8)^2 + (5 - 8)^2 + (6 - 8)^2 + (8 - 8)^2 + (10 - 8)^2 + (12 - 8)^2 + (18 - 8)^2}{8}$

$= \cfrac{36 + 25 + 9 + 4 + 0 + 4 + 16 + 100}{8}$

$= \cfrac{194}{8}$

$= 24.25$

Calculate variance for the following data: $2,4,6,8$ and $10$ .

Given data:

$2,4,6,8,10$

To find:

Variance

$\sigma^2=?$

Solution:

$\bar x=\cfrac{2+4+6+8+10}{5}$

$=\cfrac{30}{5}$

$=6$

$\sigma^2=\cfrac{1}{n-1}\sum_{i=1}^{n}{(x_i-\bar x)^2}$

$=\cfrac{1}{5-1}\left\{(2-6)^2+(4-6)^2+(6-6)^2+(8-6)^2+(10-6)^2\right\}$

$=\cfrac14 (16+4+0+4+16)$

$=\cfrac{40}{4}$

$=10$

Therefore, varaince of above data

$=10$

Find standard deviation 50, 56, 59, 60, 63, 67, 68.

$\Rightarrow$

$\overline{x}=\dfrac{50+56+59+60+63+67+68}{7}=\dfrac{423}{7}=60.4$

$x_i$	$x_i-\overline{x}$	$(x_i-\overline{x})^2$
$50$	$-10.4$	$108.16$
$56$	$-4.4$	$19.36$
$59$	$-1.4$	$1.96$
$60$	$-0.4$	$0.16$
$63$	$2.6$	$6.76$
$67$	$6.6$	$43.56$
$68$	$7.6$	$57.76$
		$\sum (x_i-\overline{x})^2=237.72$

$\Rightarrow$

$\sigma=\sqrt{\dfrac{x_i-\overline{x})^2}{N}}=\sqrt{\dfrac{237.72}{7}}=5.82$

The p.m.f. of a r.v. $X$ is
$P(x) = \left\{\begin{matrix}\dfrac {2x}{n(n + 1)}, & x = 1, 2, ... n;\\ 0, & otherwise.\end{matrix}\right.$
Find $E(X)$ and $Var (X)$

$\\ The\quad expected\quad value\quad of\quad X\quad is\quad usually\quad written\quad as\quad E(X)\quad or\quad m.\\ E(X)=\sum { x\times P(X=x) } \\ E(X)=1×P(X=1)+2×P(X=2)+.........+n×P(X=n)\\ \quad \quad \quad =1×\frac { 2×1 }{ n(n+1) } +2×\frac { 2×2 }{ n(n+1) } +.........+n×\frac { 2×n }{ n(n+1) } \\ \quad \quad =\frac { 2 }{ n(n+1) } ×(1^{ 2 }+2^{ 2 }+3^{ 2 }+.....n^{ 2 })\\ =\quad \quad \frac { 2 }{ n(n+1) } ×(\frac { n(n+1)(2n+1) }{ 6 } )\\ \quad E(X)=\frac { 2n+1 }{ 3 } =m\\ The\quad variance\quad of\quad a\quad random\quad variable\quad tells\quad us\quad something\quad about\quad the\quad spread\quad of\quad the\\ possible\quad values\quad of\quad the\quad variable.For\quad a\quad discrete\quad random\quad variableX,the\quad variance\quad of\quad X\\ \quad is\quad written\quad as\quad Var(X).\quad Var(X)=E[(X–m)^{ 2 }]\quad where\quad m\quad is\quad the\quad expected\quad value\quad E(X)\\ This\quad can\quad also\quad be\quad written\quad as\quad :\quad Var(X)\quad =\quad E(X^{ 2 })–m^{ 2 }\\ E(X^{ 2 })=\quad \frac { 2 }{ n(n+1) } ×(1^{ 3 }+2^{ 3 }+3^{ 3 }+.....n^{ 3 })\\ \quad \quad \quad =\frac { 2 }{ n(n+1) } \times (\frac { n(n+1) }{ 2 } )^{ 2 }\\ \quad \quad =\frac { n(n+1) }{ 2 } \\ Var(X)\quad =\quad E(X^{ 2 })–m^{ 2 }\\ \quad \quad \quad \quad \quad \quad =\quad \frac { n(n+1) }{ 2 } -(\frac { 2n+1 }{ 3 } )^{ 2 }\\ solving\quad we\quad get,\\ \quad \quad \quad Var(X)\quad \quad =\quad \frac { (n+2)(n-1) }{ 18 } \\ \\$

Compute standard deviation for the following data.
Class: 0-4 4-8 8-12 12-16 16-20
Frequency: 6 10 12 10 8

$Class$	$Class-mark$ $x_i$	$f_i$	$f_ix_i$	$x-\overline{x}$	$(x-\overline{x})^2$	$f(x-\overline{x})^2$
$0-4$	$2$	$6$	$12$	$-8.35$	$69.7225$	$418.335$
$4-8$	$6$	$10$	$60$	$-4.35$	$18.9225$	$189.225$
$8-12$	$10$	$12$	$120$	$-0.35$	$0.1225$	$1.47$
$12-16$	$14$	$10$	$140$	$3.65$	$13.3225$	$133.225$
$16-20$	$18$	$8$	$144$	$7.65$	$58.5225$	$468.18$
$Total$		$46$	$476$			$1210.435$

$\sum f_ix_i=476$ and

$\sum f_i=46$

$\therefore$

$Mean(\overline{x})=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{476}{46}=10.35$

$\Rightarrow$ Standard deviation

$\sigma=\sqrt{\dfrac{\sum f(x-\overline{x})^2}{\sum f_i}}$

$=\sqrt{\dfrac{1210.435}{46}}$

$=\sqrt{26.3138}$

$=5.13$

If the mean of the data: $7, 8, 9, 7, 8, 7, \lambda, 8$ is $8$ , then the variance of this data is?

$\Rightarrow$ Given numbers are

$7,8,9,7,8,7,\lambda,8$

$\Rightarrow$

$\overline{x}=8$

$\therefore$

$\overline{x}=\dfrac{7+8+9+7+8+7+\lambda+8}{8}$

$\therefore$

$8=\dfrac{54+\lambda}{8}$

$\therefore$

$\lambda=10$

$Variance=\dfrac{(7-8)^2+(8-8)^2+(9-8)^2+(7-8)^2+(8-8)^2+(7-8)^2+(10-8)^2+(8-8)^2}{8}$

$=\dfrac{1+0+1+1+0+1+4}{8}$

$=\dfrac{8}{8}$

$=1$

$\therefore$

$Variance=1$

In binomial distribution, the sum and the product of mean and the variance are $\dfrac{25}{3}$ and $\dfrac{50}{3}$ respectively. Determine the distribution.

Mean

$=np$

Variance

$=npq$

$np+npq=\dfrac {25}{3}$

$n^2 p^2 q=\dfrac {50}{3}$

Also

$q=1-p$

on solving we get

$n=15\ p=\dfrac {1}{3}\ q=\dfrac {2}{3}$

$B$ in rational distribution is at

$15$ distribute trails in which probability of success

$=\dfrac {1}{3}$

Obtain the mean and standard deviation for the following data set:
X 10 20 30 40 50 60 70
Y 05 15 30 35 25 27 08

$x$	$y$	$xy$	$x-\overline{x}$	$(x-\overline{x})^2$
$10$	$05$	$50$	$-32$	$1024$
$20$	$15$	$300$	$-22$	$484$
$30$	$30$	$900$	$-12$	$144$
$40$	$35$	$1400$	$-2$	$4$
$50$	$25$	$1250$	$8$	$64$
$60$	$27$	$1620$	$18$	$324$
$70$	$08$	$560$	$28$	$784$
	$\sum y=145$	$\sum xy=6080$		$\sum(x-\overline{x})^2=2828$

$\Rightarrow$

$Mean(\overline{x})=\dfrac{\sum xy}{\sum y}=\dfrac{6080}{145}=41.93\approx 42$

$\sigma=\sqrt{\dfrac{\sum (x-\overline{x})^2}{\sum y}}$

$=\sqrt{\dfrac{2828}{145}}$

$=\sqrt{19.50}$

$=4.46$

Write the formula of mode for grouped data and explain terms in it.

$Mode=Li+[f1-f\dfrac { 0 }{ 2 } f1-f0-f2]\times h$

$Li=lower\quad limit.$

$F1=first\quad frequency$

$F0=upper\quad frequency\quad of\quad first\quad frequency.$

$F2=second\quad frequency.$

$H=difference\quad between\quad class\quad interval.$

Given $N = 10 , \Sigma x_i = 60$ and $\Sigma x_i^2= 1000 .$ The standard deviation is

Given that

$N = 10$

$\sum x_i = 60$

$\sum x_i^2 = 1000$

we know

Standard Deviation

$(\sigma) = \sqrt{\dfrac{\sum x_i^2}{N} - \left(\dfrac{\sum x_i}{N} \right)^2}$

Substitute above values

$= \sqrt{\dfrac{1000}{10} - \left(\dfrac{60}{10} \right)^2}$

$= \sqrt{100 - \dfrac{(60)^2}{10^2}}$

$= \sqrt{100 - \dfrac{3600}{100}}$

$= \sqrt{100 - 36}$

$= \sqrt{64}$

Standard Deviation

$= 8$

The following table given the daily wages of workers in a factory. Compute the standard deviation and the coefficient of variation of the wages of the workers.
Wages (Rs) $125-175$ $175-225$ $225-275$ $275-325$ $325-375$ $375-425$ $425-475$ $475-525$
Number of workers $2$ $22$ $19$ $14$ $3$ $4$ $6$ $1$

$Wages$	$x_i$	$f_i$	$f_ix_i$	$(x-\overline{x})$	$(x-\overline{x})^2$
$125-175$	$150$	$2$	$300$	$-124.65$	$15537.6225$
$175-225$	$200$	$22$	$4400$	$-74.65$	$5572.6225$
$225-275$	$250$	$19$	$4750$	$-24.65$	$607.6225$
$275-325$	$300$	$14$	$4200$	$25.35$	$642.6225$
$325-375$	$350$	$3$	$1050$	$75.5$	$5677.6225$
$375-425$	$400$	$4$	$1600$	$125.35$	$15712.6225$
$425-475$	$450$	$6$	$2700$	$175.35$	$30747.6225$
$475-525$	$500$	$1$	$500$	$225.35$	$50782.6225$
		$\sum f_i=71$	$\sum f_ix_i=19500$		$\sum(x-\overline{x})^2=125280.98$

$\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{19500}{71}=274.65$

$\sigma=\sqrt{\dfrac{\sum(x-\overline{x})^2}{\sum f_i}}$

$=\sqrt{\dfrac{125280.98}{71}}$

$=\sqrt{1764.5208}$

$=42.0061\approx 42$

$\Rightarrow$

$CV=\dfrac{\sigma}{\overline{x}}\times 100$

$=\dfrac{42}{274.65}\times 100$

$=15.3$

If a coin is tossed three times and $X=$ number of heads, then find $E(X)$ and $V(X)$ .

1092251_1190511_ans_7941cbd6dcb14e618c50e1c290370abe.png

The ages of ten students of a group are given below. The ages have been recorded in years and months:
$8-6,9-0,8-4,9-3,7-8,8-11,8-7,9-2,7-10,8-8$
What is the lowest age?
What is the highest age?
Determine the range?

According to the given data,

(i) Lowest age

$= 7-8$

(ii) Highest age

$= 9-3$

(iii) Range

$= (9-3) - (7-8) = 1-7$

Find the expected value, variance and standard deviation of a random variable whose $p.m.f$ is.
$X=x$ $1$ $2$ $3$
$p(X=x)$ $\frac{1}{5}$ $\frac{2}{5}$ $\frac{2}{5}$

$X=x$	1	2	3
$P(X=x)$	0.2	0.4	0.4

$\rightarrow$ PMF (probability mass function)

Expected value

$=E[X]=0.2\times 1+0.4\times 2+0.4\times 3$

$=2.2$

Variance

$=E[X^{2}]-(E[X])^{2}$

$E[X^{2}]=0.2\times 1^{2}+0.4\times 2^{2}+0.4\times 3^{2}=5.4$

Variance

$=5.4-(2.2)^{2}=0.56$

Standard deviation

$=\sqrt{variance}=\sqrt{0.56}=0.748$

On situation where mode would be an appropriate representative value.

A footballer's performance can be rated in terms of how often he scores goals in calendar year.

The variance of first $50$ even natural numbers is

$\textbf{Step -1: Find the mean of first 50 even natural numbers.}$

$\text{First }50\text{ even natural nos. are:}$

$2,4,6,8,\ldots,100$

$\text{Sum of observations}=2+4+6+8+\ldots+100$

$=2(1+2+3+4+\ldots+50)$

$=2\left(\dfrac{50\times51}{2}\right)$ ${\left[\because\text{Sum of n natural nos.}=\dfrac{n(n+1)}{2}\right]}$

$=50\times51$

$=2550$

$\text{Number of observations, }n=50$

$\text{Mean, }\overline{x}=\dfrac{2550}{50}$

$=51$

$\textbf{Step -2: Find the required variance.}$

$\sum_{i=1}^{50} x_i^2=2^2+4^2+6^2+\ldots+100^2$

$=\sum_{k=1}^{50} (2k)^2$

$=4\sum_{k=1}^{50}k^2$

$=4\times\dfrac{50(50+1)(2(50)+1)}{6}$ ${\left[\because\sum_{k=1}^{n} k^2=\dfrac{n(n+1)(2n+1)}{6}\right]}$

$=\dfrac{2\times50\times51\times101}{3}$

$=171700$

$\because\text{Variance, }\sigma^2=\dfrac{\sum x_i^2}{n}-\overline{x}^2$

$\therefore\sigma^2=\dfrac{171700}{50}-51^2$

$=\dfrac{171700-130050}{50}$

$=\dfrac{41650}{50}$

$=833$

$\textbf{Hence, the variance of first 50 even natural numbers is 833.}$

Find the mode of $10$ observations, if their mean is $5$ and median is $4$ .

We proved that,

$Mode=Mean-3(mean-mean)$

$Mode=3 median-2 mean$

Given that,

$mean=5 \ \ meadian=4$

now,

$Mode=3(4)-2(5)$

$=12-10=2$

$mode=2$

Hence, thgis is the answer.

Find the median of:
$17, 23, 36, 12, 18, 23, 40$ and $20$ .

1100941_1192552_ans_5f89e9bdfd714b7ca46423f948ad6b77.jpg

The mean deviation from mean of the observation $a,a+d,a+2d,....,a+2nd$ is

$a, a+d,.....a+2nd$

no. of terms - k

$a+(k-1)d=a+2nd$

no. of terms

$\Rightarrow k=(2n+1)$

sum of all terms

$=\dfrac{k}{2}[a+l]$

$=\dfrac{2n+1}{2}[a+a+2nd]$

sum

$=(2n+1)[a+nd]$

Mean

$=\dfrac{sum}{k}=\dfrac{(2n+1)[a+nd]}{(2n+1)}=(a+nd)$

mean

$=(a+nd)$

mean deviation

$=\dfrac{1}{k}\left \{[mean-a]+[mean -(a+d)]+...….[mean -(a+2nd)]\right\}$

$=\dfrac{1}{k}[nd+(n-1)d+(n-2)d+....0+1d+2d+...nd]$

$=\dfrac{d}{(2n+1)}[1+2+3+...…n+0+1+2+3+.....n]$

$=\dfrac{d}{(2n+1)}\left[\dfrac{n(n+1)}{2}+\dfrac{n(n+1)}{2}\right]$

Mean deviation

$=\dfrac{d}{(2n+1)}[n(n+1)]$ .

Calculate the range and coefficient of range with following data
Marks 0-10 10-20 20-30 30-40 40-50
No. of student 02 05 07 08 03

$x_{m}$ = Mid value of highest class =

$45$

$x_{o}$ = Mid value of lowest class =

$5$

Therefore,

Range =

$x_{m}-x_{o} = 45-5 = 40$

Coefficient of range =

$\dfrac{x_{m}-x_{o}}{x_{m}+x_{o}}$

$\dfrac{45-5}{45+5} = \dfrac{40}{50} =\dfrac{4}{5} = 0.8$

Find the mean deviation from the median for the following data:
$x_1^{}$
6 9 3 12 15 13 21 22
$f_1^{}$ 4 5 3 2 5 4 4 3

xi	6	9	3	12	15	13	21	22
fi	4	5	3	2	5	4	4	3

First calculating cummulative frequency

xi	fi	cf
6	4	4
9	5	4+5=9
3	3	9+3=12
12	2	12+2=14
15	5	14+5=19
13	4	19+4=23
21	4	23+4=27
22	3	27+3=30

$\therefore N= 30$ (even)

Median = Mean of

$(\frac{N}{2})^{th}$ term is

$(\frac{N}{2}+1)^{th}$ term

$= \frac{15th\, term + 16^{th}\, term}{2}$

As both observations lie in the

c.f 19

Median = 15 = M

xi 6	fi 4	$\|xi-M\|$ $\|6-15\|=\|-9\|=9$	$fi\|xi-M\|$ $4\times 9=36$
9	5	$\|9-15\|=\|-6\|=6$	$5 \times 6=30$
3	3	$\|3-15\|=\|-12\|=12$	$3 \times 12=36$
12	2	$\|12-15\|=\|-3\|=3$	$2 \times 3=6$
15	5	$\|15-15\|=0$	$5 \times 0=0$
13	4	$\|13-15\|=\|-2\|=2$	$4 \times 4=8$
21	4	$\|21-15\|=\|6\|=6$	$4 \times 6=24$
22	3	$\|22-15\|=\|7\|=7$	$3 \times 7=21$
	$\sum fi=30$		$\sum fi\|xi-M\| = 161$

$\therefore$ Mean deviation

from median

$=\frac{\sum fi\left | xi-M \right |}{\sum fi}$

$\sum fi\left | xi-M \right |= 161$

$\sum fi=30$

$M.D (M) = \frac{161}{30}$

$= 5.37$

1198571_1249831_ans_73c6e6bb978b433ab573e342431891bc.jpg

Determine (a) the mode (b) the median , and (c) the mean for each of the following set of data :
(i) 1,16,11,6,11,8,14,11,21

Mean=(1+16+11+6+11+8+14+11+21)/9=11

Mode= max no of number in data set by arranging in ascending/ descending order=11

Median= middle term by arranging in ascending/ descending order=11

Mean=Median=Mode=11

Find the standard deviation of $10$ observations $111,\ 211,\ 311,...,1011$

The given numbers are,

$111,211,311,.....,1011$

$\Rightarrow$

$\overline{x}=\dfrac{111+211+311+411+511+611+711+811+911+1011}{10}=\dfrac{5610}{10}=561$

$\Rightarrow$ Standard deviation

$=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{10}}$

$=\sqrt{\dfrac{(111-561)^2+(211-561)^2+(311-561)^2+......+(911-561)+(1011-561)^2}{10}}$

$=\sqrt{\dfrac{(-450)^2+(-350)^2+(-250)^2+(-150)^2+(-50)^2+(50)^2+(150)^2+(250)^2+(350)^2+(450)^2}{10}}$

$=\sqrt{\dfrac{202500+122500+62500+22500+2500+2500+22500+62500+122500+202500}{10}}$

$=\sqrt{\dfrac{825000}{10}}$

$=\sqrt{82500}$

$=50\sqrt{33}$

Hence, this is the answer.

The smallest and largest numbers in the data are 21 and 60 respectively. Find the range of the data.

Smallest number = 21

Largest number = 60

Range = Largest - Smallest

= 60 - 21

= 39.

1156353_1278769_ans_130cbf876f26468d822b485e59436f52.jpg

Define both of these:- median, altitude

Median = A median of a triangle is a line segment from a vertex of the triangle to the midpoint of the side opposite that vertex.

Altitude = The perpendicular height of a triangle is known as its altitude

Find the mean deviation about the median for the data
$36,72,46,42,60,45,53,46,51,49$

Given =36,72,46,42,60,45,53,46,51,49

n = 10

median

$=\frac{(\frac{n}{2})^{th}obs+(\frac{n}{2}+1)^{th}}{2}obs$

$\frac{5^{th}obs+6^{th}obs}{2}$

$=\frac{46+49}{2}=\frac{95}{2}=47.5$

mean deviation above median =

$\frac{\epsilon \left | x:-m \right |}{10}$

$\frac{(136-47.51+142-47.51+145-47.51+146-47.51+146-47.51+149-47-51+151-47.51+153-47.51+160-47.51+172-47.51)}{10}$

$\frac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}$

$\frac{70}{10}=7$

1195133_1296398_ans_59197026eb99405784e60113d1e5bea3.jpg

A random variable X has the following probability distribution.
X=x -2 -1 0 1 2 3
P(X=x) 0.1 k 0.2 2k 0.3 k
Find the value of k and the variance of X.

Sum of total probability

$\begin{array}{l} 0.1+k+0.2+2k+0.3+k=1 \\ 4k=1-0.6 \\ k=\dfrac { { 0.4 } }{ 4 } =0.1 \\ Mean\, \bar { x } =\left( { -2\times 0.1 } \right) +\left( { -1\times 0.1 } \right) +0+\left( { 1\times 0.2 } \right) +\left( { 2\times 0.3 } \right) +\left( { 3\times 0.1 } \right) \\ =-0.2-0.1+0+0.2+0.6+0.3 \\ =0.8 \\ Variance=\sqrt { \sum { { { \left( { x-\bar { x } } \right) }^{ 2 } }{ P_{ 1 } } } } =\sqrt { 0.144+0.004+0.128+0.008+0.432+0.484 } \\ =\sqrt { 102 } =10.954 \end{array}$

Find the mean deviation about the median for the data
$13,17,16,14,11,13,10,16,11,18,12,17$

Given: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 19

n = 12

median =

$\frac{6^{th}obs+7^{th}obs}{2}$

$=\frac{13+14}{2}=\frac{27}{2}=13.5$

mean deviation about median

$\frac{(110-13.51+111.13.51+111-13.51+112-13.51+113-13.51+113-13.51+114-13.51+116-13.5+116-13.51+117-13.51+117-13.51+118-13.51)}{12}$

$\frac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5}{12}$

$=\frac{28}{12}=2.23$

1195132_1296396_ans_61a7130f710b4002b3148ea24b5ae569.jpg

In a traffic survey of $125$ cars, the number of people in each car was recorded.
Number of people in each car $1$ $2$ $3$ $4$ $5$
Frequency $50$ $40$ $10$ $20$ $5$
Find the range.

1487020_1306878_ans_718aa708020045e282723374f8393fbf.jpg

Make a frequency table each of the following sets of data.
(i) Ages of students in years in class VI B of a school:
12,13,13,14,14,14,14,13,13,14,13,14,13,13,12,14,13,14,12,13
(ii) The median used by students of class VI B of a school to listen to music:

1244939_1326057_ans_793e0f76210048c6a0abca23cbcba2bb.jpg

If the mean is $54$ and mode $52$ , find the median.

Mean =

$54$

Mode =

$52$

Mean - Mode =

$3 ~$ (Mean

$-$ Median)

$54 - 52$ =

$3(54-$ median

$)$

$2 = 3$ (

$54-$ median)

$2 = 162 - 3$ median

$3$ median

$= 162 - 2$

$3$ median

$= 160$

median

$= \dfrac{160}{3}$

median

$= 53.3$

On the annual day of school age-wise participation of students is given in the following frequency distribution table. Find the median of the students

Age of student	$C.I$	$c.f$	$f$
Less than $6$	$4-6$	$2$	$2$
Less than $8$	$6-8$	$6$	$4$
Less than $10$	$8-10$	$12$	$6$
Less than $12$	$10-12$	$22$	$10$
Less than $14$	$12-14$	$42$	$20$
Less than $16$	$14-16$	$67$	$25$
Less than $18$	$16-18$	$76$	$9$

$N=76$

Median

$={\dfrac{N}{2}}^{th}$ term

$=\dfrac{76}{2}$

$={38}^{th}$ term

$\therefore$ Median

$=38$

Median class

$=12-14$

Find the median of variants $5, 7, 8, 11, 13, 17$ with frequencies $3, 7, 7, 12, 13, 10$ respectively.

Given that variants

$5, 7, 8, 11, 13, 17$ with frequencies

$3, 7, 7, 12, 13, 10$ respectively.

x	f	Cumulative frequency
$5$	$3$	$3$
$7$	$7$	$10=3+7$
$8$	$7$	$17 = 10+7$
$11$	$12$	$29=17+12$
$13$	$13$	$42=29+13$
$17$	$10$	$52=42 + 10$

$\sum f = N = 52$

$N = 52$ -even

$\dfrac{N}{2} = \dfrac{52}{2}$

$= 26^{th}$ term

the cumulative frequency just gleater than

$\dfrac{N}{2}$ value is

$29$ , and the value of x corresponding to

$29$ is

$11$ .

therefore median is

$11$ .

Find the median of the following data:
$3, 1, 5, 6, 3, 4, 5$

Given the data is

$3, 1, 5, 6, 3, 4, 5$ .

If we arrange them in ascending order, then we get,

$1,3,3,4,5,5,6$ .

As there are

$7$ data so the median will be

$\dfrac{7+1}{2}=4$ the term.

So the median is

$4$ .

Find the median of the following data:
$3, 1, 5, 6, 3, 4, 5, 6$

Given data = 3, 1, 5, 6, 3, 4, 5, 6

After arranging data in ascending order, we get

$1, 3, 3, 4, 5, 5, 6, 6$

Here the number of terms is 8 which is even

Median when terms are even= Average of

$\dfrac{n}{2}th$ term and

$\left ( \dfrac{n}{2} + 1 \right)th$ term

$\dfrac{n}{2}th$ term =

$\dfrac{8}{2}$ = 4th term = 4

$\left (\dfrac{n}{2} + 1\right )th$ term = 5th term = 5

$\therefore Median = \dfrac{4 + 5}{2} = \dfrac{9}{2} = 4.5$

The mean and standard deviation of $50$ items were found to be $75$ and $10$ respectively. At the time of checking it was found that one item $65$ was incorrect. Calculate the mean and standard deviation, if the wrong item is omitted.

$Mean=75$

$\Rightarrow \dfrac{\sum x_i}{n}=75\Rightarrow \sum x_i=75\times 50=3750$

$\Rightarrow 3750-65=3685$

New mean

$=\dfrac{3685}{49}=75.2041$

Standard deviation

$\left( \sigma \right)=10$

Variance

$\left( \sigma ^2\right) =100$

$\therefore \dfrac{\sum x_1^2}{n}-\left( \dfrac{\sum x_1}{n}\right)^2=100$

$\Rightarrow \dfrac{\sum x_1^2}{50}-\left( 75\right)^2=100$

$\Rightarrow \sum x_1^2=286250$

New

$\sum x_1^2=286250-\left( 65\right)^2$

$=282025$

New variance

$=\dfrac{282025}{49}-\left( 75.2041\right)^2=99.9556$

New standard deviation

$=\sqrt{99.9556}=9.99778$

Hence, solved.

Write the first $10$ prime numbers and find their median.

First

$10$ prime numbers

$2, 3, 5, 7, 11, 13, 17, 19, 23, 29$

Here number of prime numbers

$(n) = 10$

So median is average of

$\dfrac{n}{2} ^{th}$ term &

$\left(\dfrac{n}{2} + 1\right)^{th}$ term

median

$= \dfrac{\dfrac{n}{2}^{th} term + \left(\dfrac{n}{2} + 1 \right)^{th} term}{2}$

$= \dfrac{\dfrac{10}{2}^{th} terms + \left(\dfrac{10}{2} + 1\right)^{th} term}{2}$

$= \dfrac{5^{th} term + 6^{th} term}{2}$

median

$= \dfrac{11 + 13}{2}$

$= \dfrac{24}{2}$

Median

$= 12$

Is the mode always at the center of data?

A mode is that value among the observations which occurs most frequently.

When a data is arranged in ascending order, then the middle most term is called median .

Thus, mode is not always at the center of data.

The median of $43,75,15,86,30$ is

Given data is

$43,75,15,86,30$

The ascending order is

$15,30,43,75,86$

The median is the middle most term which is

$43$

Find the median of the observations: $59, 75, 68, 70, 74, 75, 80$ .

The observations in ascending order:

$59, 68, 70, 74, 75, 75, 80$ .
Since, the number of observations,

$n = 7$ , which is odd.

$= Median =\Bigg(\dfrac{n+1}{2}\Bigg)th=\Bigg(\dfrac{7+1}{2}\Bigg)th=\Bigg(\dfrac{8}{2}\Bigg)th= 4th$ Observation

$74$
So, the median of the observations is

$74$ .

Find the mean deviation about the mean for the data:
$4,7,8,9,10,12,13,17$

$Mean(\overline{x})=\dfrac{4+7+8+9+10+12+13+17}{8}=\dfrac{80}{8}=10$

$x_i$	$\|x_i-\overline{x}\|$
$4$	$6$
$7$	$3$
$8$	$2$
$9$	$1$
$10$	$0$
$12$	$2$
$13$	$3$
$17$	$7$
	$\sum\|x_i-\overline{x}\|=24$

$\Rightarrow$ Mean deviation about mean

$=\dfrac{24}{8}=3$

Find the median
61-70 5
81-80 15
81-90 20
91-100 30
101-110 10
111-120 8

1484624_1536496_ans_4bd7df9104154f55a5ceb924e4fa7290.jpg

More than ogive and less than ogive for a given data with mean $23$ meet at $(20,22)$ . find its mode.

1230293_1463472_ans_3bdf578bc2164ea585c947c1bab98678.jpg

The marks obtained by $11$ students of a class in a test are given below:-
$23,2,15,38,21,19,23,23,26,34,23$
Find the mode of the above data.

We know mode means which observation is most frequently occured

In above data observations

we observe that

$23$ is most frequently occured

$\therefore$ mode

$= 23$

Find the mode for the following data:
$7.5, 7.3, 7.2, 7.2, 7.4, 7.7, 7.7, 7.5, 7.3, 7.2, 7.6, 7.2$

Given data is,

$7.5,7.3,7.2,7.2,7.4,7.7,7.7,7.5,7.3,7.2,7.6,7.2$

We know that mode means which is the most frequently occurred in a series.

In the above series, we observe that

$7.2$ most occurred data

$4$ times hence, the mode of the given data is

$7.2$ .

Find the mean, variance and standard deviation for first six odd natural numbers.

1982554_1710892_ans_51beb2511ac848bbac13ceef97bac08e.jpg

Find the median of:
$3, 11, 7, 2, 5, 9, 9, 2, 10$

The value of the middle-most observation is called the median of the data.
First arrange the given numbers in ascending order,

$2, 2, 3, 5, 7, 9, 9, 10, 11$
Here

$n = 9,$ which is odd.
Where, n is the number of the given number.

$\therefore$ median = value of

$\dfrac{1}{2} (n + 1)^{th}$ observation.

$= \dfrac{1}{2} (9 + 1)$

$= \dfrac{1}{2} (10)$

$= 10 / 2$

$= 5$
Then, value of

$5^{th}$ term

$= 7$
Hence, the median is

$7.$

Using short ad method.find the mean, variance and standard deviation for the data:
Class 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Frequency 3 7 12 15 8 3 2

2037926_1710898_ans_0b71000b3f154d9ebadbe7549cf16ba6.jpg

Find the median of:
$21, 15, 6, 25, 18, 13, 20, 9, 16, 8, 22$

The value of the middle-most observation is called the median of the data.
First arrange the given numbers in ascending order,

$6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25$
Here

$n = 11$ , which is odd.
Where, n is the number of the given number.

$\therefore$ median = value of

$\dfrac{1}{2} (n + 1)^{th}$ observation.

$= \dfrac{1}{2} (11 + 1)$

$= \dfrac{1}{2} (12)$

$= 12 / 2$

$= 6$
Then, value of

$6^{th}$ term

$= 16$
Hence, the median is

$16.$

Find the mean, variance and standard deviation for the numbers 4, 6, 10, 12, 7, 8, 13, 12.

1982552_1710891_ans_0d75820d67414b03b1d95c33a75df4bf.jpg

Find the median of:
$10, 32, 17, 19, 21, 22, 9, 35$

The value of the middle-most observation is called the median of the data.
First arrange the given numbers in ascending order,

$9, 10, 17, 19, 21, 22, 32, 35$
Here

$n = 8,$ which is even.
Where, n is the number of the given number.

$\therefore \,median = 1/2 {(n/2)^{th} term + ((n/2) + 1)^{th} term}$

$= 1/2 {(8/2)^{th} term + ((8 / 2) + 1)^{th} term}$

$= 1/2 {(4)^{th} bterm + ((4) + 1)^{th} term}$

$= 1/2 {(4)^{th} term + (5)^{th} term}$

$= 1/2 (19 + 21)$

$= 1/2 (40)$

$= (40 / 2)$

$= 20$
Hence, the median is

$20.$

Using short ad method.find the man, variance and standard deviation for the data:
$x_i$ 6 10 14 18 24 28 30
$f_i$ 2 4 7 12 8 4 3

2016749_1710894_ans_1f2b81e0ab4448549e42343f8848d1ed.png

Using short ad method.find the man, variance and standard deviation for the data:
$x_i$ 4 8 11 17 20 24 32
$f_i$ 3 5 9 5 4 3 1

1982555_1710893_ans_99fdde488f2c45d0849055d6540dfc30.jpg

Find the median of first $15$ odd numbers.

The first 15 odd numbers are

$1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27$ and

$29$
Here

$n = 15,$ which is odd.
Where, n is the number of the given number.

$\therefore \,median = value \,of \,\dfrac{1}2 (n +1)^{th}$ observation.

$= \dfrac{1}2 (15 + 1)$

$= \dfrac{1}2 (16)$

$= \dfrac{16}2$

$= 8$
Then, value of

$8^{th}$ term

$= 15$
Hence, the median of first

$15$ odd numbers is

$15.$

The age (in years) of $11$ cricket players are given below:
$28, 34, 32, 41, 36, 32, 32, 38, 32, 40, 31$
Find the mode of the ages.

Mode is the value of the variable which occurs most frequently.

Arranging the given numbers in ascending order, we get

$28, 31, 32, 32, 32, 32, 34, 36, 38, 40, 41$

Clearly,

$32$ occurs maximum number of times.

Hence, mode of the ages (in years) of

$11$ cricket players is

$32$ years.

Find the median for the following distribution:

Wages per day in Rs. $38$ $45$ $48$ $55$ $62$ $65$
No. of workers $14$ $8$ $7$ $10$ $6$ $2$

We write the distribution in the cumulative frequency table.

Wages per day in Rs.	Number of worker $(f)$	Cumulative frequency
$38$	$14$	$14$
$45$	$8$	$22$
$48$	$7$	$29$
$55$	$10$	$39$
$62$	$6$	$45$
$65$	$2$	$47$

Here the total number of observations are $47$ which is odd.

So median $=((n+1)/2)^{th}term$

$= ((47+1)/2 )^{th} term$

$= (48/2)^{th} term$

$= 24^{th} term$

$= 48$ [Since $23^{rd}$ to $29^{th}$ observation is $48$ ]

Hence the median of the given data is $48$ .

For the following set of numbers, find the median: $10, 75, 3, 81, 17, 27, 4, 48, 12, 47, 9, 15$ .

Arranging the numbers in ascending order :

$3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81$

Median is the middle term of the series, when the series is arranged in ascending/ descending order.

Here, $n = 12$ which is even. For a sequence with even number of terms, the median is given by the following formula:

$\text{Median} = \dfrac{1}{2}\left( \frac{n}{2}^{th}\ \text{term} + \left(\frac{n}{2}+1\right)^{th}\ \text{term}\right)$

$=\dfrac{1}{2}\left( \frac{6}{2}^{th}\ \text{term} + \left(\frac{6}{2}+1\right)^{th}\ \text{term}\right)$

$= \dfrac{1}{2}(6^{th} \text{term} + (6+1)^{th} \text{term})$

$= \dfrac{1}{2}(6^{th} \text{term} + 7^{th} \text{term})$

$= \dfrac{1}{2}(15+17)$

$= \dfrac{1}{2}\times32$

$= 16$

Hence, the median is $16$ .

Find the median of first $50$ whole numbers.

The first 50 whole numbers are

$0, 1, 2, 3, 4, …, 49.$
Here

$n = 50,$ which is even.
Where, n is the number of the given number.

$\therefore$ median

$= 1/2 {(50/2)^{th} term + ((50/2) + 1)^{th} term}$

$= 1/2 {(25)^{th} term + ((25) + 1)^{th} term}$

$= 1/2 {(25)^{th} term + (26)^{th} term}$

$= 1/2 (24 + 25)$

$= 1/2 (49)$

$= (49 / 2)$

$= 24.5$
Hence, the median of first 50 whole numbers is 24.5.

A student scored the following marks in $11$ questions of a question paper: $3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7$ Find the median marks.

Arranging the data in the ascending order

$1,2,2,3,4,5,5,6,7,7,8$

Here number of terms, $n = 11$

Here n is odd.

So median $= [(n+1)/2 ]^{th}$ observation

$= (11+1)/2$

$= 12/2$

$=6^{th}$ observation

Here $6^{th}$ observation is $5$ .

Hence the median is $5$ .

Find the median of:
$55, 60, 35, 51, 29, 63, 72, 91, 85, 82$

The value of the middle-most observation which is arranged in increasing order is called the median of the data.
First arrange the given numbers in ascending order,

$29, 35, 51, 55, 60, 63, 72, 82, 85, 91$

Here

$n = 10,$ which is even.

Where, n is the number of the given number.

$\therefore \,median = \dfrac{1}{2} [{(n / 2)^{th} \text {term} + ((n / 2) + 1)^{th} \text {term}}]$

$= \dfrac{1}{2}[{(10/ 2)^{th} term + ((10 / 2) + 1)^{th} term}]$

$= \dfrac{1}{2}[{(5)^{th}term + ((5) + 1)^{th} term}]$

$= \dfrac{1}{2}[{(5)^{th} term + (6)^{th} term}]$

$= \dfrac{1}{2} [(60 + 63)]$

$= \dfrac{1}{2}[(123)]$

$= 61.5$

Hence, the median is

$61.5.$

Find the mode of the data:
$27, 23, 39, 18, 27, 21, 27, 27, 40, 36, 27$

Mode is the value of the variable which occurs most frequently.

Arranging the given numbers in ascending order, we get

$18, 21, 23, 27, 27, 27, 27, 27, 36, 39, 40$

Clearly,

$27$ occurs maximum number of times.

Hence, mode of the given number is

$27$ .

Find the mode of the data:
$10, 8, 4, 7, 8, 11, 15, 8, 6, 8$

Mode is the value of the variable which occurs most frequently.

Arranging the given numbers in ascending order, we get

$4, 6, 7, 8, 8, 8, 8, 10, 11, 15$

Clearly,

$8$ occurs maximum number of times.

Hence, mode of the given number is

$8$ .

Find the median of the following set of numbers: $9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7$ .

Arranging the numbers in ascending order :

$0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9$

Here, $n = 12$ which is even

$Median = \frac{1}{2}( n/2^{th} term + ((n/2)+1)^{th} term)$

$= \dfrac{1}{2}(12/2)^{th} term + ((12/2)+1)^{th} term)$

$= \dfrac{1}{2} (6^{th} term + (6+1)^{th} term)$

$=\dfrac{1}{2} (6^{th} term + 7^{th} term)$

$= \dfrac{1}{2} (4+5)$

$= 9/2$

$= 4.5$

Hence the median is $4.5$ .

At a shooting competition, the scores of a competitor were as given below :
Score
0
1
2
3
4
5
No. of shots
0
3
6
4
7
5
What was his median score?

We write the marks in cumulative frequency table.

Score $x$	No. of shots (f)	fx	Cumulative frequency
0	0	0	0
1	3	3	3
2	6	12	9
3	4	12	13
4	7	28	20
5	5	25	25
Total	$\Sigma f = 25$	$\Sigma fx = 80$

Here the number of observations, $n = 25$ which is odd.

$Median = ((n+1)/2)^{th} term$

$= (25+1)/2$

$= 26/2$

$= 13^{th} term$

$= 3$

Hence his median score is $3$ .

Marks obtained by $70$ students are given below:
Marks
20
70
50
60
75
90
40
No. of students
8
12
18
6
9
5
12
Calculate the median marks.

We write the marks in ascending order in cumulative frequency table.

Marks	No. of students (f)	Cumulative frequency
20	8	8
40	12	20
50	18	38
60	6	44
70	12	56
75	9	65
90	5	70

Here total number of observations, $n = 70$ which is even.

$Median =\dfrac{1}{2}( n/2^{th} term + ((n/2)+1)^{th} term)$

$= \dfrac{1}{2}(70/2^{th} term + ((70/2)+1)^{th} term)$

$= \dfrac{1}{2}(35^{th} term + (35+1)^{th} term)$

$= \dfrac{1}{2} (35^{th} term + 36^{th} term)$

$= \dfrac{1}{2} (50+50)$ [Since all observations from $21^{st}$ to $38^{th}$ are $50$ ]

$= \dfrac{1}{2} \times 100$

$= 50$

Hence the median is $50$ .

Find the mode of the following observations
$5, 7, 6, 8, 9, 0, 6, 8, 1, 8$

Given set is $5, 7, 6, 8, 9, 0, 6, 8, 1, 8$ .

In this set, $8$ occurs maximum number of times.

Hence the mode is $8$ .

Find the mode of the following sets of numbers: $3, 2, 0, 1, 2, 3, 5, 3$ .

Mode is the number that appears most often in a set of numbers.

(i)Given set is $3, 2, 0, 1, 2, 3, 5, 3$ .

In this set, $3$ occurs the maximum number of times.

Hence the mode is $3$ .

Find the median for the following distribution.
Marks
35
45
50
64
70
72
No. of students
3
5
8
10
5
5

We write the distribution in the cumulative frequency table.

Marks	No. of students (f)	Cumulative frequency
35	3	3
45	5	8
50	8	16
64	10	26
70	5	31
72	5	36

Here total number of observations, $n = 36$ which is even.

$Median = \dfrac{1}{2}( n/2^{th} term + ((n/2)+1)^{th} term)$

$= \dfrac{1}{2}(36/2^{th} term + ((36/2)+1)^{th} term)$

$=\dfrac{1}{2}(18^{th} term + (18+1)^{th} term)$

$=\dfrac{1}{2}(18^{th} term + 19^{th} term)$

$=\dfrac{1}{2}(64+64)$ [Since $17^{th}$ to $26^{th}$ observation is $64$ ]

$= \dfrac{1}{2}\times128$

$= 64$

Hence the median is $64$ .

The daily wages in (rupees of) $19$ workers are
$41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35$ .
Find the median?

Arranging the observations in ascending order,

$21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53$

As there are $19$ observations which is odd so, $\left(\dfrac{n+1}2\right)^{th}$ observation will be the median.

$\left(\dfrac{n+1}2\right)=\dfrac{19+1}2$

$=\dfrac{20}2$

$=10$

Hence, $10^{th}$ observation that is $31$ is the median of the data set.

A boy scored the following marks in various class tests during a term each test being marked out of $20$ : $15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16$ . What are his median marks?

Given: Marks scored by the boy in 11 subjects $: 15, 17,16,7,10,12,14,16,19,12,16$

Ascending order of the marks $:\ 7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19$

Here the number of observations, $n = 11$ which is odd.

Median $= \bigg( \dfrac{n+1}{2} \bigg)^{th} \mathrm{term}$

$= \dfrac{11+1}{2}$

$= \dfrac{12}{2}$

$= 6^{th}\ \mathrm{term}$

$= 15$

Hence the median marks of the boy is $15$

Find the mode of the following sets of numbers: $9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7$

Given set is $9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7$ .

In this set, $5$ occurs the maximum number of times.

Hence the mode is $5$ .

Find the median of first nine even natural numbers.

The first nine even natural numbers are

$2,4,6,8,10,12,14,16,18$

The total number of observations

$=9$ , which is odd.

Hence, the median is

$10$ , the 5th observation.

Find the
mean deviation about the median for the data
$x_i$ 5 7 9 10 12 15
$f_i$ 8 6 2 2 2 6

The given
observations are already in ascending order.

Adding a column corresponding to cumulative
frequencies of the given data, we obtain the

$x_i$	$f_i$	c.f.
5	8	8
7	6	14
9	2	16
10	2	18
12	2	20
15	6	26

Here $\mathrm{N}=26$
which is even

Median is
the mean of 13th and 14th observations. Both of these observations lie in the
cumulative frequency 14, for which the corresponding observation is 7

$\therefore Median=\dfrac{13^{\text {th }} \text { observation }+14^{\text {th }} \text { observation }}{2}=\dfrac{7+7}{2}=7$

The absolute values of the deviations from median,
ie. $\left|x_{i}-M\right|$ , are

$\left\|\mathrm{x}_{i}-\mathrm{M}\right\|$	2	0	2	3	5	8
$\mathrm{f}_{i}$	8	6	2	2	2	6
$\mathrm{f}_{i}\left\|\mathrm{x}_{i}-\mathrm{M}\right\|$	16	0	4	6	10	48

$\sum \limits _{i=1}^{6} f_{i}=26$ and $\sum \limits _{i=1}^{6} f_{i}\left|x_{i}-M\right|=84$

M.D. $(M)=\dfrac{1}{N} \sum \limits _{i=1}^{6} f_{i}\left|x_{i}-M\right|=\dfrac{1}{26} \times 84=3.23$

Enter 1 if it is true else enter 0.
In a set of observations, the observation that occurs the most often is called Mode.

1900251_1790872_ans_8dc6bae0c3694b769bd58b40a5cdeac7.jpg

Enter 1 if it is true else enter 0.
In a given data, arranged in ascending or descending order, the middle most observation is called Median.

The correct answer is 1.

1900253_1790875_ans_24a02717084f4bb3a4d5c26aeea0d05b.jpg

Observe the data and answer the questions that follow:
$16,15,16,16,8,15,17$
(a) Which data value can be put in the data so that the mode remains the same?
(b) At least how many and which value(s) must be put in to change the mode to 15?
(c) What is the least number of data values that must be put in to change the mode to 17? Name them.

Arranging the given data in ascending order, we get

$8,15,15,16,16,16,17$

(a) The observation

$16$ occurs most frequently 3 times in the data

$\therefore$ Mode

$=16$

If we place

$15$ in the data, the mode will change as

$15$ already occurs two times.

Hence,

$16, 8$ or

$17$ can be put in the data so that the mode will remain the same.

(b) The mode will be

$15$ if it occurs 4 times at least.

Hence, two more

$15$ should be put in the data to make the mode

$15$ .

(c)

$17$ comes only once in the data and to be the mode it should occur at least

$4$ times.

Therefore,

$17$ should be put in the data at least

$3$ times to change the mode to

$17$ .

Enter 1 if it is true else enter 0.
Mean, Median, Mode are the measures of Central tendency.

The correct answer is 1.

1900256_1790879_ans_1fbdce3456c346b2867a621530f25923.jpg

Find the mean deviation about the mean for the data.
$x$ 5 10 15 20 25
$f_{i}$ 7 4 6 3 5

$\mathrm{N}=\sum \limits _{i=1}^{5} \mathrm{f}_{i}=25$

$\sum \limits _{i=1}^{5} f_{i} x_{i}=350$

$\therefore \bar{x}=\dfrac{1}{N} \sum_{i=1}^{5} f_{i} x_{i}=\dfrac{1}{25} \times 350=14$

$\therefore \mathrm{MD}(\overline{\mathrm{x}})=\dfrac{1}{\mathrm{N}} \sum \limits_{i=1}^{5} \mathrm{f}_{i}\left|\mathrm{x}_{i}overline{\mathrm{x}}\right|=\dfrac{1 }{25} \times 158=6.32$

1904913_1798775_ans_ad40c78cd73d4ca59a08687249e5a6ed.PNG

Find the mode of the given data:
$10,8,4,7,8,11,15,8,4,2,3,6,8$

Rearranging the given observation into ascending order

$2,3,4,4,6,7,8,8,8,8,10,11,15$

The mode is

$8$ as it occurs most frequently i.e., 4 times in the data.

Age (in years) of 6 children of two groups are recorded as below:
Group A Group B
7 7
7 9
9 11
8 12
10 12
10 12

(a) Find the mode and range for each group.
(b) Find the range and mode if the two groups are combined together

(a) For group A;

Mode is 7 and 10 because both come two times.

Range

$=10-7=3$ years

For group B;

Mode is 12 as it occurs most frequently 3 time.

Range

$= 12 - 7=5$ years

(b) Combining the group A and the group B and arranging them in ascending order, we get

$7,7,7,8,9,9,10,10,11,12,12,12,$

As 7 and 12 occur most frequently 3 times in the above given data.

The mode is 7 and 12 years

Range

$=$ Highest age

$-$ Lowest age

$=12-7=5$ years

Find the median of
$26,33,41,18,30,22,36,45,$ and $24$

Given

Numbers are

$26,33,41,18,30,22,36,45,$ and

$24$

Total numbers

$=9$

Total number of numbers is odd.

Hence, the median will be the middle number.

Here, middle number is

$30$ .

Therefore, the median is

$30$ .

Find the mean deviation about the median for the data
$\mathrm{x_i}$ 15 21 27 30 35
$\mathrm{f}_{\mathrm{i}}$ 3 5 6 7 8

The given observations are already in ascending order.

Adding a column corresponding to cumulative frequencies of the given data, we obtain the table.

Here,

$N=29$ which is odd

Median =

$=\left(\dfrac{7+7}{2}\right)^{\text {th }}$ observation

$=15^{th }$ observation

This observation lies in the cumulative frequency 21, for which the corresponding observation is 30

$\therefore$ Median

$=30$

$\sum\limits_{i=1}^{5} f_{i}=29, \sum \limits _{i=1}^{5}f_{i}\left|x_{i}-M\right|=148$

$\therefore \mathrm{M} . \mathrm{D} .(\mathrm{M})=\dfrac{1}{\mathrm{N}} \sum \limits _{i=1}^{5} \mathrm{f}_{i}\left|\mathrm{x}_{i} \mathrm{M}\right|=\dfrac{1 }{29} \times 148=5.1$

1904772_1798938_ans_5828d1d442934173864cf7d264ee5eb8.PNG

A student got the following marks in $9$ questions of a question paper.
$3, 5, 7, 3, 8, 0, 1, 4$ and $6.$
Find the median of these marks.

The following points need to be followed in order to find the median of a data:

1. Arrange the given data in descending order:

$8, 7, 6, 5, 4, 3, 3, 1, 0$

2. The middle term of the sequence so obtained is the median of the data. Here,

$4$ is the middle term, which is the

$5^{th}$ term.
Hence, median

$= 4$

Find the mean and variance for the data
$x_i$ 6 10 14 18 24 28 30
$f_i$ 2 4 7 12 8 4 3

Here

$N=40, \sum \limits _{i=1}^{7} f_{i} x_{i}=760$

$\therefore \bar{x}=\dfrac{\sum \limits _{i=1}^{7} f_{i} x_{i}}{N}=\dfrac{760} {40}=19$

$\mathrm{Variance}=(\sigma)^{2}=\dfrac{1}{\mathrm{N}} \sum \limits _{i=1}^{7} \mathrm{f}_{i}\left(x_{i}-\bar{x}\right)^{2}=\dfrac{1}{40}\times 1736=43.4$

1905489_1799240_ans_25229291075d40e49f7fc64ef80b1b8c.PNG

Find the mean and variance for the data
$x_i$ 92 93 97 98 102 104 109
$f_i$ 3 2 3 2 6 3 3

$\mathrm{x_i}$

$\mathrm{f}_{\mathrm{i}}$

$\mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}$

$\mathrm{x}_{1}-\bar{x}$

$\left(\mathrm{x}_{1}-\bar{x}\right)^{2}$

$\mathrm{f}_{\mathrm{i}}\left(\mathrm{x}_{\mathrm{i}}-\bar{x}\right)^{2}$

92 3 276 -8 64 192

93 2 186 -7 49 98

97 3 291 -3 9 27

98 2 196 -2 4 8

102 6 612 2 4 24

104 3 312 4 16 48

109 3 327 9 81 243

22 2200 640

Here

$N=22, \sum \limits _{i=1}^{7} f_{i} x_{i}=2200$

$\therefore \bar{x}=\dfrac{\sum \limits _{i=1}^{7} f_{i} x_{i}}{N}=\dfrac{1}{22} \times 2200 =100$

$\mathrm{Variance}=(\sigma)^{2}=\dfrac{1}{\mathrm{N}} \sum \limits _{i=1}^{7} \mathrm{f}_{i}\left(x_{i}-\bar{x}\right)^{2}=\dfrac{1}{22} \times 640=29.09$

Find the mean and variance for the first n natural numbers

The mean of first n natural numbers is calculated as follows.

Mean

$=\dfrac{\text { Sum of all observations }}{\text { Number of observations }}$

$\therefore$ Mean

$=\dfrac{\dfrac{n(n+1)}{2}}{n}=\dfrac{n+1}{2}$

Variance

$\left(\sigma^{2}\right)=\dfrac{1}{n} \sum \limits _{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}$

$=\dfrac{1}{n} \sum \limits _{i=1}^{n}\left[x_i-\left(\dfrac{n+1}{2}\right)\right]^{2}$

$=\dfrac{1}{n} \sum \limits _{n=1}^{n} x_i^{2}-\dfrac{1}{n} \sum \limits _{i=1}^{n}2\left(\dfrac{n+1}{n}\right) x+\dfrac{1}{n} \sum \limits _{i=1}^{n}\left(\dfrac{n+1}{2}\right)^{2}$

$=\dfrac{1}{n} \dfrac{n(n+1)(2 n+1)}{6}-\left(\dfrac{n+1}{n}\right)\left[\dfrac{n(n+1)}{2}\right]+\dfrac{(n+1)^{2}}{4n} \times n$

$=\dfrac{(n+1)(2 n+1)}{6}-\dfrac{(n+1)^{2}}{2}+\dfrac{(n+1)^{2}}{4}$

$=\dfrac{(n+1)(2 n+1)}{6}-\dfrac{(n+1)^{2}}{4}$

$=(n+1)\left[\dfrac{4 n+2 -3 n-3}{12}\right]$

$=\dfrac{(n+1)(n-1)}{12}$

$=\dfrac{n^{2}-1}{12}$

Find the mean and the median of :
$2,4,5,8,10,13$ and $14$

Given
Numbers are

$2,4,5,8,10,13$ and

$14$
The mean can be calculated as below :
Mean

$=$ (Sum of numbers)

$/$ (Number of numbers)

$=(2+4+5+8+10+13+14)/7$

$=56/7$

$=8$
Hence, the mean is

$8$

Total numbers are

$7$
Since, total number of numbers is odd, the median will be the middle number.
Here, middle number is

$8$
Therefore, the median is

$8$

Find the mode of the following distribution:
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Number of students 15 20 24 24 12 31 71 52

Here, the maximum number of students
i.e. 71 have got marks in the interval 60-70.
So, the modal class is 60-70
Now, modal class = 60-70,
lower limit (I) of the modal class= 60, class size(h)=10
frequency

$({ f }_{ 1 })$ of the modal class=71
frequency

$({ f }_{ 0 })$ of class preceding the modal class=31
frequency

$({ f }_{ 2 })$ of class preceding the modal class=52
Now, let us substitute these values in the formula

$Mode = 1+\left( \dfrac { { f }_{ 1 }-{ f }_{ 0 } }{ 2{ f }_{ 1 }-{ f }_{ 0 }-{ f }_{ 2 } } \right) \times h$

$=60+\left( \dfrac { 71-31 }{ 2X71-31-52 } \right) \times10$

$=60+\dfrac { 40 }{ 59 } \times10$

$=60+6.78$

$=66.78$

Fill in the blanks:
A data arranged in ascending or descending order is called an........ data.

A data arranged in ascending or descending order is called an arrayed data.

Find the mode of the following data: $9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17$ and $8$

Arrange in ascending order

$3,5,8,8,9,9,11,11,11,11,16,17$

It’s seen that

$11$ occurs

$4$ times in the given data.
Hence, mode

$= 11$

From the following data, find: Median
$25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38$ and $83$

Arranging the given data in ascending order, we have:

$0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95$
Median is the mean of

$8^{th}$ and

$9^{th}$ term
Thus, median

$= (40 + 45)/ 2 = 85/2 = 42.5$

The marks obtained by $19$ students of a class are given below:
$27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35$ and $28$ . Find: median

Arranging in ascending order:

$22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37$
The middle term is

$10^{th}$ term i.e.

$29$
Hence, median

$= 29$

The following table shows the expenditure of $60$ boys on books. Find the mode of their expenditure:

Clearly,
Mode is in

$30-35$ because it has the maximum frequency.
1796705_1834252_ans_3fa51fd4e5bb4d1d938518d99a36b566.png

241, 243, 347, 350, 327, 299, 261, 292, 271, 258 and 257.

Firstly arrange the numbers in ascending order
241, 243, 257, 258, 261, 271, 292, 299, 327, 347, 350
Now since n = 11 (odd)
Median = value of

$\left ( \frac{n + 1}{2}\right )^{th}$ term

$= 6^{th}$ term

$= 271$
Thus median is 271.

63, 17, 50, 9, 25, 43, 21, 50, 14 and 34.

Firstly arranging the numbers in ascending order

$9,14,17,21,25,34,43,50,63$

Since,

$n = 10$ (even)

We find the median as:

Median

$= \dfrac{1}{2} \left [ \mathrm{value\ of} \left ( \dfrac{n}{2} \right )^{th} \mathrm{term + value\ of} \left ( \dfrac{n}{2} + 1 \right )^{th} \mathrm{term} \right ]$

$= \dfrac{1}{2} \left [ \mathrm{value\ of} \left ( \dfrac{10}{2} \right )^{th} \mathrm{term + value\ of} \left ( \frac{10}{2} + 1 \right )^{th} \mathrm{term} \right ]$

$= \dfrac{1}{2} [25 + 34]$

$= \dfrac{1}{2} [59]$

$= 29.5$

Thus the median is 29.5

Find the mode of the following data: $7, 9, 8, 7, 7, 6, 8, 10, 7$ and $6$ .

First, arrange the given set of numbers in ascending order

i.e,

$6,6,7,7,7,7,8,8,9,10$

It can be seen that

$7$ occurs

$4$ times in the given data.
Hence, mode

$= 7$

Out of 10 students, who appeared in a test, three secured less than 30 marks and 3 secured more than 75 marks. The marks secured by the remaining 4 students are 35, 48, 66 andFind the median score of the whole group.

Here, total observations = n = 10 (even)
Median

$= \frac{1}{2}$ [ value of

$\left ( \frac{10}{2} \right )^{th}$ term + value of

$\left ( \frac{10}{2} + 1 \right )^{th}$ term ]

$= \frac{1}{2}$ [vale of

$(5)^{th}$ term + value of

$(6)^{th}$ term ]

$\therefore$ Median

$= \frac{1}{2} (40 + 48) = \frac{88}{2} = 44$
Therefore, the median score of the whole group is 44.

233, 173, 189, 208, 194, 204, 194, 185, 200 and 220.

Firstly arrange the numbers in ascending order
172,185,189,194,194,200,204,208,220,223

Median

$= \frac{1}{2} \left [ value of \left ( \frac{n}{2} \right )^{th} term + value of \left ( \frac{n}{2} + 1 \right )^{th} term \right ]$

$= \frac{1}{2} \left [ value of \left ( \frac{10}{2} \right )^{th} term + value of \left ( \frac{10}{2} + 1 \right )^{th} term \right ]$

$= \frac{1}{2} [200 + 194]$

$= \frac{1}{2} [394]$

$= 197$
Thus the median is 197

The weights (in kg) of $10$ students of a class are given below:
$21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21$ and $24.$
Find the median of their weights.

Arranging the given data in descending order:

$28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5$
It’s seen that,
The middle terms are

$24$ and

$24$ that is

$5^{th}$ and

$6^{th}$ terms
Thus,

$Median = \dfrac{(24 + 24)}{2} \\= \dfrac{48}{2} \\= 24$

Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of milk.
Milk (litre) 1 - 3 3 - 5 5 - 7 7 - 9 9 - 11 11 - 13
No. of hotels 7 5 15 20 35 18

The maximum class frequency is 35.

The class corresponding to this frequency is 9 - 11.

So, the modal class is 9 - 11.

L (Lower limit of modal class) = 9

$f_{1}$ (frequency of the modal class) = 35

$f_{0}$ (frequency of the class preceding the modal
class) = 20

$f_{2}$ (frequency of the class succeeding the modal class) = 18

h (class size) = 2

Mode =

$L + \left ( \dfrac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right ) * h$

=

$9 + \left ( \dfrac{35 - 20}{2 * 35 - 20 - 18} \right ) *2$

=

$9 + 0.9375$

=

$9.9375 \approx 9.95$

Hence, the mode of supply of milk is 9.94 litres.

The following table shows the classification of number of vehicles and their speeds on Mumbai-Pune express way. Find the median of the data.
Average Speed of vehicles
(km/hr) 60 - 64 64 - 69 70 - 74 75 - 79 79 - 84 84 - 89
No. of vehicles 10 34 55 85 10 6

Here, total frequency =

$\sum fi = 200$

$\therefore$ The median class is 74.5 - 79.5

From the above table, we get

L (Lower class limit of the median class) = 74.5

N (Sum of frequencies) = 100

h (Class interval of the median class) = 5

f (frequency of the median class) = 85

cf (Cumulative frequency of the class

proceedings the median class) = 99

Now, Median $$= L + \left ( \dfrac{\dfrac{N}{2}

- {cf}}{f} \right ) * h$$

$$= 74.5 + \left ( \dfrac{\dfrac{100}{2} -

{99}}{85} \right ) * 5$$

$= 74.5 + 0.059$

$= 74..559 \approx 75$

$\therefore$ The median of the given data is 75 km/hr (approx).

1914569_1852549_ans_c376db078c354a7fbc6089f21d1911e3.PNG

Grouped frequency distribution of supply of milk to hotels and the number of hotels is given in the following table. Find the mode of the supply of mik.
Age (years) :ess than 5 5 - 9 10 - 14 15 - 19 20 - 24 25 - 29
No. of patients 38 32 50 36 24 20

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1988119_1852678_ans_a954544dfc8745aeaff49845a757d58f.jpg

Electricity used by some families is shown in the following table. Find the mode for use of electricity.
Use of electricity (Unit) $0 - 20$ $20 - 40$ $40 - 60$ $60 - 80$ $80 - 100$ $100 - 120$
No of families $13$ $50$ $70$ $100$ $80$ $17$

The maximum class frequency is

$100$ .
The class corresponding to this frequency is

$60 - 80$ .
So, the modal class is

$60 - 80$ .
L (Lower limit of modal class)

$= 60$

$f_{1}$ (frequency of the modal class)

$= 100$

$f_{0}$ (frequency of the class preceding the modal class)

$= 70$

$f_{2}$ (frequency of the class succeeding the modal class)

$= 80$
h (class size)

$= 20$
Mode

$=L + \left ( \dfrac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right ) * h$

$=60 + \left ( \dfrac{100 - 70}{2 * 100 - 70 - 80} \right ) *20$

$= 60 + 12$

$= 72$
Hence, the modal electricity is

$72$ units.

In a basket, there are $10$ tomatoes. The weight of each of these tomatoes in grams is as follows:
$60, 70, 90, 95, 50, 65, 70, 80, 85, 95$ .Find the median of the weights of tomatoes.

The weight of

$10$ tomatoes in a basket in grams in ascending order is as follows:

$50, 60, 65, 70, 70, 80, 85, 90, 95, 95$
Since, the number of observations is

$10$ , which is even.
So, median

$=\dfrac{1}{2} \times \Bigg[\Bigg(\dfrac{10}{2}\Bigg)th\,observation+\Bigg(\dfrac{10}{2}+1\Bigg)th\,observation\Bigg]$

$=\dfrac{1}{2} \times [5th\,observation+(5+1)th\,observation]$

$=\dfrac{1}{2} \times [5th\,observation+6th\,observation]$

$=\dfrac{1}{2} \times [70+80]$

$=\dfrac{150}{2}$

$=75$
So, the median of the weights of tomatoes is

$75$ grams.

The following table shows the information regarding the milk collected from farmers on a milk collection centre and the content of fat in the milk, measured by a lactometer. find the mode of fat content.
Content of Fat (%) 2 - 3 3 - 4 4 - 5 5 - 6 6 - 7
Milk collected (liter) 30 70 80 60 20

The maximum class frequency is 80.
The class corresponding to this frequency is 4 - 5.
So, the modal class is 4 - 5.
L (Lower limit of modal class) = 4

$f_{1}$ (frequency of the modal class) = 80

$f_{0}$ (frequency of the class succeeding the modal class) = 60
h (class size) = 1
Mode =

$L + \left ( \dfrac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right ) * h$
=

$4 + \left ( \dfrac{80 - 70}{2 * 80 - 70 - 60} \right ) *1$
= 4 + 0.33
= 4.33
Hence, the modal fat content is 4.33 litres.

The distance covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distance.
Distance (km) 200-210 210-220 220-230 230-240 240-250
No. of buses 40 60 80 50 20

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1988113_1853108_ans_af76cf14773d47c89fbf33f08c047dfe.jpg

The heights (in cm) of the volley-ball players from team A and team B were recorded as:
Team A: 180, 178, 176, 181, 190, 175, 187
Team B: 174, 175, 190, 179, 178, 185, 177
Find the median of team A and team B.

Median of team A:

Arranging heights in ascending order, we get

175, 176, 178, 180, 187, 190

Total number of observations = n = 7 (odd)

$\therefore$ Median

$= \left (\frac{n + 1}{2} \right )^{th}$ observation

$= \left ( \frac{7 + 1}{2} \right )^{th}$ observation =

$4^{th}$ observation = 180 cm

Median of team B:

Arranging heights in ascending order, we get

174, 175, 177, 178, 179, 185, 190

Total number of observations = n = 7 (odd)

$\therefore$ Median

$= \left (\frac{n + 1}{2} \right )^{th}$ observation

$= \left ( \frac{7 + 1}{2} \right )^{th}$ observation =

$4^{th}$ observation = 178 cm

The production of electric bulbs in different factories is shown in the following table. Find the median of the productions.
No. of bulbs produced
(Thousands) 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100
No. of factories 12 35 20 15 8 7 8

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1988125_1852565_ans_5dd377bcb0e04009b21f2c3ff1bdf91e.jpg

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
No. of Cars 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80
Frequency 7 14 13 12 20 11 15 8

Class interval which has mode is 40 - 50
Maximum frequency is 20.
Here,

$l = 40, f_{1} = 20, f_{0} = 12, f_{2} = 11, h = 10$

$\therefore \textbf{Mode} = l + \left [ \dfrac{f_{1} - f_{0}}{2 \times f_{1} - f_{0} - f_{2}} \right ] \times h$

$= 40 + \left \{ \dfrac{20 + 12}{40 - 12 - 11} \right \} \times 10$

$= 4000 + \dfrac{80}{17}$

$= 4000 + \dfrac{45}{8}$

$= 40 + 4.7$

$= 44.7$

$\therefore \textbf{Mode} = 44.7$ Cars.

The following table gives the distribution of lifetime of 400 neon lamps:
Find the median lifetime of the lamp.

$\begin{aligned}&n=400 \therefore \dfrac{n}{2}=\dfrac{400}{2}=200\\&\text { Class interval having median is }=(3000-3500)\\&\begin{array}{l}\begin{aligned}l=3000, n=400, f &=86, c f=130, h=500 \\\therefore \text { Median }=& l+\left[\dfrac{\dfrac{n}{2}-c f}{f}\right] \times h \\&=3000+\left(\dfrac{200-130}{86}\right) \times 500 \\&=3000+\dfrac{35000}{86}\end{aligned} \\=3000+406.98 \\\therefore \text { Median }=3406.98 \text { hours }\end{array}\end{aligned}$
1841198_1869663_ans_e7e7a311e6eb415ca17c6103de2e463d.PNG

1841198_1869663_ans_e7e7a311e6eb415ca17c6103de2e463d.PNG

Write the formula to find median for grouped data.

Median =

$l + \left [ \dfrac{\dfrac{N}{2} - cf}{f} \right ] \times h$

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
Runs scored Number of batsmen
3000 - 4000 4
4000 - 5000 18
5000 - 6000 9
6000 - 7000 7
7000 - 8000 6
8000 - 9000 3
9000 - 10000 1
10000 - 11,000 1
Find the mode of the data.

Class interval which has mode is =4000 - 5000
Maximum frequency is
Here,

$l = 4000, f_{1} = 18, f_{0} = 4, f_{2} = 9, h = 1000$

$\therefore \textbf{Mode} = l + \left [ \dfrac{f_{1} - f_{0}}{2 \times f_{1} - f_{0} - f_{2}} \right ] \times h$

$= 4000 + \left \{ \dfrac{18 - 4}{36 - 4 - 9} \right \} \times 1000$

$= 4000 + \dfrac{14}{23} \times 1000$

$= 4000 + \dfrac{14000}{23}$

$= 4000 + 608.7$

$= 4608.7$

$\therefore$ Mode = 4608.7 runs.

Find the median of the data 15, 22 , 9 , 20 , 6, 18 , 11 ,25 ,14.

$6 , 9,11 , 14,15 , 18 , 20 , 22 , 25$ (ascending order )

$N = 9(odd)$

$= \dfrac{N +1}{2} + \dfrac{9 + 1}{2} = \dfrac{10}{2} = 5^{th}$
Median

$=15$

A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards bu less than 60 years
Age (in years) No. of policy holders
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89
Below 50 92
Below 55 98
Below 60 100
.

$n = 100 \therefore \dfrac{n}{2} = 50$
Median in which C.I. is (35 - 40)

$l = 35, n =100, f = 33, cf = 45, h = 5$

$\therefore \mathbf{Median} = l + \left [ \dfrac{\dfrac{n}{2} - cf}{f} \right ] \times h$

$= 35 + \left ( \dfrac{50 - 45}{33} \right ) \times 5$

$= 35 + \dfrac{25}{33}$

$= 35 + 0.76$

$\therefore \mathbf{Median} = 35.76$ years

1841132_1869478_ans_ea270d8df7de43e884ac907ea3dab040.PNG

Find the median for the following frequency distribution table.
Class interval 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30
Frequency 5 3 9 10 8 5

Class interval	Frequency (t)	Cumulative frequency (fc)
0 - 5	5	5
5 - 10	3	8
10 - 15	9	17
15 - 20	10	27
20 - 25	8	35
25 - 30	5	40
	N = 40

N = 40

$\dfrac{N}{2} = \dfrac{40}{2} = 20$
Median class is 15 - 20
LRL = 15
Fc = 17
Fm = 9
i = 5
Median = LRL
+

$\dfrac{\dfrac{N}{2} - f_{c}}{f_{m}} \times i$

$= 15 + \dfrac{( 20 - 17)}{9} \times 5$

$15 + \dfrac{3}{9} \times 5$

$= 15 + \dfrac{5}{3} = 15 + 1 .66$
Median = 16 .66

A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but lets than 60 years.
Age (in years) No. of policyholders (cf)
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89
Below 50 92
Below 55 98
Below 60 100

Given the cumulative frequency distribution table convert into a frequency distribution table

Cumulative greater than

$\dfrac{n}{2} = \dfrac{100}{2} = 50$ is

$78$ .

$\therefore$ Median class is 35 - 40

$l = 35, h = 5, f = 33 \;and \;cf = 45$

$\mathbf{Median} = l + \left [ \dfrac{\dfrac{N}{2} - cf}{f} \right ] \times h$

$= 35 + \left [ \dfrac{\dfrac{100}{2} - 45}{33} \right ] \times 5$

$= 35 + \dfrac{5}{33} \times 5$

$= 35 + 0.76$

$= 35.76$

$\therefore$ Median age is 35.76 years $$

Find the median of the data 22 , 28 ,34 , 49 , 44 , 57 , 18 , 10 , 33 ,41 , 66 , 59

$10 , 18 , 22 , 28 ,33 , 34 , 41 , 44 , 49 , 57 , 59 , 66$ ( Ascending order )

$N = 12(even)$
Median =

$\dfrac{6th+7th}{2}= \dfrac{34 + 41}{2} = \dfrac{75}{2} = 37 . 5$

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to $117.5-126.5,126.5-135.5$ $171.5-180.5)$

$\begin{aligned}&n=40 \therefore \dfrac{n}{2}=20\\&\text { C.l. which has median is }(144.5-153.5)\end{aligned}$

$\text{l=144.5, n=40, f=12, cf=17, h=9}$

$\begin{aligned} \therefore \quad \text { Median } &=l+\left[\dfrac{\dfrac{\mathrm{n}}{2}-\mathrm{c} f}{\mathrm{f}}\right] \times \mathrm{h} \\ &=144.5+\left(\dfrac{20-17}{12}\right) \times 9 \\ &=144.5+\left(\dfrac{20-17}{12}\right) \times 9 \\ &=144.5+\dfrac{3}{12} \times 9 \\ &=144.5+\dfrac{9}{4} \\ &=14.5+2.25 \\ \therefore \text { Median }=& 146.75 \mathrm{mm} \end{aligned}$
1841197_1869659_ans_a67b1df2250e4552871801a4b3ac1578.PNG

1841197_1869659_ans_a67b1df2250e4552871801a4b3ac1578.PNG

Find the coefficient of range of the following data -
Size 10 - 15 15 - 20 20 - 25 25 - 30
Frequency 2 4 6 8

Lower limit (S) = 10
Upper limit (L) = 30
Range =

$\dfrac{L - S}{L + S}$

$= \dfrac{30 - 10}{30 + 10}$

$=\dfrac{20}{40}$

$=\dfrac{1}{2} = 0.5$

Find the mean deviation from mean for the data
$4, 7, 8, 9, 10, 12, 13, 17$

Here ,

$N = 8$
Mean

$(\bar{x}) = \dfrac{\sum \ x_{1} - \bar{x}}{N}$

$= \dfrac{24}{8} = 3$
1954298_1874320_ans_a6540edd1dec4018862374e9e68936a3.png

Find the mean deviation from mode for the data
$2, 2, 2.5, 2.1, 2.9, 2.8, 2.5, 2.3$

In given data 2-5 occurs maximum times (3 times)
Thus Mode (z) = 2.5

$x_{i}$	2.2	2.5	2.1	2.5	2.9	2.8	2.5	2.3	N = 8
$\| x_{i} - z \|$	0.3	0	0.4	0	0.4	0.3	0	0.2	$\sum \| x_{i} - z \| = 1- 6$

Thus mean deviation

$(\delta z. )$ about mode

$\delta_{z} = \frac{1}{N} = \sum | x_{i} - z | = \dfrac{1-6}{8} = 0.2$

Find the mean deviation from mode for the data
$2, 4, 6, 4, 8, 6, 4, 10, 4, 8$

Maximum time ( 4 times ) in given

$x_{i}$	$\| x_{i} - z\|$
2	2
4	0
6	2
4	0
8	4
6	2
4	0
10	6
4	0
8	4
N = 10	$\sum \| x_{i} - z \| = 20$

Thus , Mean deviation from mode ,

$\delta _{2} = \sum | x\sum | x_{i} - z |$

$=\dfrac{20}{10} = 2$

Find the mean deviation of circles in given table .
Diameter (mm) 43-46 47-50 51-54 55-58 59-62
Number of circles 15 17 21 22 25

Here groups not continuous . To make it continuous we will subtract 0-5 from lower limit of group and add 0-5 in upper limit.
Calculate table for mean deviation and mean diameter
Mean deviation

$(\bar{x}) = A + \dfrac{\sum f_{i} y _{i}}{N} h = 52.5 + \dfrac{25}{100} \times 4 = 52.5 + 1 = 53.5 mm$
1955738_1874748_ans_a0dd8417cbf24a30a19f96095c61f51e.png

Find mean and variance for the data
$x_{i}$ 6 10 14 18 24 28 30
$f_{i}$ 2 4 7 12 8 4 3

$x_{i}$	$f_{i}$	$f_{i}x_{i}$	$(x_{i} - \bar {x})$	$(x_{i} - \bar {x})^{2}$	$f_{i} (x_{i} - \bar{x})^{2}$
6	2	12	-13	169	338
10	4	40	-9	81	324
14	7	98	-5	25	175
18	12	216	-1	1	12
24	8	192	5	25	200
28	4	112	9	81	324
30	3	90	11	121	363
	N = 40	$\sum f _{i} x _{i} = 760$		$\sum (x _{i} - \bar{x})^{2} = 503$	$\sum f _{i} (x _{i} - \bar{x}) ^{2} = 1736$

Mean

$(\bar{x})$ =

$\dfrac{\sum f_{i}x_{i}}{N} = \dfrac{760}{40} = 19$
variance

$(o^{2}) = \dfrac{\sum f_{i}(x_{i} - \bar{x})^{2}}{N}$

$=\dfrac{1736}{40} = 43.4$

Find the mode for the following data .
22 , 36 , 18 , 22 , 20 , 34 , 22 , 42, 46 , 42

1971829_1871087_ans_6fe4548244884c459e1bde0436cc955e.png

Find the mode for the following data .
4 , 3 , 1 , 5 , 3 , 7 , 9 , 6

1971827_1871084_ans_e2bd8f5c534e4f1cbf781c5311b1b2a7.png

Find the mean deviation from mean for the data
$28, 60, 38, 30, 32, 45, 53, 36, 44, 34$

Here ,

$N = 10$
Mean

$(\bar{x}) = \dfrac{\sum x_{1}}{N} = \dfrac{400}{10} = 40$
Mean deviation

$= \dfrac{\sum |x_{1} - \bar{x}}{N} = \dfrac{84}{10} = 8.4$
1954432_1874328_ans_a1c77dd36ba643bfae391bf036eb6148.png

Find the mean deviation imaginary mean 35 for following distribution.
$35, 25, 33, 50, 37, 33, 37, 30$

Term $(x_{i})$	Frequency $(f_{i})$	Deviation from Assumed A = 3 $(x_{i} - A)$	$(x_{i} - A )^{2}$	$f_{i} (x_{i} - A )^{2}$
35	2	0	0	0
25	1	-10	100	100
33	2	-2	4	8
50	1	15	225	225
37	2	2	4	8
30	1	-5	25	25
	N = 9			$\sum f_{i} (x_{i} - A)^{2} = 366$

Standard deviation

$(\sigma ) =\sqrt{\dfrac{\sum f_{i}(x_{i}-A)^{2}}{N}} = \sqrt{\dfrac{366}{9}}$

$=\sqrt{40.66} = 6.376$

If value of all value terms of a series are same, then find value of dispersion

Value of dispersion is zero because dispersion (deviation ) cannot be find for same values

Find median of the following variables values
25,34,33,13,20,26,36, 28,19,34

Arranging given values in ascending order
13, 19, 20, 25, 26, 28, 33, 34, 34, 36
In given values. No. of terms n = 10 which is even no. Thus,
Median

$= \frac{\frac{n^{th}}{2} term + (\frac{n}{2} + 1)^{th} term}{2}$

$= \frac{\frac{10^{th}}{2} term + (\frac{10}{2} + 1)^{th} term}{2}$

$= \frac{5^{th} term + 6^{th} term}{2}$

$= \frac{26 + 28}{2} = 27$
Thus, median = 27

Find median from the following frequency distribution

Here

$\frac{N}{2} = \frac{100}{2} = 50$
Cumulative frequency just above 50 is 65 whose corresponding class-interval is 40 - 50.
Thus, median class = 40 - 50

$\because$ I = 40, f = 30, C = 35, h = 10
Median

$= l + \frac{(\frac{N}{2} - C)}{f} \times h$

$= 40 + (\frac{50 - 35}{30}) \times 10 = 40 + \frac{150}{30}$

$= 40 + 5 = 45$
Thus, median class = 45
1955852_1877761_ans_9fa060f67aac4ddbb04b91efbe9d5487.png

Marks obtained by student of a class are given in the frequency distribution table. Find their median.

Preparing cumulative frequency table:

Here

$\frac{N}{2} = \frac{100}{2} = 50$
Cumulative frequency just above 50 is 74 whose corresponding class-interval is 20 - 30.
Thus, median class = 20 - 30

$\because$ I = 20, f = 42, C = 32, h = 10
Median

$= l + \frac{(\frac{N}{2} - C)}{f} \times h$

$= 20 + (\frac{50 - 32}{42}) \times 10 = 20 + \frac{180}{42}$

$= 20 + 4.29 = 24.29$
Thus, median class = 24.29
1955855_1877753_ans_a13acd76d8cf43f085330e972d75ac06.png

Find the mean and standard deviation of the following frequency distribution by step deviation method.
Class 1-5 6-10 11-15 16-20 21-25 26-30 31-35
Frequency 5 7 18 25 20 4 1

Mean

$(\bar{x}) = \dfrac{1}{N} \sum f_{i} x_{i} = \dfrac{1}{80} \times 1360 = 17$
1955779_1874972_ans_ff62c165c16845369ced83f1b36519aa.png

Find median from the frequency distribution table

Here

$\frac{N}{2} = \frac{157}{2} = 78.5$
Cumulative frequency just above 78.5 is 122 whose corresponding class-interval is 16 - 24.
Thus, median class = 16 - 24

$\because$ I = 16, f = 50, C = 72, h = 8
Median

$= l + \frac{(\frac{N}{2} - C)}{f} \times h$

$= 16 + (\frac{78.5 - 72}{50}) \times 8 = 16 + \frac{6.5 \times 8}{50}$

$= 16 + \frac{52}{50} = 16 + 1.04 = 17.04$
Thus, median class = 17.04
1955849_1877771_ans_403bf988344743b38ad381c4e70a9469.png

Find the mean deviation about mode and its coefficient from the following data-
Central size 6 7 8 9 10 11 12
Frequency 3 6 9 13 8 5 4

Here Maximum frequency is 13 and its corresponding value is 9

$\therefore$ mode (z) = 9
1955782_1874979_ans_0c3cab3d0fdc470b9540463e6063a90f.png

Marks obtained by 100 students are given in the table. Find their median.

Preparing cumulative frequency table:

Here

$\frac{N}{2} = 50$
Cumulative frequency just above 50 is 70 whose corresponding class-interval is 40 - 50.
Thus, median class = 40 - 50

$\because$ I = 40, f = 44, C = 26, h = 10
Median

$= l + \frac{(\frac{N}{2} - C)}{f} \times h$

$= 40 + (\frac{50 - 26}{44}) \times 10 = 40 + \frac{240}{44}$

$= 40 + 5.45 = 45.45$
1955857_1877751_ans_6dd7e0e8bc5d45e49519f4cfd7bc1016.png

Find the formula to find standard deviation in individual series

$\sigma = \sqrt{\dfrac{\sum x^{2} }{n}- \left ( \dfrac{\sum x}{n} \right )^{2}}$

Find the median of the data
19, 25, 59, 48, 35, 31, 30, 32, 51.
If 25 is replaces by 52 then find new median.

Arranging given data in ascending order
19, 25, 30, 31, 32, 35, 48, 51, 59
Number of terms in given data, n = 9 which is odd, thus
Median

$= (\frac{n + 1}{2})^{th}$ term

$=(\frac{9 + 1}{2})^{th}$ term

$= 5^{th}$ term = 32
In second case, when 25 is replaced by 52 then arranging data in ascending order
19, 30, 31, 32, 35, 48, 51, 52, 59
Median

$= (\frac{n + 1}{2})^{th}$ term

$=(\frac{9 + 1}{2})^{th}$ term

$= 5^{th}$ term = 35

Find the dispersion of the following data centred size -
Central size 32--38 38-44 44-50 50-56 56-62 62-68
No. of student 3 6 9 13 8 5

$= 36 [2- 0.07838] = 36 \times 1.9162 = 69.17832$
1955785_1874987_ans_599af79c203245f5b6d8501a9f109667.png

Number of children in 100 families are shown in the table.
Find their median.

Here

$\frac{N}{2} = \frac{100}{2} = 50$
C.F. just above 50 is 72 whose corresponding variable value is 2.

$\therefore$ Required median = 2
1955842_1877810_ans_495602fc35634099b0f2a4fb0cc1b13d.png

Find mode of the following distribution:
Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
Frequency 4 7 13 9 3

Here from table it is clear that maximum frequency is 13 and its corresponding class is 20 - 40
So, modal class = 20 - 30
Thus, I = 20,

$f_1 = 13, f_0 = 7, f_2 = 9$ and

$h = 10$

$\because$ Mode

$= l +\left ( \frac{f_1 - f_2}{2f_1 - f_0 - f_2} \right ) \times h$

$= 20 + \left ( \frac{13 - 7}{2 \times 14 - 7 - 9} \right ) \times 10$

$= 20 + \left ( \frac{6 \times 10}{26 - 16} \right )$

$= 20 + \frac{60}{10} = 20 + 6 = 26$
Thus, required mode = 26

Write importance of median.

Importance of median:
(i) It is only average which is used while dealing with qualitative data.
(ii) It is used for typical value in problems concerning wages, distribution of wealth etc.

Here,

$\frac{N}{2} = \frac{369}{2} = 184.5$
Cumulative frequency just above 184.5 is 212 whose corresponding is variable value is 8.0
Thus, median = 8.0
1865217_1878119_ans_6bf20131bc284037bb90d9e1c3386d49.PNG

1865217_1878119_ans_6bf20131bc284037bb90d9e1c3386d49.PNG

Find mode of the following distribution:
Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100
Frequency 3 15 24 8 5

From table it is clear that maximum frequency is 24 and its corresponding class is 40 - 60.
Thus, I = 40,

$f_1 = 24, f_0 = 15, f_2 = 8$ and

$h = 20$

$\because$ Mode

$= l + \left ( \frac{f_1 - f_2}{2f_1 - f_0 - f_2} \right ) \times h$

$= 40 \left ( \frac{24 - 15}{2 \times 24 - 15 - 18} \right ) \times 20$
Thus, required mode = 26

$= 40 + \frac{180}{25}$

$= 40 + 7.2 = 47.2$
Thus, required mode = 47.2

x 0.1 0.2 0.3 0.4 0.5 0.6 0.7
f 30 60 20 40 10 50 35

Here,

$\frac{N}{2} = \frac{245}{2} = 122.5$
Cumulative frequency just above 122.5 is 150 whose corresponding is variable value is 0.4
Thus, median = 0.4
1865213_1878115_ans_3f462d53c8ec44599b1e5312fdb77828.PNG

1865213_1878115_ans_3f462d53c8ec44599b1e5312fdb77828.PNG

Find the median of the frequency distribution:

Required median = 35
1955840_1877813_ans_1c8bf2fd125f4a8582e0fd41671ce82d.png

Write the formula to find median by grouped frequency distribution.

Median(M)

$= l + \frac{(\frac{N}{2} - C)}{f} \times h$

Where I = Lower limit of median
N = Total frequency
C = Cumulative frequency of class preceding the median class
h = Class size
f = frequency of median class

Marks obtained by students in a class are given in the table. Find median

Writing the distribution in tabulation form

$\frac{N}{2} = \frac{99}{2} = 49.5$
C.F. just above 49.5 is 52 whose corresponding variable value is 30.

$\therefore$ Required median = 30
1955847_1877793_ans_516acb65e5d4412495520b7c1c45178a.png

Find mode of the following frequency distribution
Class Frequency
10 - 15 3
15 - 20 7
20 - 25 16
25 - 30 12
30 - 35 9
35 - 40 5
40 - 45 3

Maximum frequency is 16 and its corresponding class is 20 - 25
So modal class = 20 - 25
Now, I = 20,

$f_1$ = 16,

$f_0$ = 7,

$f_2$ = 12 and h = 5

$\because$ Mode

$= l + \left ( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right ) \times h$

$= 20 + \left ( \frac{16 - 7}{(2 \times 16) - 7 - 12} \right ) \times 5$

$= 20 + \left ( \frac{9}{32 - 19} \right ) \times 5$

$= 20 + \frac{45}{13} = 20 + 3.46 = 23.46$
Thus, mode = 23.46 (approx)

Find mode of the following data :
Class-interval 0-20 20-40 40-60 60-80 80-100 100-120 120-140
Frequency 6 8 10 12 6 5 3

Here maximum frequency is 12 whose class is 60 - 80. So modal class 60 - 80

$\therefore l = 60, h = 20, f_{1} = 12, f_{0} = 10, f_{2} = 6$

$\because Mode = l + \left ( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right ) \times h$

$= 60 + \left ( \frac{12 - 10}{2 \times 12 - 10 - 6} \right ) \times 20$

$= 60 + \left ( \frac{2 \times 20}{24 - 16} \right ) = 60 + 20 \times \frac{2}{8}$

$= 60 + 5 = 65$
Thus, required mode = 65

Following data represents life period observations of electrical instruments.
Life period (in hrs) 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 100 35 52 61 38 29
Find mode of life period of instruments.

Here maximum frequency is 61 and its corresponding class-interval is 60 - 80. So modal class = 60 - 80.

$l = 60, f_{1} = 61, f_{0} = 52, f_{2} = 38,$ and

$h = 20$

$\displaystyle Mode = l + \left ( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right ) \times h = 60 + \left ( \frac{61 - 52}{2 \times 61 - 52 - 38} \right ) \times 20$

$= 60 + \frac{9}{122 - 52 - 38} \times 20 = 60 + \frac{9}{32} \times20$

$60 + \frac{180}{32} = 60 + 5.625 = 65.625$
Thus, mode of life period is 65.625 hrs.

Following distribution represents weight of 30 students in a class. Find median weight of students.
Weight (in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75
No. of students 2 3 8 6 6 3 2

Here, N = 30 and

$\frac{N}{2} = {30}{2} = 15 C.f.$ just above 15 is 9 and corresponding class-interval is 55 - 60.

$\therefore$ median class = 55 - 60

$\therefore l = 55, \frac{N}{2} = 15, C = 13, f = 6 \;and \;h = 5$

$\displaystyle Median = l + \left( \frac{\frac{N}{2}-C}{f} \right ) \times h = 55 + \left ( \frac{15 - 13}{6} \right ) \times 5$

$55 + \frac{2}{6} \times 5 = 55 + \frac{5}{3} = 55 + 1.67 = 56.67 kg (approx)$
Thus, median weight of students = 56.67 kg (approx)
1865314_1878281_ans_48c163186afa423d852cf6e4210c79ac.PNG

1865314_1878281_ans_48c163186afa423d852cf6e4210c79ac.PNG

If median of 17, 26, 60, 45, 33 is 33 and if 17 is replaced by 27 then what will be new median?

If 17 replaced by 27 then numbers will be 27, 26, 60, 45, 33 writing in ascending order 26, 27, 33, 45, 60.
Here n = 5 (odd number)
Median

$\displaystyle = \left ( \frac{n + 1}{2} \right )^{th}$

$\displaystyle = \left ( \frac{5 + 1}{2} \right )^{th} = \left ( \frac{6th}{2} term \right )$

$\displaystyle = 3rd \;term = 33$

$\therefore Median = 33$

Find mode of the following frequency distribution
Height (in cm) 52 - 55 55 - 58 58 - 61 61 - 64
No. of students 10 20 25 10

Here, Maximum frequency is 25 and its corresponding class is 58 - 61
Thus, modal class = 58 - 61
Now, I = 58,

$f_1$ = 25,

$f_0$ = 20,

$f_2$ = 10 and h = 3

$\because$ Mode

$= l + \left ( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right ) \times h$

$= 58 + \left ( \frac{25 - 20}{(2 \times 25) - 20 - 10} \right ) \times 3$

$= 58 + \left ( \frac{5 \times 3}{50 - 30} \right )$

$= 58 + \frac{15}{20} = 58 + 0.75 = 58.75$
Thus, required mode = 58.75marks (approx)

Find mode of the following frequency distribution
Marks 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70
No. of students 4 28 42 20 6

Here, Maximum frequency is 42 and its corresponding class is 40 - 50
Thus, modal class = 40 - 50
Now, I = 40,

$f_1$ = 42,

$f_0$ = 28,

$f_2$ = 20 and h = 10

$\because$ Mode

$= l + \left ( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right ) \times h$

$= 40 + \left ( \frac{42 - 28}{(2 \times 42) - 28 - 20} \right ) \times 10$

$= 40 + \frac{140}{36} = 40 + 3.88 = 43.88$
Thus, mode = 43.88 marks (approx)

Find mode of the following data :
Class-interval Frequency
80-85 33
85-90 27
90-95 85
95-100 155
100-105 110
105-110 45
110-115 15

Here maximum frequency

$= 155$
whose class is

$95 - 100$

$\therefore i = 95, f_{1} = 155, f_{0} = 85, f_{2} = 110, h = 5$

$\because Mode = l + \left ( \dfrac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right ) \times h$

$= 95 + \left ( \dfrac{155 - 85}{2 \times 155 - 85 - 110} \right ) \times 5$

$= 95 + \dfrac{70 \times 5}{310 - 195}$

$= 95 + \dfrac{70}{23}$

$= 95 + 3.04 = 98.04$
Thus, required mode

$= 98.04$

From the following data calculate standard deviation and its variability-
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
Students 19 3 2 49 24 5 2 1

2141549_2022180_ans_42de1cfb739d4924ab24146628d6c1fb.jpeg

Calculate standard deviation of the following:
X 10 20 30 40 50 60

Standard Deviation is given as:

2144414_2024580_ans_a38c9d88caf2411fa67e766a85824fc7.PNG

Find the standard deviation for the following distribution:

Class

Frequency (f)

0-10

5

10-20

8

20-30

15

30-40

16

40-50

6

Solution is given by:

2157604_2031932_ans_d593bae6dbd34b51b574d94759c8647d.PNG

In a class of $40$ students, marks obtained by the students in a class test (out of $10$ ) are given below.
Marks 1 2 3 4 5 6 7 8 9 10
Number of students 1 2 3 3 6 10 5 4 3 3
Calculate the following for the given distribution:
Median

Here,

$N=40\,and\, \dfrac{N}{2}=\dfrac{40}{2}=20$

Marks corresponding to cumulative frequency

$20$ is

$6$

Thus, the required median is

$6$

1889021_1909271_ans_31ac787044ed48de90c217fb4b1838c7.PNG

The Patel family flies from their home town, H, to Kiruna , K , in Lapland.
The Patel family has six suitcases.
The number of items in each suitcase is shown below.
$15$ $16$ $16$ $18$ $19$ $21$
Write down the mode.

Given number of items in suitcases is given as,

$15,\quad 16,\quad 16,\quad 18,\quad 19,\quad 21$

Mode of the data is the value that appears most frequently in the given data,

In given data,

$16$ has appeared 2 times in given data and remaining all values have appeared once.

Thus, mode of the data is

$16$

The Patel family flies from their home town, H, to Kiruna , K , in Lapland.
The Patel family has six suitcases.
The number of items in each suitcase is shown below.
$15$ $16$ $16$ $18$ $19$ $21$
Work out the median.

Arranging given data in ascending order, we get,

$15,\quad 16,\quad 16,\quad 18,\quad 19,\quad 21$

As number of data points is even, median is average of

${ \left( \frac { N }{ 2 } \right) }^{ th }$ and

${ \left( \frac { N+2 }{ 2 } \right) }^{ th }$ observation

$\frac { N }{ 2 } =\frac { 6 }{ 2 } =3$

$\frac { N+2 }{ 2 } =\frac { 6+2 }{ 2 } =\frac { 8 }{ 2 } =4$

Thus, median is an average of

${ 3 }^{ rd }$ and

${ 4 }^{ th }$ observation.

$\therefore Median=\frac { 16+18 }{ 2 } =\frac { 34 }{ 2 } =17$

The numbers $42, 43, 44, 44, (2x + 3), 45, 45, 46, 47$ have been put in the ascending order. If the median is $45$ , find $x$ . Hence, find the mode of the above data.

The given observations in ascending order are

$42, 43, 44, 44, (2x + 3), 45, 45, 46, 47$
Here, number of observations is

$9$ which is odd. So there will be one median class i.e.

$\left ( \frac{n + 1}{2} \right )$ th term
Median = Value of

$\left ( \frac{n + 1}{2} \right )$ th term

$\left ( \frac{9 + 1}{2} \right )$ th term = 5 th term

$\Rightarrow$ Median

$= 2x + 3$
But Median

$= 45$ (given)

$45 = 2x + 3$

$\Rightarrow 2x = 42$

$\Rightarrow x = 21$
The observations are

$42, 43, 44, 44, 45, 45, 46, 47$ .
The number

$4$ occurs most frequently i.e.

$3$ times.
Hence, Mode of the above data is

$45$ .
Hence,

$x = 21$ and Mode

$= 45$ .

In a class of $40$ students, marks obtained by the students in a class test (out of $10$ ) are given below.
Marks 1 2 3 4 5 6 7 8 9 10
Number of students 1 2 3 3 6 10 5 4 3 3
Calculate the following for the given distribution:
Mode

Here,

$N=40\,and\, \dfrac{N}{2}=\dfrac{40}{2}=20$

Clearly,

$6$ occurs

$10$ times which is maximum.

Hence, mode is

$6$ .

1889024_1909276_ans_5f3a1e5592a545e29a43e2b516220e46.PNG

Find SD and Coefficient of variation.
Roll No. 1 2 3 4 5 6 7 8 9 10
Marks 50 45 60 40 40 60 65 50 56 70

2162764_2034411_ans_f2de03a759fa4c66b673138cbf747bf1.png

Calculate the coefficient of variation:
Class
Frequency
0-10
5
10-20
8
20-30
10
30-40
15
40-50
12

Which measure of dispersion is the best and how?

Standard deviation is considered to be the best measure of dispersion and is therefore, the most widely used measure of dispersion.

It is based on all values and thus, provides information about the complete series. Because of this reason, a change in even one value affects the value of standard deviation.
It is independent of origin but not of scale.
It is useful in advance statistical calculations like comparison of variability in two data sets.
It can be used in testing of hypothesis.
It is capable of further algebraic treatment.

Find the coefficient of variation from the following data:

Class

Frequency

10-20

4

20-30

5

30-40

6

40-50

2

50-60

3

Calculate the standard deviation from the following data:
Class
Frequency (f)
10-20
4
20-30
5
30-40
6
40-50
2
50-60
3

Define standard deviation and and formula.

A standard deviation is a statistic that measures the dispersion of a dataset relative to its mean.

In a certain examination, there were 100 candidates of whom 21 failed, 6 secured distinction, 12 were placed in the first division, 43 were placed in the third division and 18 in the second division. It is known that at least 75% marks are required for distinction, 40% for passing, 50% for second division and 60% for first division calculate median.

2158196_2032310_ans_3145991947724c189abe6bbacf6580bb.PNG

Calculate the standard deviation:
Class
Frequency
0-10
5
10-20
8
20-30
10
30-40
15
40-50
12

Following are the marks obtained by 10 students of a class. Calculate standard deviation and coefficient of standard deviation.
Marks 12, 8, 17, 13, 15, 9, 18, 11, 6, 1.

2158242_2032326_ans_ba7205c2998f47ab9ae1d7aca32c94b5.PNG

Fill in the blank:
_________________ is the value of variance when the standard deviation of a given series is 2.5.

Variance = Square of standard deviation = Square of 2.5 = 6.25.

6.25 is the value of variance when the standard deviation of a given series is 2.5.

Calculate the coefficient of variation using assumed mean method from the following information:
Weeks
COVID Patients (‘000)
0-4
600
4-8
400
8-12
250
12-16
200
16-20
300
20-24
350
24-28
450
28-32
500
32-36
750
36-40
900

Calculate coefficient of variation from the following data:
Marks
No. Of students
0-20
8
20-40
12
40-60
30
60-80
20
80-100
10

The number of books bought at a book fair by $200$ students from a school are given in the following table:
No. of Books $0$ $1$ $2$ $3$ $4$
No. of Students $35$ $64$ $68$ $18$ $15$
Calculate the standard deviation.

No. of Books $x$	$f$	$d = x - 2$	$fd$	$d^{2}$
$0$	$35$	$-2$	$-70$	$140$
$1$	$64$	$-1$	$-64$	$64$
$2$	$68$	$0$	$0$	$0$
$3$	$18$	$1$	$18$	$18$
$4$	$15$	$2$	$30$	$60$
	$200$		$-86$	$282$

Standard Deviation

$= \sqrt {\dfrac {\sum fd^{2}}{\sum f} - \left (\dfrac {\sum fd}{\sum f}\right )^{2}}$

$= \sqrt {\dfrac {282}{200} - \left (\dfrac {-86}{200}\right )^{2}}$

$= \sqrt {1.41 - 0.1849}$

$= \sqrt {1.2251}$

$= 1.107$

Calculate the standard deviation of the following data :
$x$ 3 8 13 18 23
$f$ 7 10 15 10 8

Assume

$A = 13$

$x$	$f$	$d$	$fd$	$f{d}^{2}$
$3$	$7$	$-2$	$-14$	$28$
$8$	$10$	$-1$	$-10$	$10$
$13\left( A \right)$	$15$	$0$	$0$	$0$
$18$	$10$	$1$	$10$	$10$
$23$	$8$	$2$	$16$	$32$
			$2$	$80$

Standard deviation

$=\sqrt { \dfrac { \sum { f{ d }^{ 2 } } }{ \sum { f } } -{ \left( \dfrac { \sum { fd } }{ \sum { f } } \right) }^{ 2 } } \times C$

$=\sqrt { \dfrac { 80 }{ 50 } -{ \left( \dfrac { 2 }{ 50 } \right) }^{ 2 } } \times 5$

$=\sqrt { 1.6-{ \left( 0.04 \right) }^{ 2 } } \times 5$

$=\sqrt { 1.5984 } \times 5=6.32$

Calculate the standard deviation of the following data
$x$ $3$ $8$ $13$ $17$ $23$
$f$ $7$ $10$ $15$ $10$ $8$

From the given data:

$x_i$	$f_i$	$f_ix_i$	$x_i-\bar x$	$(x_i-\bar x)^2$	$f(x_i-\bar x)^2$
$3$	$7$	$21$	$-10$	$100$	$700$
$8$	$10$	$80$	$-5$	$25$	$250$
$13$	$15$	$195$	$0$	$0$	$0$
$17$	$10$	$170$	$4$	$16$	$160$
$23$	$8$	$184$	$10$	$100$	$800$
$\sum x_i = 65$	$\sum f_i = 50$	$\sum f_ix_i = 650$			$\sum f(x_i-\bar x)^2 = 1910$

Mean

$=\cfrac{\sum f_ix_i}{\sum f_i} = \cfrac{650}{50} = 13$

Standard Deviation

$= \sqrt{\dfrac{\sum f_i(x_i-\bar x)^2}{\sum f_i}} = \sqrt{\dfrac{1910}{50}} = \sqrt{\dfrac{191}{5}}$

Let $a, b, c, d$ and $e$ be the observations with mean ${ m }$ and standard deviation $S$ . The standard deviation of the observations a $+ { k } ,b+ { k } ,c+ { k } , d+ { k }$ and $e+ { k }$ is

Given that

a, b, c, d and e be the observations with mean m and standard Deviations

i.e. mean

$= \dfrac{Sum \ of \ observations}{No \ of \ observations}$

$m = \dfrac{a + b + c + d + e}{5}$ __(I)

and

$S = \sqrt{\dfrac{(a - m)^2 + (b - m)^2 + (c - m)^2 + (d - m)^2 + (e - m)^2}{5}}$

New observations are

$a + k , b + k , c + k, d + k$ and

$e + k$

Let mean

$= n$

$\Rightarrow n = \dfrac{ a+ k + b + k + c + k + d + k + e + k}{5}$

$= \dfrac{a + b + c + d + e}{5} + \dfrac{5 k}{5}$ from (I)

$n = m + k$ __(II)

from (II)

$n = m + k$

$a + k - n = a + k (m + k)$

$= a - m$

$\Rightarrow b + k - n = b - m$

$\Rightarrow c + k - n = c - m$

$\Rightarrow d + k - n = d - m$

$\Rightarrow e + k - n = e - m$

we know standard deviation

$(\sigma) = \sqrt{\dfrac{\sum_{i = 1}^{n} (x_i - \bar{x})^2}{n}}$

and reputed standard Deviation

$\delta = \sqrt{\dfrac{(a + k - n)^2 + (b + k - n)^2 + (c + k - n)^2 + (d + k - n)^2 + (e + k - n)^2}{5}}$

Substitute the values

$= \sqrt{\dfrac{(a - m)^2 + (b - m)^2 + (c - m)^2 + (d - m)^2 + (e - m)^2}{5}}$

Standard deviation of two distribution are same and their arithmetic means are 30 and 25 respectively. If their coefficient of variation is ${ v }_{ 1 }$ and ${ v }_{ 2 }$ , then $6{ v }_{ 1 }-{ 5v }_{ 2 }$ is equal to ___________________.

Given: Standard deviation of two distributions are same and their arithmetic mean are

$30$ and

$25$ and their coefficient variance are

$v_1$ and

$v_2$

$\because (SD)_1 = (SD)_2 = (SD)$ (given)

Coefficient of variation

$CV = \dfrac{SD}{mean}$

$\Rightarrow v_1 = \dfrac{SD}{30}$ and

$v_2 = \dfrac{SD}{25}$

$\therefore 6v_1 - 5v_2 = 6 \times \dfrac{SD}{30} - 5 \times \dfrac{SD}{25}$

$\Rightarrow 6v_1 - 5v_2 = \dfrac{SD}{5} - \dfrac{SD}{5}$

$\Rightarrow 6v_1 - 5v_1 = 0$

$\therefore 6v_1 - 5v_1 = 0$ is correct answer.

Find the standard deviation of 10 observations 111, 211, 311,...... 1011.

$\begin{array}{l} \sigma =\sqrt { \frac { { \sum { { d^{ 2 } } } } }{ N } } -{ \left( { \frac { { \sum d } }{ N } } \right) ^{ 2 } } \\ assume\, \, mean \\ A=511 \\ \sigma =\sqrt { \frac { { 850.000 } }{ { 10 } } -{ { \left( { \frac { { 500 } }{ { 10 } } } \right) }^{ 2 } } } \\ \sigma =287.22 \end{array}$
1215638_1321368_ans_0b48eafeb69a49a5a52645ebdbb3e06c.PNG

1215638_1321368_ans_0b48eafeb69a49a5a52645ebdbb3e06c.PNG

The following table gives the daily wages of workers in a factory. Compute the standard deviation and the coefficient of variation of the wages of the workers.
Wages (Rs) 125-175 175-225 225-275 275-325 325-375 375-425 425-475 475-525 525-575
Number of workers 2 22 19 14 3 4 6 1 1

Wages (Rs.)	Midpoint $(x_i)$	Frequency (f) (No. of workers)	$x_i^2$	$f_ix_i$	$f_ix_i^2$
125 - 175	150	2	22500	300	45,000
175 - 225	200	22	40000	4400	8,80,000
225 - 275	250	19	62500	4750	1187500
275 - 325	350	14	90000	4200	1260000
325 - 375	300	3	122500	1050	367500
375 - 425	350	4	160000	1600	640000
425 - 475	400	6	202500	2700	1215000
475 - 525	450	1	250000	500	250000
525 - 575	500	1	302500	550	302500
		72		20050	6147500

$\sum f = N = 72$

$\bar{x} = \dfrac{\sum f_ix_i}{N} = \dfrac{20050}{72} = 278.472222$

$\sum f_ix_i = 20050$

$\sum f_ix_i^2 = 6147500$

$\therefore$ Standard deviden

$\sigma = \sqrt{\dfrac{\sum f_ix_i^2}{N} - \left(\dfrac{\sum f_ix_i}{N}\right)^2}$

$=\sqrt{\dfrac{6147500}{72} - \left(\dfrac{20050}{72}\right)^2}$

$=\sqrt{7835.165895}$

$\sigma = 88.51647245$

Coefficient of veriation

$= \dfrac{\sigma}{\bar{x}}$

$= \dfrac{88.51647245}{278.4722222}$

$= 0.317864639$

$=0.32$

Find the variance and standard devaiation of the following frequency distribution.
${ x }_{ i }$ 6 10 14 18 24 28 30
${ f }_{ i }$ 2 4 7 12 8 4 3

Now we can prepare a table for calculating various & standard deviation

$x_i$	$f_i$	$x_i^2$	$f_ix_i$	$f_i x_i^2$
6 10 14 18 24 28 30	2 4 7 12 8 4 3	36 100 196 324 576 784 900	12 40 98 216 192 112 90	72 400 1372 3888 4608 3136 2700
Total	$40$		$760$	$16,176$

Here

$\sum f = N = 40$

We know mean

$\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{760}{40} = 19$

variance

$(\sigma^2) = \dfrac{1}{N} \sum f_i x_i^2 - (\bar{x})^2$

$= \dfrac{1}{40} (16176) - (19)^2$

$= 404.4 - 361$

$= 43.4$

Standard deviation

$(\sigma) = \sqrt{variance}$

$= \sqrt{43.4}$

$= 6.59$ (Rounded two decimals)

Find the standard deviation of the numbers $62,58,53,50,63,52,55$ .

First of we have to prepare table

$x_i$ mean $\bar{x}$	$x_i -\bar{x}$	$(x_i - \bar{x})^2$
$62$ $56.14$ $58$ $56.14$ $53$ $56.14$ $50$ $56.14$ $36$ $56.14$ $52$ $56.14$ $55$ $56.14$	$5.86$ $1.86$ $-3.14$ $-6.14$ $6.86$ $-4.14$ $-1.14$	$34.3396$ $3.4596$ $9.8596$ $37.6996$ $47.0596$ $17.1396$ $1.2996$
		$\sum (x_i - \bar{x})^2 = 150.8572$

We know

$\bar{x} = \dfrac{\sum x}{n}$

where n is no of observations

$n = 7$

mean

$(\bar{x}) = \dfrac{62 +58+53+50+63+52+55}{7}$

$= \dfrac{393}{7} = 56.1429$

$\cong 56.14$ rounded to 2 decimals

$\bar{x} = 56.14$

$\therefore$ we have standard deviation

$= \sqrt{\dfrac{\sum (x_i - x)^2}{n-1}}$

$= \sqrt{\dfrac{150.8572}{7-1}}$

$=\sqrt{\dfrac{150.8572}{6}} = \sqrt{25.1429}$

$= 5.01427$

$\cong 5.01$

Rounded 2 decimals

A random variable $x$ has the following probability distribution.
$x=x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(x)$ $0.1$ $k$ $0.2$ $2k$ $0.3$ $k$
Find the value of $k$ & calculate mean and variance of $x$ .

Given that a random variable x has the following probability distribution

$X = x$	-2	-1	0	1	2	3
$p(x)$	0.1	k	0.2	2k	0.3	k

We know that in a probability total probability is unity

i.e

$\displaystyle\sum p(x) = 1$

$\Rightarrow 0.1 + k + 0.2 + 2k + 0.3 + k = 1$

$\Rightarrow 0.6 + 4k = 1$

$\Rightarrow 4k = 1 - 0.6 = 0.4$

$4k = 0.4$

$k = \dfrac{0.4}{4} = 0.1$

then probability distribution is value of

$k = 0.1$

$X = x$	-2	-1	0	1	2	3
$p(x)$	0.1	0.1	0.2	0.2	0.3	0.1

we know

mean of(M)

$x = \displaystyle\sum_{i=1}^n x_i p (x_i)$

$= (-2) \times 0.1 + (-1) \times 0.1 + 0 \times 0.2 + 1 \times 0.2 + 2 \times 0.3 + 3 \times 0.1$

$= -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3 = 0.8$

mean of

$x = 0.8$

variance of

$x = \displaystyle\sum x^2 p(x) - \mu^2$

$= (-2)^2 \times 0.1 + (-1)^2 \times 0.1 + 0^2 \times 0.2 + 1^2 \times 0.2 + 2^2 \times 0.3 + 3^2 \times 0.1 - (0.8)^2$

$= 0.4 + 0.1 + 0 + 0.2 + 1.2 + 0.9 - 0.64$

variance of

$x = 2.16$

The mean of $9$ variates is $11$ . If eight of them are $7, 12, 9, 14, 21, 3, 8$ and $15$ , find the $9th$ variate.

1874842_1498187_ans_f05f7aa6e4fc4d07ae5235025cf6ecb5.jpg

Using short ad method.find the man, variance and standard deviation for the data:
Class 0-10 10-20 20-20 30-40 40-50
Frequency 5 8 15 16 6

1982556_1710897_ans_ce6fcc93f87a4bc2b47ea62fccce7a68.jpg

Calculate standard deviation from the following data:

Marks (Less than)

Number of students (f)

20

8

40

20

60

50

80

70

100

80

Calculate coefficient of variation using step deviation method from the following information:
Class Interval
Frequency
0-5
15
5-10
10
10-15
11
15-20
12
20-25
6
25-30
6
30-35
10
35-40
15
40-45
10
45-50
5

Using short ad method.find the mean, variance and standard deviation for the data:
Class 25-35 35-45 45-55 55-65 65-75
Frequency 64 132 153 140 51

2037927_1710899_ans_63650b46b06d4116be4f0d01de252181.jpg

Find the variance of the following data $6, 8, 10, 12, 14, 16 , 18, 20, 22, 24$ .

Given Data

$6,8,10,12, 14, 16,18,20,22,24$

$x_1$	$\bar{x}$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
$6$ $8$ $10$ $12$ $14$ $16$ $18$ $20$ $22$ $24$	$15$ $15$ $15$ $15$ $15$ $15$ $15$ $15$ $15$ $15$	$-9$ $-7$ $-5$ $-3$ $-1$ $1$ $3$ $5$ $7$ $9$	$81$ $49$ $25$ $9$ $1$ $1$ $9$ $25$ $49$ $81$

We know

Mean

$= \dfrac{\displaystyle \sum x}{n}$

where n is no" of observations

$\displaystyle \sum (x_i - \bar{x})^2$

$= 330$

$\therefore$ mean

$(\bar{x}) = \dfrac{\displaystyle\sum x}{n} = \dfrac{6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24}{10}$

$= \dfrac{150}{10}$

$\bar{x} = 15$

$\therefore$ variance

$= \dfrac{\displaystyle\sum (x_i - \bar{x})^2}{n - 1}$

$= \dfrac{330}{10-1}$

$= \dfrac{330}{9}$

variance

$= 36.67$

The frequency distribution table shows the number of mango trees in a groove and their yield of mangoes. Find the median of data.
No. of Mangoes 50 - 100 100 - 150 150 - 200 200 - 250 250 - 300
No. of trees 33 30 90 80 17

1960804_1852509_ans_83d8b3964693418599510316868de8d5.jpg

Find out standard deviation of the following data:
Size 1 2 3 4 5 6 7 8
Frequency 5 10 15 20 25 30 35 40

The solution is as follows:
2144744_2024928_ans_de330a4a02914ecba0836d7b80f8fa26.PNG

Find out standard deviation of the following data:
Size 4 6 8 10 12 14 16
Frequency 1 2 3 5 3 2 1

2158241_2032325_ans_f8550081f4784982a325195890877bfe.PNG

Heights in cm.	$120-124$	$125-129$	$130-134$	$135-139$	$140-144$
No. of students	$6$	$9$	$10$	$12$	$8$

No. of wards	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
No. of houses	$32$	$57$	$28$	$96$	$138$	$90$	$66$	$58$

Income (in rupees)	500-600	400-500	300-400	200-300	100-200
Number of persons	17	20	30	18	15

Age in (years)	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	8	14	25	28	25	10	5

Marks	No. of Students
0 - 10	5
10 - 20	12
20 - 30	14
30 - 40	10
40 - 50	8
50 - 60	6

Classes	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequencies	2	3	5	10	3	5	2

Height in cms	No. of children
70-75	3
75-80	4
80-85	7
85-90	7
90-95	15
95-100	9
100-105	6
105-110	6
110-115	3

Subject	Mathematics	Physics	Chemistry
Mean	42	32	40.9
Standard deviation	12	15	20

Diameters	No. of circles
$33-36$	$15$
$37-40$	$17$
$41-44$	$21$
$45-48$	$22$
$49-52$	$25$

Marks	Up to $5$	Up to $6$	Up to $7$ $	Up to $8$	Up to $9$	Up to $10$
No of students	$5$	$11$	$19$	$31$	$40$	$45$

Variate	15	18	20	22	25	27	30
Frequency	4	6	8	9	7	8	6

$98.1$	$98.6$	$99.2$	$90.3$	$86.5$	$95.3$	$92.9$	$96.3$	$94.2$	$95.1$
$89.2$	$92.3$	$97.1$	$93.5$	$92.7$	$95.1$	$97.2$	$93.3$	$95.2$	$97.3$
$96.0$	$92.1$	$84.9$	$90.0$	$95.7$	$98.3$	$97.3$	$96.1$	$92.1$	$89$

Marks $(x)$	$10$	$20$	$30$	$40$	$50$	$60$
No. of students $(f)$	$8$	$12$	$20$	$10$	$7$	$3$

No. of books	$0$	$1$	$2$	$3$	$4$
No. of students	$35$	$64$	$68$	$18$	$15$

Wages in Rs	$30-34$	$34-38$	$38-42$	$42-46$	$46-50$	$50-54$
No. of workers	$4$	$7$	$9$	$11$	$6$	$3$

Class-interval	Frequency
$1-5$	$1$
$6-10$	$2$
$11-15$	$3$
$16-20$	$4$

Wages (Rs)	$125-175$	$175-225$	$225-275$	$275-325$	$325-375$	$375-425$	$425-475$	$475-525$
Number of workers	$2$	$22$	$19$	$14$	$3$	$4$	$6$	$1$

$X=x$	$1$	$2$	$3$
$p(X=x)$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{2}{5}$

Number of people in each car	$1$	$2$	$3$	$4$	$5$
Frequency	$50$	$40$	$10$	$20$	$5$

Milk (litre)	1 - 3	3 - 5	5 - 7	7 - 9	9 - 11	11 - 13
No. of hotels	7	5	15	20	35	18

Average Speed of vehicles (km/hr)	60 - 64	64 - 69	70 - 74	75 - 79	79 - 84	84 - 89
No. of vehicles	10	34	55	85	10	6

Statistics - Class 11 Engineering Maths - Extra Questions

Find the median of the following data: 2,7,3,15,12,172, 7, 3, 15, 12, 17 and 55.

Find the mode of 14,25,14,28,18,17,18,14,23,22,14,1814, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.

Calculate the coefficient of range for the following data:Heights in cm.120−124120-124125−129125-129130−134130-134135−139135-139140−144140-144No. of students66991010121288

Calculate the coefficient of range for the following data.No. of wards1122334455667788No. of houses3232575728289696138138909066665858

Calculate the mean deviation from the mean for the scores given below:15,11,13,20,26,18,2115, 11, 13, 20, 26, 18, 21

The formula for coefficient of variation (C.V.) is given by _______________.

The weight(in kg) of 1313 students in a class are 42.5,47.5,48.6,50.5,49,46.2,49.8,45.8,43.2,48,44.7,46.9,42.442.5, 47.5, 48.6, 50.5, 49, 46.2, 49.8, 45.8, 43.2, 48, 44.7, 46.9, 42.4. Find the range and coefficient of range.

The smallest value of a collection of data is 1212 and the range is 5959. Find the largest value of the collection of data.

Write two merits of mode.

The largest value in a collection of data is 7.447.44. If the range is 2.262.26, then find the smallest value in the collection.

Find the mode of the following data. 12, 24, 36, 46, 25, 38, 72, 36, 25, 38, 12, 24, 46,25, 12, 24, 46, 25, 72, 12, 24, 36,25, 38 and 36.

Write the relation between standard deviation of a set of scores and its variance.

What is the range of the data : 48,65,27,23,44,41,25,70,49 48, 65, 27, 23, 44, 41, 25, 70, 49 ?

Find the mean S.DS.D of 1,2,3,4,5,61,2,3,4,5,6.

Find the sum of deviation of all observations of the data 5, 8, 10, 15, 22 from their mean.

Find the mode of this data2,5,4,3,5,2,5,2,4,52,5,4,3,5,2,5,2,4,5

The rainfall ( in mm) in a city on 77 days of a certain week was recorded as follows: Days Mon Tue Wed Thurs Fri Sat Sun Rainfall ( in mm) 0.0 12.2 2.1 0.0 20.5 5.5 1.0 i) Find the range of the rainfall in the above data.

The runs scored in a cricket match by 1111 players is as follows:6,15,120,50,100,80,10,15,8,10,156, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15Find the mean, mode and median of this data. Are the three same?

Calculate the variance and standard deviation of the following continuous frequency distributionClass interval 30−4030-4040−5040-5050−6050-6060−7060-7070−8070-8080−9080-9090−10090-100Frequency 337712121515883322

Find the standard deviation of 9,16,23,30,37,44,519,16,23,30,37,44,51.

Find the median of:26,33,41,18,30,22,36,4526, 33, 41, 18, 30, 22, 36, 45 and 2424

The S. D of 1515 items is 66 and if each item is decreased by 11 , then standard deviation will be

Find the median of:3.2,4.8,5.6,7.3,8.93.2, 4.8, 5.6, 7.3, 8.9 and 9.19.1

Let a,b,c,da, b, c, d and ee be the observations with mean m{ m } and standard deviation ss. The standard deviation of the observations aa +k,b+k,c+k,d+k+ { k } , b+k ,c+k , d+k and e+ke+k is

Find the median of:80, 48, 66, 61, 75, 52, 4580, 48, 66, 61, 75, 52, 45 and 7070

Find the coefficient of range for the following data.Size10-1510-1515-2015-2020-2520-2525-3025-30Frequency22446688

Following tare the lives in hours of 1515 pieces of the components of aircraft engine.. Find the median:: 715,724,725,710,729,745,694,699,696,712,734,728,716,705,719715,724,725,710,729,745,694,699,696,712,734,728,716,705,719

Find the median of 8, 15, 7, 19, 16, 128, 15, 7, 19, 16, 12 and 1010

What is mode ?

If the median of \cfrac{a}{3},\cfrac{a}{2},\cfrac{a}{4},\cfrac{{2a}}{5},\cfrac{a}{6}\cfrac{a}{3},\cfrac{a}{2},\cfrac{a}{4},\cfrac{{2a}}{5},\cfrac{a}{6} is 1212 then find the value of a\left( {a > 0} \right)a\left( {a > 0} \right).

The median of the given data 6,8,9,15,x,29,25,21,22 is 18, find value of x ?

Calculate variance and standard deviation of the following data :10, 12, 8, 14, 16

Find the median of the following :33,31,48,45,41,92,78,51,61

A die is tossed thrice. Getting an even number is considered as success. What is the variance of the binomial distribution ?

Find the mean deviation from the mean of the following data, using the step deviation method:MarksNo. of students0-10610-20520-30830-401540-50750-60660-703

From the following frequency distribution, find :(i) the median(ii) lower quartile(iii) upper quartile(iv) inter quartile rangeVariate15182022252730Frequency4689786

A data has 1313 observations arranged in descending order. Which observation represents the median of data?

The distribution below gives weights of 3030 students of a class. Find the median weight of the students.

The length of 4040 leaves of a plant are measured correctly of the nearest millimeter, and the data obtained is represented in the following table:Find the median length of the leaves.

The average weight of 4040 Students is 3232. If the Heaviest and Lightest are excluded the average weight reduces by 11. If only the Heaviest is excluded then the average is 3131. Then what is the weight of the Lightest?

The exam scores of all 500 students were recorded and it was determined that these scores were normally distributed. If Jane's score is 0.8 standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?

25, 16, 26, 16, 32, 31, 19, 28 and 35

Find the mode of the following distribution:Income (in rupees)500-600400-500300-400200-300100-200Number of persons1720301815

Find the mode of the following data:Marks below1020304050Number of students38172022

Find the mode of the following distribution:Marks more than01020304050Number of students50464020103

Find the mode of the following distribution:Marks45678910Frequency 381012161210

Find the mode of the following distribution:Age in (years)0-1010-2020-3030-4040-5050-6060-70Frequency 814252825105

Find the mode of the following data:Class interval1112131415161718Frequency29348783

Find the median of3.2,4.8,5.6,5.6,7.3,8.93.2,4.8,5.6,5.6,7.3,8.9 and 9191

The marks (out of 100100) obtained by 77 students in Mathematics examination are given below. Find the mode for these marks.99, 100, 95, 100, 100, 60, 9099, 100, 95, 100, 100, 60, 90

2.5 2.5 2.1 2.5 2.7 2.8 2.5

2 4 6 2 6 6 7 8

Find the mode of the following data.Marks353637383940No of Student 090709040402

Find mode of the following frequency distribution :x345678f246321

A data has 19 observations arranged in descending order which observation represents the median data?

Find mode of the following frequency distribution :x1.11.21.31.41.51.6f20508060158

2 5 7 5 3 1 5 8 7 5

The weights of 1010 students (in kg) are given below :40, 35, 42, 43, 37, 35, 37, 37, 42, 3740, 35, 42, 43, 37, 35, 37, 37, 42, 37Find the mode of the data.

The number of members in 30 families of a village are according to following table.Find their mode.No. of members2345678No. of families12461035

Marks obtained by students are given below, find the mode of marks.Marks1020304050607080No. of Students281626201674

Age in years of 20 students of a class are as follows:15 16 13 14 1413 15 14 13 1314 12 15 14 1613 14 14 13 15Find mode by expressing them in frequency distribution:

Find the median of :66, 98, 54, 92, 87, 63, 7266, 98, 54, 92, 87, 63, 72

Find mode of the following frequency distributionMarksNo. of Students0 - 10510 - 201220 - 301430 - 401040 - 50850 - 606

Find the mode of 0, 5, 2, 7, 2, 1, 1, 3, 2, 4, 5, 7, 5, 10, 5, 2, 7, 2, 1, 1, 3, 2, 4, 5, 7, 5, 1 and 22

Find the mean and variance for the data 6, 7, 10 , 12, 13, 4, 8, 126, 7, 10 , 12, 13, 4, 8, 12

Find the mean and variance for the data \displaystyle X_{i} \displaystyle X_{i} 6101418242830 \displaystyle f_{i} \displaystyle f_{i} 24712843

Find the mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 174, 7, 8, 9, 10, 12, 13, 17

Find the mean and standard deviation using short-cut method \displaystyle X_{i} \displaystyle X_{i} 606162636465666768 \displaystyle f_{i} \displaystyle f_{i} 21122925121045

Find the mean and variance for the following frequency distrubutionClasses0-1010-2020-3030-4040-50Frequencies5815166

Find the mean and variance for the following frequency distribution Classes0-3030-6060-9090-120120-150150-180180-210Frequencies23510352

Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 4438, 70, 48, 40, 42, 55, 63, 46, 54, 44

Find the mean and variance for the data \displaystyle X_{i} \displaystyle X_{i} 92939798102104109 \displaystyle f_{i} \displaystyle f_{i} 3232633

Find the mean variance and standard deviation using short-cut methodHeightin cmsNo. of children70-75375-80480-85785-90790-951595-1009100-1056105-1106110-1153

Find the mean deviation about the median for the data 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 1713, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.

The mean and variance of eight observation are 99 and 9.25,9.25, respectively. If six of the observation are 6, 7, 10, 12, 126, 7, 10, 12, 12 and 1313, find the remaining two observations.

The mean and standard deviation of a group of 100100 observations were found to be 2020 and 33 respectively. Later on it was found that three observations were incorrect which were recorded as 21, 2121, 21 and 1818. Find the mean and standard deviation if the incorrect observation are omitted

From the data given below state which group is more variable AA or BB Marks10-2020-3030-4040-5050-6060-7070-80Group A917323340109Group B1020302543157

The diameters of circles (in mmmm) drawn in a design are given below :Diameters No. of circles33-3633-36151537-4037-40171741-4441-44212145-4845-48222249-5249-522525

From the prices of shares XX and YY below find out which is more stable in value :X35545253565852505149Y108107105105106107104103104101

Find the median of the following data: $2, 7, 3, 15, 12, 17$ and $5$ .

Find the mode of $14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18$ .

Calculate the coefficient of range for the following data:
Heights in cm. $120-124$ $125-129$ $130-134$ $135-139$ $140-144$
No. of students $6$ $9$ $10$ $12$ $8$

Calculate the coefficient of range for the following data.
No. of wards $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
No. of houses $32$ $57$ $28$ $96$ $138$ $90$ $66$ $58$

Calculate the mean deviation from the mean for the scores given below:
$15, 11, 13, 20, 26, 18, 21$

The weight(in kg) of $13$ students in a class are $42.5, 47.5, 48.6, 50.5, 49, 46.2, 49.8, 45.8, 43.2, 48, 44.7, 46.9, 42.4$ . Find the range and coefficient of range.

The smallest value of a collection of data is $12$ and the range is $59$ . Find the largest value of the collection of data.

The largest value in a collection of data is $7.44$ . If the range is $2.26$ , then find the smallest value in the collection.

What is the range of the data : $48, 65, 27, 23, 44, 41, 25, 70, 49$ ?

Find the mean $S.D$ of $1,2,3,4,5,6$ .

Find the mode of this data
$2,5,4,3,5,2,5,2,4,5$

The rainfall ( in mm) in a city on $7$ days of a certain week was recorded as follows:

Days

Mon

Tue

Wed

Thurs

Fri

Sat

Sun

Rainfall

( in mm)

0.0

12.2

2.1

0.0

20.5

5.5

1.0

i) Find the range of the rainfall in the above data.

The runs scored in a cricket match by $11$ players is as follows:
$6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15$
Find the mean, mode and median of this data. Are the three same?

Calculate the variance and standard deviation of the following continuous frequency distribution
Class interval $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$
Frequency $3$ $7$ $12$ $15$ $8$ $3$ $2$

Find the standard deviation of $9,16,23,30,37,44,51$ .

Find the median of:
$26, 33, 41, 18, 30, 22, 36, 45$ and $24$

The S. D of $15$ items is $6$ and if each item is decreased by $1$ , then standard deviation will be

Find the median of:
$3.2, 4.8, 5.6, 7.3, 8.9$ and $9.1$

Let $a, b, c, d$ and $e$ be the observations with mean ${ m }$ and standard deviation $s$ . The standard deviation of the observations $a$ $+ { k } , b+k ,c+k , d+k$ and $e+k$ is

Find the median of:
$80, 48, 66, 61, 75, 52, 45$ and $70$

Find the coefficient of range for the following data.
Size $10-15$ $15-20$ $20-25$ $25-30$
Frequency $2$ $4$ $6$ $8$

Following tare the lives in hours of $15$ pieces of the components of aircraft engine $.$ Find the median $:$ $715,724,725,710,729,745,694,699,696,712,734,728,716,705,719$

Find the median of $8, 15, 7, 19, 16, 12$ and $10$

If the median of $\cfrac{a}{3},\cfrac{a}{2},\cfrac{a}{4},\cfrac{{2a}}{5},\cfrac{a}{6}$ is $12$ then find the value of $a\left( {a > 0} \right)$ .

Calculate variance and standard deviation of the following data :
10, 12, 8, 14, 16

Find the median of the following :
33,31,48,45,41,92,78,51,61

Find the mean deviation from the mean of the following data, using the step deviation method:
Marks No. of students
0-10 6
10-20 5
20-30 8
30-40 15
40-50 7
50-60 6
60-70 3

From the following frequency distribution, find :
(i) the median
(ii) lower quartile
(iii) upper quartile
(iv) inter quartile range
Variate
15
18
20
22
25
27
30
Frequency
4
6
8
9
7
8
6

A data has $13$ observations arranged in descending order. Which observation represents the median of data?

The distribution below gives weights of $30$ students of a class. Find the median weight of the students.

The length of $40$ leaves of a plant are measured correctly of the nearest millimeter, and the data obtained is represented in the following table:
Find the median length of the leaves.

The average weight of $40$ Students is $32$ . If the Heaviest and Lightest are excluded the average weight reduces by $1$ . If only the Heaviest is excluded then the average is $31$ . Then what is the weight of the Lightest?

Find the mode of the following distribution:
Income (in rupees) 500-600 400-500 300-400 200-300 100-200
Number of persons 17 20 30 18 15

Find the mode of the following data:
Marks below 10 20 30 40 50
Number of students 3 8 17 20 22

Find the mode of the following distribution:
Marks more than 0 10 20 30 40 50
Number of students 50 46 40 20 10 3

Find the mode of the following distribution:
Marks 4 5 6 7 8 9 10
Frequency 3 8 10 12 16 12 10

Find the mode of the following distribution:
Age in (years) 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 8 14 25 28 25 10 5

Find the mode of the following data:
Class interval 11 12 13 14 15 16 17 18
Frequency 2 9 3 4 8 7 8 3

Find the median of
$3.2,4.8,5.6,5.6,7.3,8.9$ and $91$

The marks (out of $100$ ) obtained by $7$ students in Mathematics examination are given below. Find the mode for these marks.
$99, 100, 95, 100, 100, 60, 90$

Find the mode of the following data.
Marks 35 36 37 38 39 40
No of Student 09 07 09 04 04 02

Find mode of the following frequency distribution :
x 3 4 5 6 7 8
f 2 4 6 3 2 1

Find mode of the following frequency distribution :
x 1.1 1.2 1.3 1.4 1.5 1.6
f 20 50 80 60 15 8

The weights of $10$ students (in kg) are given below :
$40, 35, 42, 43, 37, 35, 37, 37, 42, 37$
Find the mode of the data.

The number of members in 30 families of a village are according to following table.
Find their mode.
No. of members 2 3 4 5 6 7 8
No. of families 1 2 4 6 10 3 5

Marks obtained by students are given below, find the mode of marks.
Marks 10 20 30 40 50 60 70 80
No. of Students 2 8 16 26 20 16 7 4

Age in years of 20 students of a class are as follows:
15 16 13 14 14
13 15 14 13 13
14 12 15 14 16
13 14 14 13 15
Find mode by expressing them in frequency distribution:

Find the median of :
$66, 98, 54, 92, 87, 63, 72$

Find mode of the following frequency distribution
Marks No. of Students
0 - 10 5
10 - 20 12
20 - 30 14
30 - 40 10
40 - 50 8
50 - 60 6

Find the mode of $0, 5, 2, 7, 2, 1, 1, 3, 2, 4, 5, 7, 5, 1$ and $2$

Find the mean and variance for the data $6, 7, 10 , 12, 13, 4, 8, 12$

Find the mean and variance for the data
$\displaystyle X_{i}$ 6 10 14 18 24 28 30
$\displaystyle f_{i}$
2 4 7 12 8 4 3

Find the mean deviation about the mean for the data
$4, 7, 8, 9, 10, 12, 13, 17$

Find the mean and standard deviation using short-cut method
$\displaystyle X_{i}$ 60 61 62 63 64 65 66 67 68
$\displaystyle f_{i}$ 2 1 12 29 25 12 10 4 5

Find the mean and variance for the following frequency distrubution
Classes 0-10 10-20 20-30 30-40 40-50
Frequencies 5 8 15 16 6

Find the mean and variance for the following frequency distribution
Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210
Frequencies 2 3 5 10 3 5 2

Find the mean deviation about the mean for the data
$38, 70, 48, 40, 42, 55, 63, 46, 54, 44$

Find the mean and variance for the data
$\displaystyle X_{i}$ 92 93 97 98 102 104 109
$\displaystyle f_{i}$
3 2 3 2 6 3 3

Find the mean variance and standard deviation using short-cut method
Height
in cms No. of children
70-75 3
75-80 4
80-85 7
85-90 7
90-95 15
95-100 9
100-105 6
105-110 6
110-115 3

Find the mean deviation about the median for the data $13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$ .

The mean and variance of eight observation are $9$ and $9.25,$ respectively. If six of the observation are $6, 7, 10, 12, 12$ and $13$ , find the remaining two observations.

The mean and standard deviation of a group of $100$ observations were found to be $20$ and $3$ respectively. Later on it was found that three observations were incorrect which were recorded as $21, 21$ and $18$ . Find the mean and standard deviation if the incorrect observation are omitted

From the data given below state which group is more variable $A$ or $B$
Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Group A 9 17 32 33 40 10 9
Group B 10 20 30 25 43 15 7

The diameters of circles (in $mm$ ) drawn in a design are given below :
Diameters No. of circles
$33-36$ $15$
$37-40$ $17$
$41-44$ $21$
$45-48$ $22$
$49-52$ $25$

From the prices of shares $X$ and $Y$ below find out which is more stable in value :
X 35 54 52 53 56 58 52 50 51 49
Y 108 107 105 105 106 107 104 103 104 101

The mean and standard deviation of six observation are $8$ and $4$ respectively. If each observation is multiplied by $3$ , find the new mean and new standard deviation of the resulting observations.

The following number of goals were scored by a team in a series of $10$ matches:
$2, 3, 4, 5, 0, 1, 3, 3, 4, 3$
Find the mean, median and mode of these scores.

The following observations have been arranged in ascending order. If the median of the data is $63$ , find the value of $x$ .
$29, 32, 48, 50, x, x+2, 72, 78, 84, 95$

Write the marks wise frequencies in the following frequency distribution table.
Marks Up to $5$ Up to $6$ Up to $7$ $ Up to $8$ Up to $9$ Up to $10$
No of students $5$ $11$ $19$ $31$ $40$ $45$

Find the mean deviation about mean for the following data:
$14, 21, 28, 21, 18$

The marks scored by $60$ students in a science test are given below.
Marks $(x)$ $10$ $20$ $30$ $40$ $50$ $60$
No. of students $(f)$ $8$ $12$ $20$ $10$ $7$ $3$
Calculate the variance and standard deviation

Calculate the range and coefficient of range from the following data:
Number of trees planted in $6$ months: $186, 234, 465, 361, 290, 142$

The number of books bought by $200$ students in a book exhibition is given below.
No. of books $0$ $1$ $2$ $3$ $4$
No. of students $35$ $64$ $68$ $18$ $15$
Find the variance and standard variation

Find the mean deviation about median for the following data.
$18, 23, 9, 11, 26, 4, 14, 21$

The daily wages of $40$ workers of a factory are given in the following table.
Wages in Rs $30-34$ $34-38$ $38-42$ $42-46$ $46-50$ $50-54$
No. of workers $4$ $7$ $9$ $11$ $6$ $3$
Calculate i) Mean (ii) Variance and (iii) Standard deviation of wages and interpret the findings

Calculate the range and coefficient of range from the following data:
Marks scored by $12$ students in a test: $31, 18, 27, 19, 25, 28, 49, 14, 41, 22, 33, 13$

Find the mean deviation about mean for the following data.
$15, 18, 13, 16, 12, 24, 10, 20$

If ' $x$ ' varies inversely as ' $y$ ' and $x=7$ when $y=9$ :
(a) Find constant of variation $(k)$ .
(b) Write equation of variation.
(c) Find ' $y$ ' when $x=9$ .

The coefficient of range of the data set $\{3,1,4,2,5,9,8\}$ can be represented as $\dfrac{p}{q}$ where $p$ and $q$ are co-primes. Then the value of $(q-p)^2$ is:

Find the range and the coefficient of range of $43, 24, 38, 56, 22, 39, 45$

Age (years)	:ess than 5	5 - 9	10 - 14	15 - 19	20 - 24	25 - 29
No. of patients	38	32	50	36	24	20

Use of electricity (Unit)	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$	$100 - 120$
No of families	$13$	$50$	$70$	$100$	$80$	$17$

Content of Fat (%)	2 - 3	3 - 4	4 - 5	5 - 6	6 - 7
Milk collected (liter)	30	70	80	60	20

Distance (km)	200-210	210-220	220-230	230-240	240-250
No. of buses	40	60	80	50	20

No. of bulbs produced (Thousands)	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
No. of factories	12	35	20	15	8	7	8

No. of Cars	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
Frequency	7	14	13	12	20	11	15	8

Runs scored	Number of batsmen
3000 - 4000	4
4000 - 5000	18
5000 - 6000	9
6000 - 7000	7
7000 - 8000	6
8000 - 9000	3
9000 - 10000	1
10000 - 11,000	1

Age (in years)	No. of policy holders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Age (in years)	No. of policyholders (cf)
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Diameter (mm)	43-46	47-50	51-54	55-58	59-62
Number of circles	15	17	21	22	25