If the mean of following frequency distribution is $$188$$ find the missing frequencies $$\displaystyle f_{1}\: and\: f_{2}$$ C.I. 0-80 80-160 160-240 240-320 320-400 Total Freq. 20 25 $$\displaystyle f_{1}$$ $$\displaystyle f_{2}$$ 10 100

$$\displaystyle f_{1}$$ = 15, $$\displaystyle f_{2}$$ = 30

$$\displaystyle f_{1}$$ = 51, $$\displaystyle f_{2}$$ = 30

$$\displaystyle f_{1}$$ = 20, $$\displaystyle f_{2}$$ = 40

$$\displaystyle f_{1}$$ = 15, $$\displaystyle f_{2}$$ = 20

MCQ Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 9

Find the mean marks from the following data:

Marks	No. of students
Below $10$	$4$
Below $20$	$10$
Below $30$	$18$
Below $40$	$28$
Below $50$	$40$
Below $60$	$70$

0%
$40\dfrac {5}{7} marks$
0%
$40\dfrac {4}{7} marks$
0%
$30\dfrac {5}{7} marks$
0%
$20\dfrac {4}{7} marks$

Explanation

For the grouped data, when class interval is given we find class marks by using the formula

Class mark

$(x_i)=\dfrac { \left( Lower\ limit\ + Upper\ limit \right) }{ 2 }$

Find

$x_{i} f_{i}$ and then

Mean

$=\dfrac { \sum { x }_{ i }{ f }_{ i } }{ \sum { f }_{ i } } =\dfrac {2850}{70} =\dfrac {285}{7} =40\dfrac {5}{7}$ marks

The following data shows that the age distribution of patients of malaria in a village during a particular month. Find the average age of the patients.

Age (in years)	No. of cases
5-14	6
15-24	11
25-34	21
35-44	23
45-54	14
55-64	5
65-74	3

0%
$36.12$ years
0%
$36.13$ years
0%
$13.36$ years
0%
$23.36$ years

Explanation

Age (in years)	No. of Cases (Fi)	Mid Point (Xi)	$FiXi$
5-14	6	9.5	57
15-24	11	19.5	214.5
25-34	21	29.5	619.5
35-44	23	39.5	908.5
45-54	14	49.5	693
55-64	5	59.5	2975.
65-74	3	69.54	208.5
	$\ Sigma Fi=83$		$\Sigma FiXi =2998.5$

For grouped data when class interval is given, we find the class mark.

Class mark

$=$ midpoint of an interval

$=\dfrac {(Lower \quad limit + Upper \quad limit)}{2}$

Class mark is taken as

$x_i$ for each class interval, find

$x_if_i$ .

Now mean age

$=\dfrac {\Sigma x_if_i}{\Sigma f_i}=\dfrac {2998.5}{83}=36.1265=36.13$

The mean of first five prime numbers is

0%
$5.6$
0%
$3.6$
0%
$6.83$
0%
$5.2$

Explanation

First five prime numbers are:

$2,3,5,7,11$
Mean

$= \cfrac{\text{Sum}}{\text{Count of Numbers}}$
Mean

$= \cfrac{2 + 3 + 5 + 7 + 11}{5}$
Mean

$= \cfrac{28}{5}$
Mean

$= 5.6$

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute	Number of women
65-68	2
68-71	4
71-74	3
74-77	8
77-80	7
80-83	4
83-86	2

0%
75.92
0%
35.92
0%
60.92
0%
80.92

Explanation

No. heart beats per min (Class Interval) ( $f_i$ )	no. of women	$x_i$	$x_if_i$
65-68	2	66. 5	133
68-71	4	69.5	278
71-74	3	72.5	217.5
74-77	8	75.5	604
77-80	7	78.5	549.5
80-83	4	81.5	326
83-86	2	84.5	169
Total	30		2277

$Mean\ heart\ beats=\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\frac { 2277 }{ 30 } =75.92$

Using empirical formula, calculate the mode for the following data:

$17$ ,

$16$ ,

$25$ ,

$23$ ,

$22$ ,

$23$ ,

$28$ ,

$25$ ,

$23$

0%
$14.61$
0%
$19.6$
0%
$23.6$
0%
$28.3$

Explanation

Given data set is

$17,16,25,23,22,23,28,25,25,23$

Arranging the data set in ascending order we get

$16,17,22,23,23,23,25,25,25,28$

Total number of values is

$10$

Mean is

$\dfrac{16+17+22+23+23+23+25+25+25+28}{10}=\dfrac{227}{10}=22.7$

Median is the middle value i.e.,

$\dfrac{5^{th}+6^{th}}{2}$ value is the median.

Therefore median is

$\dfrac{23+33}{2}=23$

The empirical formula is

$mode =3median-2mean$

$\implies mode=3(23)-2(22.7)=69-45.4=23.6$

The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
No. of patients	6	11	21	23	14	5

Find the mean of the data given above.

0%
$33.42$ years
0%
$35.37$ years
0%
$37.15$ years
0%
None of these

Explanation

Let us find the mean of the data:

Age (in years)	Number of patients $f_i$	Class mark $x_i$	$u_i=\frac {x_i- 30}{10}$	$f_i\times u_i$
5-15	6	10	-2	-12
15-25	11	20	-1	-11
25-35	21	$30-a$	0	0
35-45	23	40	1	23
45-55	14	50	2	28
55-65	5	60	3	15
Total	$n=80$			43

$a=30, h=10, n=80$ and

$\sum f_iu_i=43$
Mean

$=a+h\times \dfrac {1}{n}\times \sum f_iu_i=30+10\times \dfrac {1}{80}\times 43$

$=30+5.37=35.37$ years
Thus mean

$=35.37$ years.

The following table gives the literacy rate (in percentage) of

$35$ cities. Find the mean literacy rate.

Literacy rate (in $\%$ )	$45-55$	$55-65$	$65-75$	$75-85$	$85-95$
No. of cities	$3$	$10$	$11$	$8$	$3$

0%
$69.43$
0%
$70.43$
0%
$68.33$
0%
$67.23$

To find out the concentration of

$SO_2$ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of $SO_2$ (in ppm)	Frequency
0.00-0.04	4
0.04-0.08	9
0.08-0.12	9
0.12-0.16	2
0.16-0.20	4
0.20-0.24	2

Find the mean concentration of

$SO_2$ in the air.

0%
0.099
0%
0.090
0%
0.089
0%
0.990

Explanation

Concentration of SO2 (in ppm) Class Interval.	No. of localities Frequency(Fi)	Class Mark(Xi)	Fi*Xi
0.00-0.04	4	0.02	0.08
0.04-0.08	9	0.06	0.54
0.08-0.12	9	0.1	0.9
0.12-0.16	2	0.14	0.28
0.16-0.20	4	0.18	0.72
0.20-0.24	2	0.22	0.44
Total	30		2.96

$Mean\quad concentration\quad of\quad { SO }_{ 2 }\quad in\quad air=\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\frac { 2.96 }{ 30 } =0.09866\quad =0.099$

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

No. of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
No. of students	11	10	7	4	4	3	1

0%
12.475
0%
12.405
0%
12.005
0%
12.950

Explanation

No. of days (class interval)	No. of students	Classmark( $X_i$ )	$F_iX_i$
0-6	11	3	33
6-10	10	8	80
10-14	7	12	84
14-20	4	17	68
20-28	4	24	96
28-38	3	33	99
38-40	1	39	39

Total	40		499

We know that Class mark

$= \dfrac{upper limit +lower limit}{2}$

Mean number of days

$= \dfrac{\Sigma F_iX_i}{\Sigma F_i} = \dfrac {499}{40} = 12.475$ Days

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs.)	No. of households
100-150	4
150-200	5
200-250	12
250-300	2
300-350	2

Find the mean daily expenditure on food by a suitable method.

0%
211
0%
201
0%
215
0%
209

Explanation

The mean daily expenditure on food can be calculated using direct method,

Daily [Expenditure]	No. of households [Frequency]	$x_i$	$f_ix_i$
100-150	4	125	500
150-200	5	175	875
200-250	12	225	2700
250-300	2	275	550
300-350	2	325	650
Total	25		5275

$\text{Mean daily expenditure on food}=\dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } }$

$=\dfrac { 5275 }{ 25 }$

$=211\\$

The mean of the following frequency distribution is

Class Interval	Frequency
0-10	4
10-20	6
20-30	10
30-40	16
40-50	14

0%
25
0%
35
0%
30
0%
31

Explanation

Class interval	Mid point	Freq.	Diff. from (A=25)(d)	fd
0-10	5	4	-20	-80
10-20	15	6	-10	-60
20-30	25	10	0	0
30-40	35	16	10	160
40-50	45	14	20	280
Total	$\sum f = 50$			$\sum fd = 300$

$(\bar x) = A +\displaystyle \frac{\sum fd}{\sum f} = 25 + \frac{300}{50} = 31$

Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs.)	Number of workers
100-120	12
120-140	14
140-160	8
160-180	6
180-200	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

0%
145.2
0%
155.2
0%
135.2
0%
180.2

Explanation

Daily wages (in Rs.) [Class Intervals]	Number of workers [ Fi]	Xi Xi*Fi
100-120	12	110	1320
120-140	14	130	1820
140-160	8	150	1200
160-180	6	170	1020
180-200	10	190	1900
Total	40		5360

$Mean\quad daily\quad wage=\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\frac { 7260 }{ 50 } =Rs.145.2$

In a group of students, the mean weight of boys is

$65$ kg. and mean weight of girls is

$55$ kg. If the mean weight of all students of group is

$61$ kg, then the ratio of the number of boys and girls in the group is

0%
2 : 3
0%
3 : 1
0%
3 : 2
0%
4 : 3

Explanation

Let the no. of boys and girls are

$\displaystyle n_{1}$ and

$\displaystyle n_{2}$ respectively.
Mean weight

$=\dfrac{n_1w_1+n_2w_2}{n_1+n_2}$

$\Rightarrow \displaystyle \frac{65n_{1}+55n_{2}}{n_{1}+n_{2}}=61$

$\Rightarrow 4n_{1}=6n_{2}$

$\Rightarrow n_{1}:n_{2}=3:2$

The median of a set of eight numbers is

$4.5$ . Given that seven of the numbers are

$7, 2,3, 4,8,2$ and

$1$ find the eighth number and write down the mode of eight numbers.

0%
$4$
0%
$7$
0%
$2$
0%
$1$

Explanation

Let the eighth number be

$x$ .

Then the numbers arranged in ascending order:

$1,2,2,4,7,8,13,x$
Since these are

$8$ in number.
Median value

$\displaystyle =\frac { \left( \frac { 8 }{ 2 } \right) th\quad value+\left( \frac { 8 }{ 2 } +1 \right) th\quad value }{ 2 }$

$\displaystyle =\frac { 4th\quad value+5th\quad value }{ 2 }$

$\displaystyle =\frac { 4+7 }{ 2 } =\frac { 11 }{ 2 } =5.5$

$\displaystyle \therefore$ Mode

$= 2$

The mean of

$1, 7, 5, 3, 4$ and

$4$ is

$m$ . The observations

$3, 2, 4, 2, 3, 3$ and

$p$ have the mean

$(m - 1)$ . Find the median of the second set of data.

0%
$4$
0%
$2.5$
0%
$3$
0%
$5$

Explanation

Mean

$=\displaystyle m=\frac { 1+7+5+3+4+4 }{ 6 } =\frac { 24 }{ 6 } =4$
Now mean of

$3,2, 4, 2, 3, 3$ and

$p$ is

$(m-1)=(4 -1) = 3$

$\displaystyle \therefore \frac { 3+2+4+2+3+3+p }{ 7 } =3$

$\displaystyle \Rightarrow \frac { 17+p }{ 7 } =3\Rightarrow 17+p=21\Rightarrow p=4$

$\displaystyle \therefore$ The observations are

$3, 2, 4, 2, 3, 3$ and

$4$ .
Arranged in ascending order, the numbers are

$2,2,3,3,3,4,4$ .
Median

$\displaystyle =\left( \frac { 7+1 }{ 2 } \right)^{th} \:\text{value}= 4^{th} \:\text{value}= 3$

The arithmetic mean of first ten natural numbers is

0%
$5.5$
0%
$6$
0%
$7.5$
0%
$10$

Explanation

Formula used:

$\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}$

So, by using the above formula we get,

$A.M=\dfrac{1+2+3+...+10}{10}$

$=\dfrac{55}{10}$

$=5.5$

Hence, option

$A$ is correct.

In order to make the computation of the arithmetic mean of a set of

$50$ numbers simpler each observation is subtracted from

$53$ and the arithmetic mean of the set of differences is found to be

$-3.5$ . The arithmetic mean of the set of given numbers is

0%
$53.07$
0%
$52.93$
0%
$56.50$
0%
$49.50$

Explanation

Mean is the sum of the values divided by the total number of values.

Given that arithmetic mean of the

$50$ numbers when each observation is subtracted by

$53$ is

$-3.5$

Let the

$50$ numbers be

$x_1,x_2,...,x_{50}$

Therefore,

$-3.5=\dfrac{(x_1-53)+(x_2-53)+...+(x_{50}-53)}{50}$

$\implies {x_1+x_2+...+x_{50}-50\times53}{=-3.5\times50}$

$\implies x_1+x_2+...+x_{50}=2475$ ------(1)

Mean of

$50$ numbers is

$\dfrac{x_1+x_2+...+x_{50}}{50}$

$\implies mean = \dfrac{x_1+x_2+...+x_{50}}{50}$

$\implies mean = \dfrac{2475}{50}$ (from (1))

$\implies mean =49.5$

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	$x$	15	16	6

The missing frequency marked

$\displaystyle x$ of the above distribution whose mean is 27 is :

0%
$7$
0%
$8$
0%
$9$
0%
$10$

Explanation

Class	Frequency (f)	Class Mark (x)	fx
0-10	5	5	25
10-20	x	15	15x
20-30	15	25	375
30-40	16	35	560
40-50	6	45	270
Total	42 + x		1230 + 15x

Given, Mean $= \cfrac { \sum { fx } }{ \sum { f} } = 27$

$=> \cfrac {1230 + 15x}{42 + x} = 27$

$=> 1230 + 15x = 1134 + 27x$

$96 = 12x$

$x = 8$

In a class of

$19$ students, seven boys failed in a test. Those who passed scored

$12,15,17,15,16,15,19, 19, 17, 18, 18$ and

$19$ marks. The median score of the

$19$ students in the class is

0%
$15$
0%
$16$
0%
$17$
0%
$18$

Explanation

The seven boys who failed the test scored the marks less than these given marks. Let the marks be

$\displaystyle { x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },{ x }_{ 4 },{ x }_{ 5 },{ x }_{ 6 },{ x }_{ 7 }$
Then the marks arranged in ascending order are

$\displaystyle { x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },{ x }_{ 4 },{ x }_{ 5 },{ x }_{ 6 },{ x }_{ 7 },12,15,15,15,16,17,17,18,18,19,19,19$

$\displaystyle \therefore \quad Median\quad score\quad =\left( \frac { 19+1 }{ 2 } \right)$ th item

$\displaystyle =10$ th item=

$15$ .

The average income of a group of persons is

$\displaystyle \bar{x}$ and that of another group is

$\displaystyle \bar{y}.$ If the number of persons of both group are in the ratio 4 : 3, then average income of combined group is

0%
$\displaystyle \frac{\bar{x}+\bar{y}}{7}$
0%
$\displaystyle \frac{3\bar{x}+4\bar{y}}{7}$
0%
$\displaystyle \frac{4\bar{x}+3\bar{y}}{7}$
0%
None of these

Explanation

Here weight ratio is,

$w_1 :w_2 =\dfrac{ 4}{3}$ (i)
Hence required average
income of the combined group is

$=\cfrac{\sum w_ix_i}{\sum w_i}=\cfrac{\bar{X}w_1+\bar{Y}w_2}{w_1+w_2}=\cfrac{\bar{X}\frac{w_1}{w_2}+\bar{Y}}{1+\frac{w_1}{w_2}}=\cfrac{4\bar{X}+3\bar{Y}}{7}$

using (i)

Calculate the mode

$17, 13, 16, 17, 16, 21, 21, 9, 12, 17$

0%
$17$
0%
$16$
0%
$21$
0%
$13$

Explanation

Here the frequency of

$17$ is highest.
So, the mode =

$17$

If the mean of following frequency distribution is

$188$ find the missing frequencies

$\displaystyle f_{1}\: and\: f_{2}$

C.I.	0-80	80-160	160-240	240-320	320-400	Total
Freq.	20	25	$\displaystyle f_{1}$	$\displaystyle f_{2}$	10	100

0%
$\displaystyle f_{1}$ = 51, $\displaystyle f_{2}$ = 30
0%
$\displaystyle f_{1}$ = 20, $\displaystyle f_{2}$ = 40
0%
$\displaystyle f_{1}$ = 15, $\displaystyle f_{2}$ = 30
0%
$\displaystyle f_{1}$ = 15, $\displaystyle f_{2}$ = 20

Explanation

Given : Mean = 188

C.L.	$\displaystyle f_{i}$	$\displaystyle x_{i}$	$\displaystyle f_{i}x_{i}$
0 - 80	20	40	800
80 - 160	25	120	3000
160 - 240	$\displaystyle f_{1}$	200	200 $\displaystyle f_{1}$
240 - 320	$\displaystyle f_{2}$	280	280 $\displaystyle f_{2}$
320 - 400	10	360	3600
Total	$\displaystyle \sum f_{i}=100=55+f_{1}+f_{2}$		$\displaystyle \sum x_{i}f_{i}=7400+200f_{1}+2800f_{2}$

$\displaystyle \overline{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}\: \: \Rightarrow 188=\frac{7400+200f_{1}+280f_{2}}{100}$

$\displaystyle 18800=7400+200f_{1}+280f_{2}\Rightarrow 18800-7400=200f_{1}+280f_{2}$

$\displaystyle 11400=200f_{1}+280f_{2}\Rightarrow 1140=20f_{1}+28f_{2}$

$\displaystyle 285=5f_{1}+7f_{2}............(i)$
Also

$\displaystyle f_{1}+f_{2}+55=100\: \: \Rightarrow f_{1}+f_{2}=100-55\: \: \Rightarrow f_{1}+f_{2}=45....................(ii)$
Solving equation (i) and (ii)

$\displaystyle f_{1}=15,f_{2}=30$

Consider the table given below

Marks	0-10	10-20	20-30	30-40	40-50	50-60
Number of Students	12	18	27	20	17	6

The arithmetic mean of the marks given above is

0%
$18$
0%
$28$
0%
$27$
0%
$6$

Explanation

$x_i = \cfrac{\text{upper limit + lower limit}}{2}$

Marks	No. of students $\left( f_i \right)$	$x_i$	$x_i f_i$
0-10	12	5	60
10-20	18	15	270
20-30	27	25	675
30-40	20	35	700
40-50	17	45	765
50-60	6	55	330
	$\Sigma{f_i} = 100$		$\Sigma{x_i f_i} = 2800$

$\therefore \; Mean=\cfrac { \Sigma { x_{ i }f_{ i } } }{ \Sigma { f_{ i } } } = \cfrac{2800}{100} = 28$

Hence, the option

$(B)$ is correct

A distribution consists of three components with frequencies

$45, 40$ and

$15$ having their means

$2, 2.5, 2$ respectively. The mean of the combined distribution is

0%
$2.1$
0%
$2.2$
0%
$2.3$
0%
$2.4$

Explanation

Given,

$f_1 = 45$ ,

$f_2 = 40$ ,

$f_3 = 15$ ,

$x_1 = 2$ ,

$x_2 = 2.5$ and

$x_3 = 2$
Now, mean

$=$

$\dfrac{x_1 f_1 + x_2 f_2 + x_3 f_3}{f_1 + f_2 + f_3}$

$=$

$\dfrac{45 \times 2 + 40 \times 2.5 + 15 \times 2}{45 + 40 + 15}$

$=$

$\dfrac{90 + 100 + 30}{100}$

$=$

$\dfrac{220}{90} = 2.2$

Find the upper quartile for the data:

$9,11,15,19,17,13,7.$

0%
$17$
0%
$16$
0%
$12$
0%
None of these

Explanation

Upper quartile is the median of the upper half of the data set.

Given data set is

$9,11,15,19,17,13,7$

Arranging the data set in the ascending order we get

$7,9,11,13,15,17,19$

Number of values in the data set is

$n=7$

Upper quartile is given by

$Q_3=\dfrac 34(n+1)=\dfrac 34(7+1)=\dfrac 34(8)=6^{th} value$

Therefore, the upper quartile is

$17$

Find the mode of the following data:

$25, 16, 19, 48, 19, 20, 34, 15, 19, 20, 21, 24, 19, 16, 22, 16, 18, 20, 16, 19$

0%
$19$
0%
$18$
0%
$17$
0%
$25$

Explanation

Frequency table for the given data as given below

Value $\displaystyle f_{i}$ :	15	16	18	19	20	21	22	24	25	34	48
Frequency $\displaystyle f_{i}$ :	1	4	1	5	3	1	1	1	1	1	1

The observation with the maximum frequency is the mode of the series.
Here, 19 has the maximum frequency of 5.
So, Mode = 19

Which of the following does not change for the observation

$23, 50, 27, 2x, 48, 59, 72, 89, 5x, 100, 120$ when

$x$ lies between

$15$ and

$20$ ?

0%
Arithmetic mean
0%
Range
0%
Median
0%
Quartile deviation

Explanation

$23, 50, 27, 2x, 48, 59, 72, 89, 5x, 100, 120$

Now, when x lies in

$15$ and

$20$ , the numbers

$2x$ and

$5x$ may range between

$30, 75$ and

$40, 100$ respectively.
Thus, there will be no effect on the minimum value and maximum value. Hence, the range of the series will not be affected.

Mean of the following frequency distribution is

Class	0-5	5-10	10-15	15-20	20-25
Frequency	2	4	8	6	10

0%
$15.5$
0%
$16.5$
0%
$14.5$
0%
$13.5$

Explanation

Class	Frequency(f)	Class Mark (x)	fx
0-5	2	2.5	5
5-10	4	7.5	30
10-15	8	12.5	100
15-20	6	17.5	105
20-25	10	22.5	225
Total	30		465

Mean

$= \cfrac { \sum { fx } }{ \sum { f }} = \cfrac {465}{30} = 15.5$

C.I.	0-10	10-20	20-30	30-40	40-50
Frequency	7	5	3	4	6

Find the mean of the following continuous distribution

0%
$23.8$
0%
$22.8$
0%
$25.8$
0%
$24.8$

Explanation

C.I.	Frequency (f)	Class Mark (x)	fx
0-10	7	5	35
10-20	5	15	75
20-30	3	25	75
30-40	4	35	140
40-50 Mean $= \dfrac {\sum { fx} }{\sum { f }} = \dfrac { 595}{25} = 23.8$	6 $\sum { f} = 25$	45	270 $\sum { fx } = 595$

The mean of the following distribution is

Class	$0-5$	$5-10$	$10-15$	$15-20$	$20-25$	$25-30$
Frequency	$4$	$5$	$7$	$12$	$7$	$5$

0%
$15$
0%
$16$
0%
$17$
0%
$18$

Explanation

Answer:-

$x_i=\cfrac{\text{upper limt + lower limit}}{2}$

Class	Frequency	$x_i$	$x_i.f_i$
$0-5$	$4$	$2.5$	$6$
$5-10$	$5$	$7.5$	$37.5$
$10-15$	$7$	$12.5$	$87.5$
$15-20$	$12$	$17.5$	$210$
$20-25$	$7$	$22.5$	$157.5$
$25-30$	$5$	$27.5$	$137.5$
	$\Sigma f_i=40$		$\Sigma f_i.x_i=640$

Mean =

$\cfrac { \Sigma f_{ i }.x_{ i } }{ \Sigma f_{ i } } =\cfrac { 640 }{ 40 } =16$

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 9 - MCQExams.com

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CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 9 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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