The variance of the data: x: $$1$$ $$a$$ $${ a }^{ 2 }$$ ...... $${ a }^{ n }$$ f: $${ _{ }^{ n }{ C } }_{ 0 }$$ $${ _{ }^{ n }{ C } }_{ 1 }$$ $${ _{ }^{ n }{ C } }_{ 2 }$$ ...... $${ _{ }^{ n }{ C } }_{ n }$$ is

$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{2n }$$

$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{ n }$$

$${ \left( \cfrac { 1+a }{ 2 } \right) }^{ 2n }-{ \left( \cfrac { 1+{ { a }^{ 2 } } }{ 2 } \right) }^{ n }$$

MCQ Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 5

3 weeks

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

Explanation

In the given graph, The dotted line represents the growth of the Plant

$A$ .

From the graph, we can observe that after

$3$ weeks Plant

$A$ has grown

$9\space cm$

2 weeks

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

Explanation

In the given graph, The dotted line represents the growth of the Plant

$A$ .

From the graph, we can observe that after

$2$ weeks Plant

$A$ has grown

$7\space cm$

What was the maximum forecast temperature during the week?

0%
$10$
0%
$20$
0%
$25$
0%
$35$

How much did Plant B grow from the end of the

$2^{nd}$ week to the end of the

$3^{rd}$ week?

0%
$2 cm$
0%
$3 cm$
0%
$4 cm$
0%
$5 cm$

Explanation

In the given graph, The straight line represents the growth of the Plant

$B$ .

From the graph, we can observe that after

$2$ weeks Plant

$B$ has grown

$7\space cm$ and

after

$3$ weeks Plant

$B$ has grown

$10\space cm$

Therefore, from the end of the

$2^{nd}$ week to the end of the

$3^{rd}$ week Plant

$B$ has grown

$10-7=3\space cm$

For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph.
How much did Plant A grow during the 3rd week?

0%
$2cm$
0%
$3cm$
0%
$4cm$
0%
$5cm$

Explanation

In the given graph, The dotted line represents the growth of the Plant

$A$ .

From the graph, we can observe that after

$2$ weeks Plant

$A$ has grown

$7\space cm$ and

after

$3$ weeks Plant

$A$ has grown

$9\space cm$

Therefore, during

$3^{rd}$ week Plant

$A$ has grown

$9-7=2\space cm$

What was the minimum actual temperature during the week?

0%
$5$
0%
$10$
0%
$15$
0%
$20$

Explanation

In the given graph, the straight line represents the actual temperature.

From the graph, we can observe that the minimum temperature was

$15^0C$ which was observed on Thursday and Friday.

The value of third quartile

$Q_3$ for the following distribution is

Marks Group	$5-10$	$10-15$	$15-20$	$20-25$	2 $5-30$	$30-35$	$35-40$	$40-45$
No. of Students	$5$	$6$	$15$	$10$	$5$	$4$	$2$	$1$

0%
$15$
0%
$21.5$
0%
$25$
0%
$25.5$

Explanation

The cumulative frequency distribution is as given below

Marks Groups (class)	No. of Students	Cumulative frequency
5-10	5	5
10-15	6	11
15-20	15	26
20-25	10	36
25-30	5	41
30-35	4	45
35-40	2	47
40-45	1	48
$N = 48$

We have

$N = 48 \quad \Rightarrow \displaystyle\frac{3N}{4} = 36$
The cumulative frequency just greater than

$\displaystyle\frac{3N}{4}$ is

$41$ . The corresponding class is

$25-30$ . It is the upper quartile class.

$\therefore \quad l = 25, \space f = 5, \space h = 5, \space F = 36$

$\therefore \quad Q_3 = l + \displaystyle\frac{\displaystyle\frac{3N}{4} - F}{f}\times h = 25 + \displaystyle\frac{36-36}{5}\times 5 = 25$

For the following information of wages of

$30$ workers in a factory, the value of

$Q_1 + Q_2 + Q_3$ is

S. No.	Wages (Rs.)	S. No.	Wages (Rs.)
1	330	16	240
2	320	17	330
3	550	18	420
4	470	19	380
5	210	20	450
6	500	21	260
7	270	22	330
8	120	23	440
9	680	24	480
10	490	25	520
11	400	26	300
12	170	27	580
13	440	28	370
14	480	29	380
15	620	30	350

0%
$1187.50$
0%
$797.50$
0%
$705$
0%
$872.50$

Explanation

Arranging the wages in ascending order, we obtain the following table:

S. No	Wages (Rs.)	S. No.	Wages (Rs.)
1	120	16	400
2	170	17	420
3	210	18	440
4	240	19	440
5	260	20	450
6	270	21	470
7	300	22	480
8	320	23	480
9	330	24	490
10	330	25	500
11	330	26	520
12	330	27	550
13	350	28	580
14	370	29	620
15	380	30	680

We have,

$n = 30$
Lower Quartile:

$\quad Q_1 =$ Value of

$\left(\displaystyle\frac{n+1}{4}\right)^{th}$ observation

$\Rightarrow\quad Q_1 =$ Value of

$\left(\displaystyle\frac{30+1}{4}\right)^{th}$ observation

$\Rightarrow\quad Q_1 =$ Value of

$\left(7.75\right)^{th}$ observation

$\Rightarrow\quad Q_1 =$ Value of

$\left(7\right)^{th}$ observation +

$\displaystyle\frac{3}{4}$ (Value of

$8^{th}$ observation - Value of

$7^{th}$ observation)

$\Rightarrow \quad Q_1 = 300 + \displaystyle\frac{3}{4}(320 - 300) = 315$

Middle Quartile (Median):

$Q_2$

= A.M. of

$\left(\displaystyle\frac{n}{2}\right)^{th}$ and

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

$\Rightarrow \quad Q_2 = \displaystyle\frac{\mbox{value of } 15^{th} \mbox{ observation} + \mbox{value of } 16^{th} \mbox{ observation}}{2}$

$\Rightarrow \quad Q_2 = \displaystyle\frac{380 + 400}{2} = 390$

Upper Quartile:

$\quad Q_3 =$ Value of

$3\left(\displaystyle\frac{n+1}{4}\right)^{th}$ observation

$\Rightarrow\quad Q_3 =$ Value of

$3\left(\displaystyle\frac{30+1}{4}\right)^{th}$ observation

$\Rightarrow\quad Q_3 =$ Value of

$23.25^{th}$ observation

$\Rightarrow\quad Q_3 =$ Value of

$23^{rd}$ observation +

$\displaystyle\frac{1}{4}($ Value of

$24^{th}$ Observation - Value

of

$23^{rd}$ observation

$)$

$\Rightarrow\quad Q_3 = 480 + \displaystyle\frac{1}{4}(490 - 480) = 482.50$

$\therefore \quad Q_1+Q_2+Q_3 = 315+390+482.50 = 1187.50$

If the quartile deviation of a set of observations is

$10$ and the third quartile is

$35$ , then the first quartile is :

0%
$24$
0%
$30$
0%
$17$
0%
$15$

Explanation

Quartile Deviation (

${ Q }_{ D }$ ) means the semi variation between the upper quartile or third quartile (

${ Q }_{ 3 }$ ) and lower quartile or first quartile (

${ Q }_{ 1 }$ ) in a distribution.
Formula =

${ Q }_{ D }=\cfrac { { Q }_{ 3 }-{ Q }_{ 1 } }{ 2 }$
Given,

${ Q }_{ D }$

$= 10$ and

${ Q }_{ 3 }$

$= 35$

$10=\cfrac { 35-{ Q }_{ 1 } }{ 2 }$

${ Q }_{ 1 }$

$= 15$

For the following data, the value of

$Q_1 + Q_3 - Q_2$ is :

Age in years:	20	30	40	50	60	70	80
No. of members:	3	61	132	153	140	51	3

0%
$Q_1$
0%
$Q_2$
0%
$Q_3$
0%
$2Q_2$

Explanation

The cumulative frequency table is as given below

Age in years	No. of students	Cumulative frequency
20	3	3
30	61	64
40	132	196
50	153	349
60	140	489
70	51	540
81	3	543
N $=$ 543

Lower Quartile: We have,

$\displaystyle\frac{N}{4} = \displaystyle\frac{543}{4} = 135.75$
Cumulative frequency just greater than

$N/4$ is

$196$ and the corresponding value of the variable is

$40$ .

$\therefore\quad Q_1 =$ Lower Quartile =

$40$ years
Middle Quartile (Medium):
We have,

$\displaystyle\frac{N}{2} = \displaystyle\frac{543}{2} = 271.5$
Cumulative frequency just greater than

$N/2$ is

$349$ and the corresponding value of the variable is

$50$ .

$\therefore\quad Q_2 =$ Middle quartile =

$50$ years
Upper Quartile:
We have

$\displaystyle\frac{3N}{4} = \displaystyle\frac{3\times543}{4} = 407.25$
Cumulative frequency just greater than

$407.25$ is

$489$ and the corresponding value of the variable is

$60$ .

$\therefore\quad Q_3=$ Upper quartile

$=60$ years

$\therefore\quad Q_1 + Q_3 - Q_2 = 40+60-50 = 50 = Q_2$

The variance of

$20$ observations is

$5 .$ If each observation is multiplied by

$- 1$ and added by

$2 ,$ then new variance is

0%
$-5$
0%
$5$
0%
$5.1$
0%
$7$

For the following data:

Weekly Income (in Rs.):	58	59	60	61	62	63	64	65	66
No. of Workers:	2	3	6	15	10	5	4	3	1

The value of

$\displaystyle\frac{Q_1+Q_3}{2}$ is equal to

0%
$Q_1$
0%
$Q_2$
0%
$Q_3$
0%
None of these

Explanation

The cumulative frequency of the table is a given below:

Weekly income ( in Rs.)	No. of workers	Cumulative frequency
58	2	2
59	3	5
60	6	11
61	15	26
62	10	36
63	5	41
64	4	45
65	3	48
66	1	49
N = 49

Lower Quartile:
We have

$\displaystyle\frac{N}{4} = \displaystyle\frac{49}{4} = 12.25$
The cumulative frequency just greater than

$\displaystyle\frac{N}{4}$ is

$26$ and the corresponding value of the variable is

$61$

$\therefore\quad Q_1 = Rs.\space 61$

Middle Quartile (median):
We have

$\displaystyle\frac{N}{2} = \displaystyle\frac{49}{2} = 24.5$
The cumulative frequency just greater than

$\displaystyle\frac{N}{2}$ is

$26$ and the corresponding value of the variable is

$61$

$\therefore\quad Q_2 = Rs.\space 61$

Upper Quartile:
We have

$\displaystyle\frac{3N}{4} = \displaystyle\frac{3\times49}{4} = 36.75$
The cumulative frequency just greater than

$\displaystyle\frac{3N}{4}$ is

$41$ and the corresponding value of the variable is

$63$

$\therefore\quad Q_3 = Rs.\space 63$

$\therefore\quad \displaystyle\frac{Q_1+Q_3}{2} = \displaystyle\frac{61+63}{2} = 62$

The actual temperature differs the most from the forecast temperature on:

0%
Tuesday
0%
Wednesday
0%
Thursday
0%
Friday

Explanation

In the given graph, the dotted line represents the forecast temperature and the straight line represents the actual temperature.

From the graph, we can observe that the actual temperature and the forecast temperature differs mostly on Thursday i.e., the maximum difference between actual and forecast temperature (

$23.5-15=8.5$ )

How far is the place of the merchant from the town?

0%
$16$
0%
$18$
0%
$20$
0%
$22$

Explanation

It is given that the courier person travels from town to a neighboring suburban area.

From the graph, we can observe that the destination place is

$22\space k.m.$ from the town.

When did the person stop on his way?

0%
9:00 a.m - 9.30 a.m
0%
9:30 a.m - 10.00 a.m
0%
10:00 a.m - 10.30 a.m
0%
10:30 a.m - 11.00 a.m

Explanation

The person had stopped between the time

$10$ a.m to

$10.30$ a.m.

This can be easily concluded since the curve on the graph between these times was horizontal to the x-axis which means no distance travelled.

Hence, option C is correct.

Fastest ride is from interval :

0%
8 a.m to 9 a.m
0%
9 a.m to 10 a.m
0%
10 a.m to 11 a.m
0%
11 a.m to 12 noon

Explanation

Time Period	Distance covered
8am-9am	$10 - 0 = 10 km$
9am-10am	$16 - 10 = 6 km$
10am-10.30am	$0 km$
10.30am-12am	$22 - 16= 6 km$

The person ride fastest during 8am to 9am.

The variance of first

$50$ even natural numbers is

0%
$\displaystyle \frac{833}{4}$
0%
$833$
0%
$437$
0%
$\displaystyle \frac{437}{4}$

Explanation

${\textbf{Step-1: Use mean formula and find the mean.}}$

$\Rightarrow \sigma^2 = \dfrac{1}{n} \sum x_i^2 - \bar{(x)}^2$

$\Rightarrow n = 50, \sum x_i = 2 +4+6+8+...+100$

${\text{We know that,}}$

$\Rightarrow \bar {x} = \dfrac{\sum x_i}{n}$

$\Rightarrow \bar {x} = \dfrac{2 +4+6+8+...+100}{50}$

$\Rightarrow \bar {x} = \dfrac{50 \times 51}{50}$

$[\because \sum 2n = n(n+1)]$

$\Rightarrow \bar {x} = 51$

${\textbf{Step-2: Put the values in variance formula.}}$

$\Rightarrow \sigma^2 = \dfrac{1}{n} \sum x_i^2 - \bar{(x)}^2$

$= \dfrac{1}{50} (2^2 +4^2+6^2+8^2+...+100^2) - {(51)}^2$

$= \dfrac{1}{50} [({2.1})^2 +({2.2})^2+({2.3})^2+({2.4})^2+...+({2.50})^2] - {(51)}^2$

$= \dfrac{1}{50} 2^2[({1})^2 +({2})^2+({3})^2+({4})^2+...+({50})^2] - {(51)}^2 ...(1)$

${\text{But,}}$

$\Rightarrow ({1})^2 +({2})^2+({3})^2+({4})^2+...+({n})^2 = \dfrac{n(n+1)(2n+1)}{6}$

$\Rightarrow ({1})^2 +({2})^2+({3})^2+({4})^2+...+({50})^2 = \dfrac{50(50+1)(2 \times 50+1)}{6}$

$\Rightarrow ({1})^2 +({2})^2+({3})^2+({4})^2+...+({50})^2 = \dfrac{50(51)(101)}{6}$

${\text{Equation (1) become,}}$

$= \dfrac{1}{50} 2^2[\dfrac{50(51)(101)}{6}] - {(51)}^2$

$= 34 \times 101 -2601$

$= 3434-2601$

$=833$

${\textbf{Thus, option B is correct.}}$

The quartile deviation of the income of a certain person given in rupees for 12 months in a year:139, 150, 151,151,157,158,160,161,162,162,173,175

0%
4.5
0%
5.5
0%
6.2
0%
none of these

Explanation

S.NO.	Income(Rs)
1	139
2	150
3	151
4	151
5	157
6	158
7	160
8	161
9	162
10	162
11	173
12	175

$N=12$

${ Q }_{ 1 }=$ Size of

$\displaystyle\frac { N+1 }{ 4 } th=3.25th$ item

$=$ size of

$3rd$ item

$+0.25$ (size of

$4th$ item)

$-$ (size of

$3rd$ item)

$=151+0.25(151-151)=$ Rs.

$151$

${ Q }_{ 3 }=$ Size of

$\displaystyle\frac { 3(N+1) }{ 4 } th=9.75th$ item

$=$ size of

$9th$ item

$+0.75$ (size of

$10th$ item)

$-$ (size of

$9th$ item)

$=162+0.75(162-162)=$ Rs.

$162$

$\therefore$ Quartile Deviation

$\displaystyle=\frac { 1 }{ 2 } \left( { Q }_{ 3 }-{ Q }_{ 1 } \right) =\frac { 11 }{ 2 } =$ Rs.

$5.5$

Standard deviation for first 10 natural numbers is

0%
$5.5$
0%
$3.87$
0%
$2.97$
0%
$2.87$

Explanation

Standard deviation of first

$n$ natural number is

$\displaystyle =\sqrt{\frac{n^2-1}{12}}=\sqrt{\frac{10^2-1}{12}}=2.87$

The quartile deviation of daily wages (in Rs.) of 7 persons given below:

$12,7,15,10,17,17,25$ is

0%
$4.5$
0%
$9$
0%
$5$
0%
$14.5$

Explanation

Writing the series in ascending order

$7,10,12,15,17,19,25$

$\displaystyle{ Q }_{ 1 }=\frac { 15+7 }{ 2 } =11,{ Q }_{ 3 }=\frac { 25+15 }{ 2 } =20$

$\therefore$ Quartile Deviation

$\displaystyle=\frac { { Q }_{ 3 }-{ Q }_{ 1 } }{ 2 } =\frac { 20-11 }{ 2 } =4.5$

Let

${ x }_{ 1 },{ x }_{ 2 },.....{ x }_{ n }$ be

$n$ observations with mean

$m$ and standard deviation

$s$ . Then the standard deviation of the observations

$a{ x }_{ 1 },a{ x }_{ 2 },.....a{ x }_{ n }$ , is

0%
$a+s$
0%
$s/a$
0%
$\left| a \right| s$
0%
$as$

Explanation

$S.D(ax) = \sqrt{E[ax-a\bar{x}]^2}=\sqrt{E[a(x-\bar{x})^2]}=\sqrt{a^2.E[(x-\bar{x})^2]}=|a|.\sqrt{E[(x-\bar{x})^2]}=|a|.S.D(x)=|a| s$

Standard deviation for first 10 even natural numbers is

0%
$11$
0%
$7.74$
0%
$5.74$
0%
$11.48$

Explanation

We know standard deviation of first

$n$ even natural number is

$, \sigma =\sqrt{\cfrac{n^2-1}{3}}$
Here

$n =10$

$\therefore \sigma = \sqrt{\cfrac{100-1}{3}}=\sqrt{33}=5.74$

If a variable

$X$ takes values

$0,1,2,....,n$ with frequencies

${ _{ }^{ n }{ C } }_{ 0 },{ _{ }^{ n }{ C } }_{ 1 },{ _{ }^{ n }{ C } }_{ 2 },......{ _{ }^{ n }{ C } }_{ n }\quad$ respectively, then S.D. is equal to :

0%
$\cfrac { n }{ 4 }$
0%
$\cfrac { n }{ 2 }$
0%
$\cfrac { \sqrt { n } }{ 2 }$
0%
$\cfrac { \sqrt { n } }{ 4 }$

Explanation

Here,

$\displaystyle \mu1'$

$=\frac{\sum r\frac{n}{r}^{n-1}C_{r-1}}{\sum ^{n}C_{r}}$

$\displaystyle$

and

$\displaystyle \mu 2'$

$\displaystyle=\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )^{n}C_{r}+\frac{n}{2}$

$\displaystyle =\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )\frac{n\left ( n-1 \right )}{r\left ( r-1 \right )}^{n-2}C_{r-2}+\frac{n}{2}$

$\displaystyle =\frac{n\left ( n-1 \right)}{2^{n}}.2^{n-2}+\frac{n}{2}$

$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}$

$\therefore$ Variance

$\displaystyle \sigma^2=\mu 2'-\left (\mu 1' \right )^{2}$

$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}-\left ( \frac{n}{2} \right )^{2}=\frac{n}{4} \mbox{or} S.D = \sigma = \sqrt{\sigma^2} = \frac{\sqrt{n}}{2}$

If a variable

$x$ takes values

$0,1,2,....n$ with frequencies proportional to the binomial coefficients

${ _{ }^{ n }{ C } }_{ 0 },{ _{ }^{ n }{ C } }_{ 1 },{ _{ }^{ n }{ C } }_{ 2 },......{ _{ }^{ n }{ C } }_{ n }$ , then mean of distribution is

0%
$\dfrac {n(n+1)}{2}$
0%
$\dfrac { n }{ 2 }$
0%
$\dfrac { 2 }{ n }$
0%
$\dfrac {n(n-1)}{2}$

Explanation

Here,

$\displaystyle \mu1 '=\frac{\sum r\frac{n}{r}^{n-1}C_{r-1}}{\sum ^{n}C_{r}}$

$\displaystyle =\frac{n.2^{n-1}}{2^{n}}=\frac{n}{2}$
and

$\displaystyle \mu 2'=\frac{1}{2^{n}}\sum_{0}^{n}\left\{ \left ( r-1 \right )+r \right \}^{n}C_{r}$

$\displaystyle=\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )^{n}C_{r}+\frac{n}{2}$

$\displaystyle =\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )\frac{n\left ( n-1 \right )}{r\left ( r-1 \right )}^{n-2}C_{r-2}+\frac{n}{2}$

$\displaystyle =\frac{n\left ( n-1 \right)}{2^{n}}.2^{n-2}+\frac{n}{2}$

$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}$

$\therefore$ Variance

$\displaystyle \sigma^2=\mu 2'-\left (\mu 1' \right )^{2}$

$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}-\left ( \frac{n}{2} \right )^{2}=\frac{n}{4}$

Consider the frequency distribution

Class interval:	0-6	6-12	12-18
Frequency:	2	4	6

The variance of the above frequency distribution, is

0%
$24$
0%
$12$
0%
$20$
0%
$25$

Explanation

Class interval	$f_i$	$x_i$	$d_i={x_i-A}$	$d_i^2$	$f_id_i^2$	$f_id_i$
0-6	2	3	-6	36	72	-12
6-12	4	9	0	0	0	0
12-18	6	15	6	36	216	36

Here,

$N=\sum {f_i }=12$

$\sum {f_id_i^2}=288$

$\sum {f_id_i}=24$

Now, variance

$\displaystyle { \sigma }^{ 2 }=\frac { \sum { f_{ i }d_{ i }^{ 2 } } }{ N } -{ \left( \frac { \sum { f_{ i }d_{ i } } }{ N } \right) }^{ 2 }$

$\Rightarrow \displaystyle { \sigma }^{ 2 }=\frac { 288 }{ 12 } -{ \left( \frac { 24 }{ 12 } \right) }^{ 2 }$

$\Rightarrow { \sigma }^{ 2 }=20$

The quartile deviation for the data

x :	2	3	4	5	6
f :	3	4	8	4	1

0%
$\cfrac { 1 }{ 2 }$
0%
$1$
0%
$\cfrac { 1 }{ 4 }$
0%
$0$

Explanation

$x$	$f$	Cumulative Frequency
2	3	3
3	4	7
4	8	15
5	4	19
6	1	20
Total	20

Here,

$n=20$

$Q_1=(\dfrac{20+1}{4})^{th}observation$

$=\dfrac{21}{4}=5.25^{th}observation$
So,

$Q_1=3$

$Q_3=\dfrac{3(n+1)}{4})^{th} observation$

$=\dfrac{63}{4}=15.75^{th} observation$
So,

$Q_3=5$

Quartile deviation

$=\dfrac{Q_3-Q_1}{2}$

$=\dfrac{5-3}{2}=1$

The variance of the data:

x:	$1$	$a$	${ a }^{ 2 }$	......	${ a }^{ n }$
f:	${ _{ }^{ n }{ C } }_{ 0 }$	${ _{ }^{ n }{ C } }_{ 1 }$	${ _{ }^{ n }{ C } }_{ 2 }$	......	${ _{ }^{ n }{ C } }_{ n }$

0%
${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{ n }$
0%
${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{2n }$
0%
${ \left( \cfrac { 1+a }{ 2 } \right) }^{ 2n }-{ \left( \cfrac { 1+{ { a }^{ 2 } } }{ 2 } \right) }^{ n }$
0%
none of these

Explanation

$\displaystyle \bar { X } =\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } \\ \displaystyle =\frac { ^{ n }C_{ 0 }+a.^{ n }C_{ 1 }+{ a }^{ 2 }.^{ n }C_{ 2 }+....{ a }^{ n }.^{ n }C_{ n } }{ ^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }+....^{ n }C_{ n } } \\ \displaystyle =\frac { { \left( 1+a \right) }^{ n } }{ { 2 }^{ n } }$
.

$\displaystyle S.D.=\sqrt { \frac { \sum { { f }_{ i }{ { x }_{ i } }^{ 2 } } }{ \sum { { f }_{ i } } } -{ \left( \frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } \right) }^{ 2 } } \\ \displaystyle =\sqrt { \frac { ^{ n }C_{ 0 }+{ a }^{ 2 }.^{ n }C_{ 1 }+{ a }^{ 4 }.^{ n }C_{ 2 }+....{ a }^{ 2n }.^{ n }C_{ n } }{ ^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }+....^{ n }C_{ n } } -{ \left[ \frac { { \left( 1+a \right) }^{ n } }{ { 2 }^{ n } } \right] }^{ 2 } } \\ \displaystyle \sqrt { \frac { { \left( 1+{ a }^{ 2 } \right) }^{ n } }{ { 2 }^{ n } } -{ \left( \frac { 1+a }{ 2 } \right) }^{ 2n } }$

$Variance = (S.D)^2$

How many less letters were collected on Wednesday than on Tuesday?

0%
$40$
0%
$30$
0%
$20$
0%
$10$

Explanation

Given that one square box represents

$10$ letters.

On Tuesday there are 5 boxes.

Therefore, number of letters collected on Tuesday are

$10\times5=50$

On Wednesday there are 4 boxes.

Therefore, number of letters collected on Wednesday are

$10\times4=40$

There are

$50-40=10$ letters less on Wednesday than on Tuesday.

The S.D. of the following frequency distribution is

Class	0-10	10-20	20-30	30-40
$\displaystyle f_{i}$	1	3	4	2

0%
$7.8$
0%
$9$
0%
$8.1$
0%
$0.9$

Explanation

Class	$x_i$	$f_i$	$u_i=\dfrac{x_i-a}{h}$	$f_iu_i$	$f_iu_i^2$
0-10	5	1	-2	-2	4
10-20	15	3	-1	-3	3
20-30	25	4	0	0	0
30-40	35	2	1	2	2
		$N=10$		$\sum f_iu_i=-3$	$\sum f_iu_i^2=9$

$\displaystyle \sigma =h\sqrt{\frac{ \Sigma f_{1}u_{1}^{2}}{N}-\left ( \frac{\Sigma f_{1}u_{1}}{N} \right )^{2}}$

$\displaystyle =10\sqrt{\frac{9}{10}-\left ( \frac{-3}{10} \right )^{2}}=9$

The adjoining diagram illustrates the information collected about the number
of brothers of the boys in a certain group. What is the total number of boys in
this group?

0%
$81$
0%
$22$
0%
$73$
0%
$34$

Explanation

From the figure,
there are

$2$ boys who have

$1$ brother, thus total

$2$ boys
there are

$5$ boys who have

$2$ brothers, thus total

$10$ boys
there are

$4$ boys who have

$3$ brothers, thus total

$12$ boys
there are

$7$ boys who have

$4$ brothers, thus total

$28$ boys
there are

$3$ boys who have

$5$ brothers, thus total

$15$ boys
there are

$1$ boys who have

$6$ brothers, thus total

$6$ boys
Hence, total number of boys

$= 2 + 10 + 12 + 28 + 15 + 6 = 73$

The incomplete pictograph shows the number of stamps(1 ticket = 4 stamps) collected by four pupils:
The missing tickets represents the number of stamps collected by Anjali.

Then how many tickets collected by Anjali?

0%
$5$
0%
$7$
0%
$6$
0%
$8$

Explanation

From the given pictograph, the total tickets coolected by Lata, Kalyan, Vaishnavi are

$13$ .

Given that one ticket symbol represents

$4$ stamps.

Therefore, total number of stamps collected by Lata, Kalyan, Vaishnavi are

$13\times4=52$ .

Given that, total number of stamps collected by four pupils is

$80$ .

Therefore the number of stamps collected by Anjali = Total number of stamps collected by four pupil - number of stamps collected by three pupils (excluding Anjali)

$=80-52=28$

Hence, the number of tickets collected by Anjali are

$\dfrac{28}4=7$

Answer the following question by reading the given pictograph.

Number of TV sets sold in the year 2005 is

0%
$1000$
0%
$500$
0%
$50$
0%
$2000$

Explanation

From the given pictograph,

$2005$ has one tv symbol.

Given that one symbol represents

$500$ tv sets.

Therefore the

$2005$ has

$1\times500=500$ tv sets.

The incomplete bar graph shows the number of cookies sold by Raju in

$5$ days. The total number of cookies sold was

$2,000$ . How many cookies were sold on Sunday?

0%
$750$
0%
$850$
0%
$800$
0%
$950$

Explanation

Given that the total number of cookies sold

$=2000$

From the given pictograph, Wednesday bar graph shows

$150$ cookies.

Similarly, Thursday bar graph shows

$350$ cookies.

Friday bar graph shows

$300$ cookies.

Saturday bar graph shows

$450$ cookies.

The total number of cookies sold on Wednesday, Thursday, Friday, Saturday

$= 150+350+300+450=1250$ .

Therefore the number of cookies sold on Sunday

$=2000-1250=750$

The year in which the most number of TV sets have been sold is

0%
$1990$
0%
$1995$
0%
$2008$
0%
$2000$

After the reading the pictograph,find the number of TV sets sold in the year

$1990$ is

0%
$1000$
0%
$1500$
0%
$1250$
0%
$2000$

Explanation

Given that each TV symbol represents

$500$ TV sets.

In the given pictograph,

$1990$ year has

$2$ TV symbols.

Therefore, number of TV sets sold in the year

$1990$ are

$2\times500=1000$

For a series the value of mean deviation is 15.The most likely value of its quartile deviation is

0%
12.5
0%
11.6
0%
13
0%
9.7

Explanation

use: M.D.

$=\displaystyle \frac{4}{4}\sigma ,Q.D=\frac{2}{3}\sigma$

$\displaystyle \Rightarrow \frac{M.D.}{Q.D}=\frac{6}{5}\Rightarrow Q.D.=\frac{5}{6}\left ( M.D. \right )$

$MD=\frac { 4 }{ 5 } \times \sigma \quad ,\quad QD=\frac { 2 }{ 3 } \times \sigma \quad ,\quad \\ \frac { MD }{ QD } =\frac { 6 }{ 5 } ,\quad QD\quad =\quad \frac { 5\times 15 }{ 6 } =12.5$

The year in which the most number of TV sets have been sold is

0%
$1980$
0%
$1985$
0%
$2000$
0%
$1990$

Explanation

It can be seen that in the year

$1985$ and

$2000$ there are

$5$ TV icons which is the most implying that maximum number of TV sets were sold in those years.

If one symbol of image(Ref image) Represents

$10$ bottles of apple juice then

$35$ bottles of apple juice represented by

Explanation

$3$ full and one half are required to show

$35$ .

Hence the answer is option C

The pictogram shows the number of visitors to an exhibition in a particular week. The number of visitors recorded after Wednesday is:

0%
$10$
0%
$18$
0%
$60$
0%
$70$

Explanation

Number of visitors in an exhibition after Wednesday

$=$ Number of visitors in an exhibition on Thursday + Number of visitors in an exhibition on Friday

Number of visitors in an exhibition after Wednesday

$=$

$(6\times 6) + (6\times 4)$

$=$

$36 + 24 = 60$

The bar graph shows the number of cakes sold at a shop in four days. What is the difference in number of cakes between the highest and the lowest daily sale?

0%
$20$
0%
$35$
0%
$30$
0%
$40$

Explanation

From the above bar graph, the highest number of cakes sold is

$45$ (i.e., on Sunday).

and the lowest number of cakes sold is

$25$ (i.e., on Monday).

Therefore, the difference between the highest and lowest number of cakes

$=45-25=20$ .

Hence, option

$A$ is correct.

The incomplete graph shows Dinesh's savings from January to April. Dinesh saved a total of Rs. 730 during the four months. The amount of money saved in February was as much as that saved in March. How much did he save in March?

0%
Rs. $245$
0%
Rs. $275$
0%
Rs. $250$
0%
Rs. $295$

Explanation

From the given graph, Dinesh saved a total of

$160$ rupees during January month and

An amount of

$80$ rupees during April month.

Let the savings in February be

$x$ rupees.

Given that, that amount of money saved in March and February are same.

Therefore, savings in March month

$=x$ rupees.

Given that, the total amount saved in 4 months

$=730$ rupees.

$\implies 160+x+x+80=730$

$\implies 240+2x=730$

$\implies 2x=730-240=490$

$\implies x=\dfrac{490}2=245$

Therefore, savings in March month

$=245$ rupees.

The bar graph shows the amount of money donated by

$4$ classes to a charity. How much more money must Class 4P donate so that their amount is the same as that of class 4S?

0%
Rs. $20$
0%
Rs. $60$
0%
Rs. $40$
0%
Rs. $100$

Explanation

From the given bargraph, class

$4P$ has donated

$60$ rupees and

class

$4S$ has donated

$100$ rupees.

Therefore, class

$4P$ has to donate

$100-60=40$ rupees so that their amount is same as that of class

$4S$ .

The following bar graph shows the rainfall at selected locations in certain months.
Which of the following statements is correct

0%
November rainfall exceeds $100$ cm in each location
0%
September rainfall exceeds April rainfall by $50$ cm in each location
0%
November rainfall is lower than April rainfall in each location
0%
None of the above

Explanation

The given bar graph shows the rainfall at six different locations 1, 2, 3, 4, 5, and 6 in 3 different months November, April, and September, and the height of each bar for three months at each location gives us the amount of rainfall.

$\Rightarrow$ From the heights of bars that rainfall in the month of November is always lower than rainfall in the months of April and September.

$\therefore$ From given options, "November rainfall is lower than April rainfall in each location" is a correct statement.

0%
5
0%
60
0%
10
0%
12

Explanation

$\Rightarrow$ One symbol represents

$5$ balloons.

$\Rightarrow$ Then,

$60$ balloons represents how many symbols.

$\therefore$ Required symbols

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

$\therefore$ The number of symbols to be drawn to represent

$60$ ballons is

$12$

The pictogram shows the number of pupils who cycle to school. What is the total number of pupils who use cycle for going school?

0%
$10$
0%
$300$
0%
$420$
0%
$450$

Explanation

Number of pupils who use cycle for going school in class 1

$= 30\times 5 = 150$

Number of pupils who use cycle for going school in class 1

$= 30\times 7 = 210$

Number of pupils who use cycle for going school in class 1

$= 30\times 3= 90$

So, Total number of pupils who use cycle for going school

$= 150+210+90 = 450$

Given above is a bar graph showing the heights of six mountain peaks. Read the above diagram and answer the following:

Write the ratio of the heights of highest peak and the lowest peak.

0%
$22 : 15$
0%
$15 : 22$
0%
$20 : 13$
0%
$13 : 22$

Explanation

From the above bar graph,

Height of highest peak

$= 8800\ m$
Height of lowest peak

$= 6000\ m$

Hence the required ratio will be,
Ratio

$=8800 : 6000$

$= 22 :15$

Hence, option

$A$ is correct.

Students at a local college were asked how many hours they slept last night. The adjoining chart shows the data . A bar graph of this data will be made on a grid that is 20 units. What scale would be most appropriate for the axis labelled "Number of Students" ?

Hours of Sleep	Number of Students
6	14
7	26
8	28
9	15
More than 10	6

0%
1 unit = 1 student
0%
1 unit = 2 students
0%
1 unit = 10 students
0%
1 unit = 28 students

Standard deviation of the distribution

$1, 3, 5,....., 13$ will be

0%
$2$
0%
$4$
0%
$7$
0%
None

Explanation

Mean of the given numbers is

$\bar x =\dfrac{1+3+5+7+9+11+13}{7}=\dfrac{49}{7}=7$

total number of values

$n=7$

Standard deviation is given by

$SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n-1}}$

$\implies SD=\sqrt{\dfrac{(1-7)^2+(3-7)^2+(5-7)^2+(7-7)^2+(9-7)^2+(11-7)^2+(13-7)^2}{7-1}}$

$\implies SD=\sqrt{\dfrac{36+16+4+0+4+16+36}{6}}$

$\implies SD=\sqrt{\dfrac{112}{6}}$

$\implies SD=\sqrt{18.6667}=4.32$

Therefore the standard deviation for the given values is

$4.32$

In the following table pass percentage of three schools from the year

$2001$ to the year

$2006$ are given. Which school students performance is more consistent?

	$2001$	$2002$	$2003$	$2004$	$2005$	$2006$
School (X)	$80$	$89$	$79$	$83$	$84$	$65$
School (Y)	$92$	$94$	$76$	$75$	$80$	$63$
School (Z)	$93$	$97$	$67$	$63$	$70$	$85$

0%
X
0%
Y
0%
Z
0%
X and Y

Explanation

Average of

$x=\cfrac { 80+89+79+83+84+65 }{ 6 } =\cfrac { 480 }{ 6 } =80$

Average of

$y=\cfrac { 92+94+76+75+80+63 }{ 6 } =\cfrac { 480 }{ 6 } =80$

Average of

$z=\cfrac { 93+97+67+63+70+85 }{ 6 } =79.16$

Average of x and y are equal and more than z.

So,their performance is consistent.

The S.D. of the following freq. dist.

Class	0-10	10-20	20-30	30-40
$f_i$	1	3	4	2

0%
$7.8$
0%
$9$
0%
$8.1$
0%
$0.9$

Explanation

Class	$x_i$	$f_i$	$u_i=\dfrac{x_i-A}{h}$	$u_i^2$	$f_iu_i$	$f_iu_i^2$
0-10	5	1	-2	4	-2	4
10-20	15	3	-1	1	-3	3
20-30	25	4	0	0	0	0
30-40	35	2	1	1	2	2
Total		10			-3	9

$S.D.$

$=h\sqrt { \cfrac { \sum { { f }_{ i }{ u }_{ i }^{ 2 } } }{ \sum { { f }_{ i } } } -{ \left( \cfrac { \sum { f } _{ i }{ u }_{ i } }{ \sum { { f }_{ i } } } \right) }^{ 2 } }$

$S.D.=10\sqrt { \cfrac { 9 }{ 10 } -{ \left( \cfrac { -3 }{ 10 } \right) }^{ 2 } } =9$

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 5 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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