NCERT Solutions for Exercise 15.2 Class 11 Maths Chapter 15 - Statistics

Edited By Sumit Saini | Updated on Jul 18, 2022 12:06 PM IST

NCERT solutions for Class 11 Maths chapter 15 exercise 15.2 discusses topics related to finding of mean, variance and standard deviation of the given data. In the NCERT book previous exercise 15.1, we have learnt to find mean deviations about mean and median. This exercise is also quite important as the questions asked from these topics are straightforward. Direct formulas are applied to solve the question. Exercise 15.2 Class 11 Maths has 10 questions and all are based on similar concepts. Solving NCERT Solutions for Class 11 Maths chapter 15 exercise 15.2 is important and it can be said that it is an easy target to enhance score in the exam. Below mentioned articles can be accessed by students for NCERT's upcoming exercises.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

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Statistics Class 11 Maths Chapter 15 -Exercise: 15.2
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Statistics Class 11 Maths Chapter 15 -Exercise: 15.2

Question:1. Find the mean and variance for each of the data.

$\small 6, 7, 10, 12, 13, 4, 8, 12$

Answer:

Mean ( $\overline{x}$ ) of the given data:

$\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{6+ 7+ 10+ 12+ 13+ 4+ 8+ 12}{8} = \frac{72}{8} = 9$

The respective values of the deviations from mean, $(x_i - \overline{x})$ are

-3, -2, 1 3 4 -5 -1 3

$\therefore$ $\sum_{i=1}^{8}(x_i - 10)^2 = 74$

$\therefore$ $\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\frac{1}{8}\sum_{i=1}^{8}(x_i - \overline{x})^2= \frac{74}{8} = 9.25$

Hence, Mean = 9 and Variance = 9.25

Question:2. Find the mean and variance for each of the data.

First n natural numbers.

Answer:

Mean ( $\overline{x}$ ) of first n natural numbers:

$\overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$

We know, Variance $\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}\left (x_i - \frac{n+1}{2} \right )^2$

We know that $(a-b)^2 = a^2 - 2ab + b^2$

$\\ \therefore n\sigma^2 = \sum_{i=1}^{n}x_i^2 + \sum_{i=1}^{n}(\frac{n+1}{2})^2 - 2\sum_{i=1}^{n}x_i\frac{n+1}{2} \\ = \frac{n(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4}\times n - 2.\frac{(n+1)}{2}.\frac{n(n+1)}{2}$

$\\ \implies \sigma^2 = \frac{(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4} - \frac{(n+1)^2}{2}$

$\\ = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} \\ = (n+1)\left [\frac{4n+2 - 3n -3}{12} \right ] \\ = (n+1).\frac{(n-1)}{12} \\ = \frac{n^2-1}{12}$

Hence, Mean = $\frac{n+1}{2}$ and Variance = $\frac{n^2-1}{12}$

Question:3. Find the mean and variance for each of the data

First 10 multiples of 3

Answer:

First 10 multiples of 3 are:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Mean ( $\overline{x}$ ) of the above values:

$\\ \overline{x} = \frac{1}{10}\sum_{i=1}^{10}x_i = \frac{3+ 6+ 9+ 12+ 15+ 18+ 21+ 24+ 27+ 30}{10} \\ = 3.\frac{\frac{10(10+1)}{2}}{10} = 16.5$

The respective values of the deviations from mean, $(x_i - \overline{x})$ are

-13.5, -10.5, -7.5, -4.5, -1.5, 1.5, 4.5, 7.5, 10.5, 13.5

$\therefore$ $\sum_{i=1}^{10}(x_i - 16.5)^2 = 742.5$

$\therefore$ $\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\frac{1}{10}\sum_{i=1}^{10}(x_i - \overline{x})^2= \frac{742.5}{10} = 74.25$

Hence, Mean = 16.5 and Variance = 74.25

Question:4. Find the mean and variance for each of the data.

$\small x_i$ 6 10 14 18 24 28 30
$\small f_i$ 2 4 7 12 8 4 3

Answer:

$x_i$	$f_i$	$f_ix_i$	$(x_i - \overline{x})$	$(x_i - \overline{x})^2$	$f_i(x_i - \overline{x})^2$
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	13	169	363
	$\sum{f_i}$ = 40	$\sum f_ix_i$ = 760			$\sum f_i(x_i - \overline{x})^2$ =1736

$N = \sum_{i=1}^{7}{f_i} = 40 ; \sum_{i=1}^{7}{f_ix_i} = 760$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{760}{40} = 19$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\implies \sigma^2 = \frac{1736}{40} = 43.4$

Hence, Mean = 19 and Variance = 43.4

Question:5. Find the mean and variance for each of the data.

$\small x_i$ 92 93 97 98 102 104 109
$\small f_i$ 3 2 3 2 6 3 3

Answer:

$x_i$	$f_i$	$f_ix_i$	$(x_i - \overline{x})$	$(x_i - \overline{x})^2$	$f_i(x_i - \overline{x})^2$
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
	$\sum{f_i}$ = 22	$\sum f_ix_i$ = 2200			$\sum f_i(x_i - \overline{x})^2$ =640

$N = \sum_{i=1}^{7}{f_i} = 22 ; \sum_{i=1}^{7}{f_ix_i} = 2200$

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{2200}{22} = 100$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\implies \sigma^2 = \frac{640}{22} = 29.09$

Hence, Mean = 100 and Variance = 29.09

Question:6 Find the mean and standard deviation using short-cut method.

$\small x_i$ 60 61 62 63 64 65 66 67 68
$\small f_i$ 2 1 12 29 25 12 10 4 5

Answer:

Let the assumed mean, A = 64 and h = 1

$x_i$	$f_i$	$y_i = \frac{x_i-A}{h}$	$y_i^2$	$f_iy_i$	$f_iy_i^2$
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
	$\sum{f_i}$ =100			$\sum f_iy_i$ = 0	$\sum f_iy_i ^2$ =286

$N = \sum_{i=1}^{9}{f_i} = 100 ; \sum_{i=1}^{9}{f_iy_i} = 0$

Mean,

$\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =64 + \frac{0}{100} = 64$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$\\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(286) - (0)^2 \right ] \\ = \frac{28600}{10000} = 2.86$

We know, Standard Deviation = $\sigma = \sqrt{Variance}$

$\therefore \sigma = \sqrt{2.86} = 1.691$

Hence, Mean = 64 and Standard Deviation = 1.691

Question:7. Find the mean and variance for the following frequency distributions.

Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210
Frequencies 2 3 5 10 3 5 2

Answer:

Classes	Frequency $f_i$	Mid point $x_i$	$f_ix_i$	$(x_i - \overline{x})$	$(x_i - \overline{x})^2$	$f_i(x_i - \overline{x})^2$
0-30	2	15	30	-92	8464	16928
30-60	3	45	135	-62	3844	11532
60-90	5	75	375	-32	1024	5120
90-120	10	105	1050	2	4	40
120-150	3	135	405	28	784	2352
150-180	5	165	825	58	3364	16820
180-210	2	195	390	88	7744	15488
	$\sum{f_i}$ = N = 30		$\sum f_ix_i$ = 3210			$\sum f_i(x_i - \overline{x})^2$ =68280

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{3210}{30} = 107$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\implies \sigma^2 = \frac{68280}{30} = 2276$

Hence, Mean = 107 and Variance = 2276

Question:8. Find the mean and variance for the following frequency distributions.

Classes 0-10 10-20 20-30 30-40 40-50
Frequencies 5 8 15 16 6

Answer:

Classes	Frequency $f_i$	Mid-point $x_i$	$f_ix_i$	$(x_i - \overline{x})$	$(x_i - \overline{x})^2$	$f_i(x_i - \overline{x})^2$
0-10	5	5	25	-22	484	2420
10-20	8	15	120	-12	144	1152
20-30	15	25	375	-2	4	60
30-40	16	35	560	8	64	1024
40-50	6	45	270	18	324	1944
	$\sum{f_i}$ = N = 50		$\sum f_ix_i$ = 1350			$\sum f_i(x_i - \overline{x})^2$ =6600

$\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{1350}{50} = 27$

We know, Variance, $\sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2$

$\implies \sigma^2 = \frac{6600}{50} = 132$

Hence, Mean = 27 and Variance = 132

Question:9. Find the mean, variance and standard deviation using short-cut method.

Hight in cms	70-75	75-80	80-85	85-90	90-95	95-100	100-105	105-110	110-115
No. of students	3	4	7	7	15	9	6	6	3

Answer:

Let the assumed mean, A = 92.5 and h = 5

Height in cms	Frequency $f_i$	Midpoint $x_i$	$\dpi{100} y_i = \frac{x_i-A}{h}$	$y_i^2$	$f_iy_i$	$f_iy_i^2$
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
	$\sum{f_i}$ =N = 60				$\sum f_iy_i$ = 6	$\sum f_iy_i ^2$ =254

Mean,

$\overline{y} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =92.5 + \frac{6}{60}\times5 = 93$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$\\ \implies \sigma^2 = \frac{1}{(60)^2}\left [60(254) - (6)^2 \right ] \\ = \frac{1}{(60)^2}\left [15240 - 36 \right ] \\ = \frac{15204}{144} = 105.583$

We know, Standard Deviation = $\sigma = \sqrt{Variance}$

$\therefore \sigma = \sqrt{105.583} = 10.275$

Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275

Question:10. The diameters of circles (in mm) drawn in a design are given below:

Diameters	33-36	37-40	41-44	45-48	49-52
No. of circles	15	17	21	22	25

Calculate the standard deviation and mean diameter of the circles.
[ Hint First make the data continuous by making the classes as $32.5-36.5,36.5-40.4,40.5-44.5,44.5-48.5,48.5-52.5$ and then proceed.]

Answer:

Let the assumed mean, A = 92.5 and h = 5

Diameters	No. of circles $f_i$	Midpoint $x_i$	$\dpi{100} y_i = \frac{x_i-A}{h}$	$y_i^2$	$f_iy_i$	$f_iy_i^2$
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
	$\sum{f_i}$ =N = 100				$\sum f_iy_i$ = 25	$\sum f_iy_i ^2$ =199

Mean,

$\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =42.5 + \frac{25}{100}\times4 = 43.5$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$\\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(199) - (25)^2 \right ]\times4^2 \\ = \frac{1}{625}\left [19900 - 625 \right ] \\ = \frac{19275}{625} = 30.84$

We know, Standard Deviation = $\sigma = \sqrt{Variance}$

$\therefore \sigma = \sqrt{30.84} = 5.553$

Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553

Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.2

The Class 11 Maths chapter 15 exercise is prepared by expert faculties.
Exercise 15.2 Class 11 Maths can be completed in less time as almost all the questions are on similar concepts.
Class 11 maths chapter 15 exercise 15.2 solutions provided here are comprehensive and step by step manner.

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Frequently Asked Questions (FAQs)

1. List out the topics which are covered in Exercise 15.2 Class 11 Maths?

Exercise 15.2 Class 11 Maths includes mean, variance and standard deviation related questions.

2. Does chapter 15 have linkage with other chapters ?

No, most of the questions are formula based and it has no linkage with other chapters.

3. What is the weightage of Exercise 15.2 Class 11 Maths in final exam?

Around 5 marks questions can be expected from this exercise.

4. Are questions asked from this exercise difficult in nature ?

Questions are more or less easier. Provided concepts are learnt well.

5. How many questions are there in the Exercise 15.2 Class 11 Maths ?

Total 10 questions are discussed in this exercise.

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