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

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.

This Story also Contains
  1. Statistics Class 11 Maths Chapter 15 -Exercise: 15.2
  2. More About NCERT Solutions for class 11 maths chapter 15 exercise 15.2
  3. Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.2
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject Wise NCERT Exemplar Solutions

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_i6101418242830
\small f_i24712843

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_i92939798102104109
\small f_i3232633

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_i606162636465666768
\small f_i21122925121045

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.

Classes0-3030-6060-9090-120120-150150-180180-210
Frequencies23510352


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.

Classes0-1010-2020-3030-4040-50
Frequencies5815166

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 cms70-7575-8080-8585-9090-9595-100100-105105-110110-115
No. of students3477159663


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



More About NCERT Solutions for class 11 maths chapter 15 exercise 15.2

The NCERT Class 11 Maths chapter Statistics is easy to understand and full marks can be scored easily if the correct formula is applied with the concept. This chapter does not have much linkage with other chapters hence if one is not good at other chapters can strengthen this chapter to score well in the exam. Exercise 15.2 Class 11 Maths deals with the questions related to finding of mean, variance and standard deviation. NCERT Solutions for Class 11 Maths chapter 15 exercise 15.2 provided here are comprehensive and is a good source of practice.

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.

Also see-

NCERT Solutions of Class 11 Subject Wise

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Subject Wise NCERT Exemplar Solutions

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Frequently Asked Question (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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