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.

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

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

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

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

First n natural numbers.

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 = $\frac{n^2-1}{12}$

First 10 multiples of 3

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

# 610141824283024712843

 $x_i$ $f_i$ $f_ix_i$ $(x_i - \overline{x})$ $(x_i - \overline{x})^2$ $f_i(x_i - \overline{x})^2$ 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

# 929397981021041093232633

 $x_i$ $f_i$ $f_ix_i$ $(x_i - \overline{x})$ $(x_i - \overline{x})^2$ $f_i(x_i - \overline{x})^2$ 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

# 60616263646566676821122925121045

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

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

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

 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

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

 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

 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

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

 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.]

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.

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## NCERT Solutions of Class 11 Subject Wise

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## Subject Wise NCERT Exemplar Solutions

Happy learning!!!

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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