# NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.2) Exercise 15.2

## NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.2) Exercise 15.2

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## Access NCERT Solutions for Class 11 Maths Chapter 15 – Statistics

Exercise – 15.2

1. Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12.

Ans: Given data: 6, 7, 10, 12, 13, 4, 8, 12.

Mean can be given as:

$\bar x = \frac{{\sum\limits_{i = 1}^8 {{x_i}} }}{{\text{n}}} = \frac{{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}}{8} = \frac{{72}}{8} = 9$.

The following table is obtained:

 ${x_i}$ $\left( {{x_i} - \bar x} \right)$ ${\left( {{x_i} - \bar x} \right)^2}$ 6 $6 - 9 = - 3$ 9 7 $7 - 9 = - 2$ 4 10 $10 - 9 = 1$ 1 12 $12 - 9 = 3$ 9 13 $13 - 9 = 4$ 16 4 $4 - 9 = - 5$ 25 8 $8 - 9 = - 1$ 1 12 $12 - 9 = 3$ 9 74

Variance can be given as:

${\sigma ^2} = \frac{1}{n}\mathop \sum \limits_{i = 1}^n {\left( {{x_i} - \bar x} \right)^2}$

${\sigma ^2} = \frac{1}{8} \times 74$

${\sigma ^2} = 9.25$.

Therefore, mean is 9 and variance is 9.25.

1. Find the mean and variance for the first $n$ natural numbers.

Ans: The mean of first $n$ natural numbers can be calculated as:

$Mean = \frac{{{\text{Sum }}of{\text{ }}observations}}{{Number{\text{ }}of{\text{ }}observations}}$

$\therefore Mean = \frac{{\frac{{n(n + 1)}}{2}}}{n} = \frac{{n + 1}}{2}$

Variance, ${\sigma ^2} = \frac{1}{n}\mathop \sum \limits_{i = 1}^n {\left( {{x_i} - \bar x} \right)^2}$

$\Rightarrow \frac{1}{n}\mathop \sum \limits_{i = 1}^n {\left( {{x_i} - \left( {\frac{{n + 1}}{2}} \right)} \right)^2}$

As we know,${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$. Hence:

$\Rightarrow \frac{1}{n}\mathop \sum \limits_{i = 1}^{\text{n}} {\text{x}}_{\text{i}}^2 - \frac{1}{{\text{n}}}\mathop \sum \limits_{{\text{i}} = 1}^{\text{n}} 2\left( {\frac{{{\text{n}} + 1}}{{\text{n}}}} \right){{\text{x}}_i} + \frac{1}{n}\mathop \sum \limits_{{\text{i}} = 1}^{\text{n}} {\left( {\frac{{{\text{n}} + 1}}{2}} \right)^2}$

$\Rightarrow \frac{1}{n}\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \left( {\frac{{n + 1}}{2}} \right)\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right] + \frac{{{{\left( {n + 1} \right)}^2}}}{{4n}} \times n$

$\Rightarrow \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{{{\left( {n + 1} \right)}^2}}}{2} + \frac{{{{\left( {n + 1} \right)}^2}}}{4}$

Taking LCM, we get:

$\Rightarrow \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} - \frac{{{{\left( {n + 1} \right)}^2}}}{4}$

Taking $(n + 1)$ as common, we get:

$\Rightarrow (n + 1)\left[ {\frac{{4n + 2 - 3n - 3}}{{12}}} \right]$

$\Rightarrow \frac{{(n + 1)(n - 1)}}{{12}}$

Now, we know, $(a + b)(a - b) = {a^2} - {b^2}$. Hence,

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

Therefore, mean is $\frac{{n + 1}}{2}$ and variance is $\frac{{{n^2} - 1}}{{12}}$.

1. Find the mean and variance for the first 10 multiples of 3.

Ans: First 10 multiples of 3 can be written as: 3, 9, 12, 15, 18, 21, 24, 27, 30. Hence, $n = 10$.

Mean, $\bar x = \frac{{\sum\limits_{i = 1}^8 {{x_i}} }}{{\text{n}}} = \frac{{3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30}}{{10}} = \frac{{165}}{{10}} = 16.5$

The following table is obtained:

 ${x_i}$ $\left( {{x_i} - \bar x} \right)$ ${\left( {{x_i} - \bar x} \right)^2}$ 3 $3 - 16.5 = - 13.5$ 182.25 6 $6 - 16.5 = - 10.5$ 110.25 9 $9 - 16.5 = - 7.5$ 56.25 12 $12 - 16.5 = - 4.5$ 20.25 15 $15 - 16.5 = - 1.5$ 2.25 18 $18 - 16.5 = 1.5$ 2.25 21 $21 - 16.5 = 4.5$ 20.25 24 $24 - 16.5 = 7.5$ 56.25 27 $27 - 16.5 = 10.5$ 110.25 30 $30 - 16.5 = 13.5$ 182.25 742.5

Variance can be given as:

${\sigma ^2} = \frac{1}{n}\mathop \sum \limits_{i = 1}^n {\left( {{x_i} - \bar x} \right)^2}$

${\sigma ^2} = \frac{1}{{10}} \times 742.5$

${\sigma ^2} = 74.25$.

Therefore, mean is $16.5$ and variance is $74.25$.

1. Find the mean and variance for the data:

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

Ans: Calculating the given data we get the following table:

 ${x_i}$ ${f_i}$ ${f_i}{x_i}$ ${x_i} - \bar x$ ${\left( {{x_i} - \bar x} \right)^2}$ ${f_i}{\left( {{x_i} - \bar x} \right)^2}$ 6 2 12 $6 - 19 = - 13$ 169 338 10 4 40 $10 - 19 = - 9$ 81 324 14 7 98 $14 - 19 = - 5$ 25 175 18 12 216 $18 - 19 = - 1$ 1 12 24 8 192 $24 - 19 = 5$ 25 200 28 4 112 $28 - 19 = 9$ 81 324 30 3 90 $30 - 19 = 11$ 121 363 40 760 1736

Here, ${\text{N}} = 40$, $\mathop \sum \limits_{i = 1}^7 {f_i}{x_i} = 760$, $\mathop \sum \limits_{i = 1}^7 {f_i}{(x - \bar x)^2} = 1736$

Mean, $\bar x = \frac{{\sum\limits_{i = 1}^7 {{f_i}{x_i}} }}{{{\text{\;N}}}} = \frac{1}{{40}} \times 760 = 19$

Variance, ${\sigma ^2} = \frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{{({x_i} - \bar x)}^2}} = \frac{1}{{40}} \times 1736 = 43.4$

Therefore, mean is $19$ and variance is $43.4$.

1. Find the mean and variance for the data:

 ${x_i}$ 92 93 97 98 102 104 109 ${f_i}$ 3 2 3 2 6 3 3

Ans: Calculating the given data we get the following table:

 ${x_i}$ ${f_i}$ ${f_i}{x_i}$ ${x_i} - \bar x$ ${\left( {{x_i} - \bar x} \right)^2}$ ${f_i}{\left( {{x_i} - \bar x} \right)^2}$ 92 3 276 $92 - 100 = - 8$ 64 192 93 2 186 $93 - 100 = - 7$ 49 98 97 3 291 $97 - 100 = - 3$ 9 27 98 2 196 $98 - 100 = - 2$ 4 8 102 6 612 $102 - 100 = 2$ 4 24 104 3 312 $104 - 100 = 4$ 16 48 109 3 327 $109 - 100 = 9$ 81 243 22 2200 640

Here, ${\text{N}} = 22$, $\mathop \sum \limits_{i = 1}^7 {f_i}{x_i} = 2200$, $\mathop \sum \limits_{i = 1}^7 {f_i}{(x - \bar x)^2} = 640$

Mean, $\bar x = \frac{{\sum\limits_{i = 1}^7 {{f_i}{x_i}} }}{{{\text{\;N}}}} = \frac{1}{{22}} \times 2200 = 100$

Variance, ${\sigma ^2} = \frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{{({x_i} - \bar x)}^2}} = \frac{1}{{22}} \times 640 = 29.09$

Therefore, mean is $100$ and variance is $29.09$.

1. Find the mean and standard deviation using short-cut method.

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

Ans: Suppose mean $A = 54$ and here, $h = 1$.

Calculating the given data, we get the following table:

 ${x_i}$ ${f_i}$ ${y_i} = \frac{{{x_i} - A}}{h}$ $y_i^2$ ${f_i}{y_i}$ ${f_i}y_i^2$ 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 80 100 220 0 286

Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{{{\text{\;N}}}} \times h$

$= 64 + \frac{0}{{100}} \times 1$

$= 64 + 0$

$= 64$

Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}y_i^2} - {{\left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)}^2}} \right]$

$= \frac{{{{(1)}^2}}}{{{{(100)}^2}}}\left[ {100(286) - {{(0)}^2}} \right]$

$= 2.86$

Hence, standard deviation, $\sigma = \sqrt {2.86} = 1.69$

Therefore, mean is $64$ and standard deviation is $1.69$.

1. Find the mean and variance for the following frequency distribution.

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

Ans: Suppose mean $A = 105$ and here, $h = 30$.

Calculating the given data, we get the following table:

 Class Frequency ${f_i}$ Mid-point ${x_i}$ ${y_i} = \frac{{{x_i} - A}}{h}$ $y_i^2$ ${f_i}{y_i}$ ${f_i}y_i^2$ 0-30 2 15 -3 9 -6 18 30-60 3 45 - 2 4 - 6 12 60-90 5 75 - 1 1 - 5 5 90-120 10 105 0 0 0 0 120-150 3 135 1 1 3 3 150-180 5 165 2 4 10 20 180-210 2 195 3 9 6 18 30 2 76

Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{{{\text{\;N}}}} \times h$

$= 105 + \frac{2}{{30}} \times 30$

$= 105 + 2$

$= 107$

Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}y_i^2} - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} \right]$

$= \frac{{{{(30)}^2}}}{{{{(30)}^2}}}\left[ {30(76) - {{(2)}^2}} \right]$

$= 2280 - 4$

$= 2276$

Therefore, mean is $107$ and variance is $2276$.

1. Find the mean and variance for the following frequency distribution.

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

Ans: Suppose mean $A = 25$ and here, $h = 10$.

Calculating the given data, we get the following table:

 Class Frequency ${f_i}$ Mid-point ${x_i}$ ${y_i} = \frac{{{x_i} - A}}{h}$ $y_i^2$ ${f_i}{y_i}$ ${f_i}y_i^2$ 0-10 5 5 -2 4 -10 20 10-20 8 15 -1 1 -8 8 20-30 15 25 0 0 0 0 30-40 16 35 1 1 16 16 40-50 6 45 2 4 12 24 50 10 68

Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{{{\text{\;N}}}} \times h$

$= 25 + \frac{{10}}{{50}} \times 10$

$= 25 + 2$

$= 27$

Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}y_i^2} - {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} \right]$

$= \frac{{{{(10)}^2}}}{{{{(50)}^2}}}\left[ {50(68) - {{(10)}^2}} \right]$

$= \frac{1}{{25}}\left[ {50 \times 68 - {{(10)}^2}} \right]$

$= \frac{{3300}}{{25}}$

$= 132$

Therefore, mean is $27$ and variance is $132$.

1. Find the mean, variance and standard deviation using short-cut method.

 Height in cm 70-75 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115 No. of children 2 1 12 29 25 12 10 4 5

Ans: Suppose mean $A = 92.5$ and here, $h = 5$.

Calculating the given data, we get the following table:

 Class Frequency ${f_i}$ Mid-point ${x_i}$ ${y_i} = \frac{{{x_i} - A}}{h}$ $y_i^2$ ${f_i}{y_i}$ ${f_i}y_i^2$ 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 60 6 254

Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{{{\text{\;N}}}} \times h$

$= 92.5 + \frac{6}{{60}} \times 5$

$= 92.5 + 0.5$

$= 93$

Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}y_i^2} - {{\left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)}^2}} \right]$

$= \frac{{{{(5)}^2}}}{{{{(60)}^2}}}\left[ {60(254) - {{(6)}^2}} \right]$

$= \frac{{25}}{{3600}}\left( {15204} \right)$

$= 105.58$

Hence, standard deviation, $\sigma = \sqrt {105.58} = 10.27$

Therefore, mean is $93$, variance is $105.58$ and standard deviation is $10.27$.

1. 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, 36.5-40.5, 40.5-44.5, 44.5 – 48.5, 48.5 – 52.5 and then proceed.)

Ans: Suppose mean $A = 42.5$ and here, $h = 4$.

Calculating the given data, we get the following table:

 Class Interval Frequency ${f_i}$ Mid-point ${x_i}$ ${y_i} = \frac{{{x_i} - A}}{h}$ $y_i^2$ ${f_i}{y_i}$ ${f_i}y_i^2$ 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 50 100 100 25 199

Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{{{\text{\;N}}}} \times h$

$= 42.5 + \frac{{25}}{{100}} \times 4$

$= 42.5 + 1$

$= 43.5$

Variance, ${\sigma ^2} = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}y_i^2} - {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} \right]$

$= \frac{{{{(4)}^2}}}{{{{(100)}^2}}}\left[ {100(19) - {{(25)}^2}} \right]$

$= \frac{{16}}{{10000}}\left( {19275} \right)$

$= 30.84$

Hence, standard deviation, $\sigma = \sqrt {30.84} = 5.55$

Therefore, mean is $43.5$, variance is $30.84$ and standard deviation is $5.55$.

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

Opting for the NCERT solutions for Ex 15.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 15.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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1. What are the topics included in Chapter 15 of NCERT Class11 Maths Textbook?

Ans. NCERT Chapter 15 of Class 11 Maths includes all the important Statistics topics that are important for the final exams. It is categorized into the following exercises for better understanding of the students:

• Chapter 15 Exercise 15.1 - Introduction

• Chapter 15 Exercise 15.2 - Measures of Dispersion

• Chapter 15 Exercise 15.3 - Range

• Chapter 15 Exercise 15.4 - Mean Deviation

• Chapter 15 Exercise 15.5 - Variance and Standard Deviation

• Chapter 15 Exercise 15.6 - Analysis of Frequency Distributions

2. Can I download the NCERT Solutions for Class 11 Maths from Vedantu site?

Ans: Yes, you can download the NCERT Solutions for Class 11 Maths from Vedantu's website. We have also provided the Class 11 Maths Chapter 15 solutions on this page in PDF format. Vedantu is an online learning platform which provides free PDF downloads of solved NCERT Solutions along with previous year papers for various state boards and CBSE exams. These solutions are solved by expert teachers on Vedantu.com.

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Ans: The NCERT Solutions offered by Vedantu for Class 11 Maths are solved by our best expert teachers who have years of valuable experience in the similar field. They provide detailed chapter-wise solutions and step-by-step instructions for easy learning. Solving these NCERT solutions regularly will also help the candidates to have a better knowledge about the topics and thus help them in boosting their confidence. For more information, follow Vedantu.

4. Which books shall I refer along with NCERT Class 11 Maths Textbook?

Ans. Although NCERT textbooks provide detailed information and solutions for every chapters, it is advisable to the students to refer to R.S. Agarwal and R.D. Sharma textbooks for better understanding. There are proper chapter-wise solutions provided in the textbook which are categorised as very short, short, long, multiple-type and high order thinking type questions. Therefore, CBSE students can refer to both R.S. Agarwal and R. D. Sharma books for gaining better concepts regarding any topic. The students can also download free PDFs of solved R.S.Agarwal's questions from the Vedantu’s website. These solutions are solved by our expert teachers who are having years of experience in this subject.