CBSE Class 11 Maths Chapter 13 Statistics – NCERT Solutions 2025–26

Exercise – 13.2

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:

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.

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

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

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

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

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

3. 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:

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 $74.25$.

4. Find the mean and variance for the data:

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

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 $43.4$.

5. Find the mean and variance for the data:

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

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 $29.09$.

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

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

Calculating the given data, we get the following table:

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 $1.69$.

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

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

Calculating the given data, we get the following table:

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 $2276$.

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

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

Calculating the given data, we get the following table:

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 $132$.

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

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

Calculating the given data, we get the following table:

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 $10.27$.

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

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:

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 13 Statistics Exercise 13.2

Opting for the NCERT solutions for Ex 13.2 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 13.2 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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