Chapter 15: Statistics Exercise 15.2

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.

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

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 $74.25$.

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 $43.4$.

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 $29.09$.

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 $1.69$.

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 $2276$.

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 $132$.

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 $10.27$.

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 $5.55$.