Chapter 15.4 - Solved Examples on Standard Deviation

In the previous section, we saw variance and standard deviation. In this section, we will see some solved examples.

Solved example 15.9
Find the variance and standard deviation of the following data:
6,8,10,12,14,16,18,20,22,24
Solution:
1. We are asked to find the variance.
• So our first aim is to find the mean $(\bar{x})$. For that, we can use the columns I, II  and III of table 15.19 below:

Table 15.19

• Based on those columns, we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{150}{10}~=~15$
2. Now we can calculate the variance. For that, we can use columns II, IV and V of the table.
• We have:
$\sigma^2~=~\frac{\sum{(x_i - \bar{x})^2}}{\sum{f_i}}~=~\frac{330}{10}~=~33$
3. Finally, we can calculate the standard deviation.
$\sigma~=~\sqrt{\sigma^2}~=~\sqrt{33}~=~5.74$

Solved example 15.10
Find the variance and standard deviation of the following data:

Table 15.20

Solution:
1. We are asked to find the variance.
• So our first aim is to find the mean $(\bar{x})$. For that, we can use the columns I, II  and III of table 15.21 below:

Table 15.21

• Based on those columns, we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{420}{30}~=~14$
2. Now we can calculate the variance. For that, we can use columns II, IV and V of the table.
• We have:
$\sigma^2~=~\frac{\sum{f_i (x_i - \bar{x})^2}}{\sum{f_i}}~=~\frac{1374}{30}~=~45.8$
3. Finally, we can calculate the standard deviation.
$\sigma~=~\sqrt{\sigma^2}~=~\sqrt{45.8}~=~6.77$
4. Note:
In this problem, we have to take frequencies also into account. This can be explained using an example. It can be written in 3 steps:
(i) Frequency of the observation 11 is nine.
• That means, the observation 11 occurs nine times.
(ii) So we have to include "$(11 - \bar{x})^2$" nine times.
• Instead of writing nine times, we can directly calculate "$9 \times (11 - \bar{x})^2$"
(iii) So we calculate "$f_i \times (x_i - \bar{x})^2$" in the column V.

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Another formula for standard deviation

This can be written in 2 steps:
1. We can derive a new formula for variance as follows:

$\begin{array}{ll}
{}&{\sigma^2}
& {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i \left(x_i - \bar{x} \right)^2} \right]} &{} \\

{}&{}
& {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i \left(x^2 _i + \bar{x}^2 - 2 x_i \bar{x} \right)} \right]} &{} \\

{}&{}
& {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i x^2 _i} + \sum{f_i \bar{x}^2} - \sum{2 f_i x_i \bar{x} } \right]} &{} \\

{}&{}
& {~=~}& {\frac{1}{\sum{f_i}}\left[\sum{f_i x^2 _i} + \bar{x}^2 \sum{f_i } - 2 \bar{x} \sum{f_i x_i } \right]~\color{green}{\text{- - - I}}} &{} \\

{}&{}
& {~=~}& {\frac{1}{\sum{f_i}}\left(\sum{f_i x^2 _i} \right) + \frac{1}{\sum{f_i}} \left(\bar{x}^2 \sum{f_i } \right) - \frac{1}{\sum{f_i}} \left(2 \bar{x} \sum{f_i x_i } \right)} &{} \\

{}&{}
& {~=~}& {\frac{1}{\sum{f_i}}\left(\sum{f_i x^2 _i} \right) + \left(\bar{x}^2 \right) - \left(2 \bar{x}^2 \right)~\color{green}{\text{- - - II}}} &{} \\

{}&{}
& {~=~}& {\frac{1}{\sum{f_i}}\left(\sum{f_i x^2 _i} \right) - \left(\bar{x}^2 \right)} &{} \\

{}&{}
& {~=~}& {\frac{1}{\sum{f_i}}\left(\sum{f_i x^2 _i} \right) - \left[\frac{1}{\sum{f_i}}\left(\sum{f_i x_i} \right) \right]^2 ~\color{green}{\text{- - - III}}} &{} \\

{}&{}
& {~=~}& {\frac{1}{\sum{f_i}}\left(\sum{f_i x^2 _i} \right) - \left[\left(\frac{1}{\sum{f_i}}\right)^2\left(\sum{f_i x_i} \right)^2 \right]} &{} \\

{}&{}
& {~=~}& {\left(\frac{1}{\sum{f_i}}\right)^2 \left[\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right) - \left(\sum{f_i x_i} \right)^2 \right]} &{} \\

\end{array}$

◼ Remarks:
(i) Line marked as I
• $\bar{x}$ is a constant. So we take it outside summations.
(ii) Line marked as II
• In the third term, $\frac{\sum{f_i x_i}}{\sum{f_i}}$ is $\bar{x}$
(iii) Line marked as III
• In this line, we write $\frac{\sum{f_i x_i}}{\sum{f_i}}$ in the place of $\bar{x}$.

2. Once we obtain the formula for variance, we can easily write the formula for standard deviation.

$\sigma~=~\sqrt{\left(\frac{1}{\sum{f_i}}\right)^2 \left[\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right) - \left(\sum{f_i x_i} \right)^2 \right]}$ 

⇒ $\sigma~=~\frac{1}{\sum{f_i}}\sqrt{\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right) - \left(\sum{f_i x_i} \right)^2}$

Solved example 15.11
Find the variance and standard deviation of the following data:

Table15.22

Solution:
1. For calculating variance, we have:
$\sigma^2~=~\left(\frac{1}{\sum{f_i}}\right)^2 \left[\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right) - \left(\sum{f_i x_i} \right)^2 \right]$
2. For calculating standard deviation, we have:
$\sigma~=~\frac{1}{\sum{f_i}}\sqrt{\left(\sum{f_i}\right) \left(\sum{f_i x^2 _i} \right) - \left(\sum{f_i x_i} \right)^2}$
3. For applying the formulas, we need the following items:
(i) $\sum{f_i}$
• This is calculated in column II of table 15.23 below:

Table 15.23

(ii) $\sum{f_i x_i}$
• This is calculated in column III of table 15.23.
(iii) $\sum{f_i (x_i)^2}$
• This is calculated in column V of table 15.23.
4. Substituting the known values in (1), we get:
$\sigma^2~=~\left(\frac{1}{48}\right)^2 \left[\left(48\right) \left(9652 \right) - \left(614 \right)^2 \right]~=~37.456$
5. So we can write:
$\sigma~=~\sqrt{37.456}~=~6.12$

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In the next section, we will see variance and standard deviation for continuous frequency distribution.

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