Dispersion (Variance and Standard Deviation) - Practice Questions & MCQ

Variance and Standard Deviation

The mean of the squares of the deviations from the mean is called the variance and is denoted by $\sigma^2$ (read as sigma square).
Variance is a quantity which leads to a proper measure of dispersion.
The variance of $n$ observations $x_1, x_2, \ldots, x_n$ is given by

$
\sigma^2=\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2
$
Standard Deviation
The standard deviation is a number that measures how far data values are from their mean.
The positive square-root of the variance is called standard deviation. The standard deviation is usually denoted by $\sigma$ and it is given by

$
\sigma=\sqrt{\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2}
$
Variance and Standard Deviation of a Ungrouped Frequency Distribution
The given data is

$
\begin{array}{rlllll}
x & : & x_1, & x_2, & x_3, & \ldots    x_n \\
f & : & f_1, & f_2, & f_3, & \ldots f_n \end{array}
$

In this case, Variance $\left(\sigma^2\right)=\frac{1}{N} \sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2$ and, Standard Deviation $(\sigma)=\sqrt{\frac{1}{N} \sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2}$ where, $\mathrm{N}=\sum_{i=1}^n f_i$

Variance and Standard deviation of a grouped frequency distribution
The formula for variance and standard deviation are the same as in the case of ungrouped frequency distribution. Here, $x_i$ is the mid point of each class.

Another formula for Standard Deviation

Variance:

$\begin{aligned}\left(\sigma^2\right) & =\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left(x_i^2+\bar{x}^2-2 \bar{x} x_i\right) \\ & =\frac{1}{N}\left[\sum_{i=1}^n f_i x_i^2+\sum_{i=1}^n \bar{x}^2 f_i-\sum_{i=1}^n 2 \bar{x} f_i x_i\right] \\ & =\frac{1}{N}\left[\sum_{i=1}^n f_i x_i^2+\bar{x}^2 \sum_{i=1}^n f_i-2 \bar{x} \sum_{i=1}^n x_i f_i\right] \\ & =\frac{1}{N}\left[\sum_{i=1}^n f_i x_i^2+\bar{x}^2 N-2 \bar{x} \cdot N \bar{x}\right] \\ & =\frac{1}{N}\left[\sum_{i=1}^n f_i x_i^2+\bar{x}^2 N-2 \bar{x} \cdot N \bar{x}\right] \\ {\left[\because \frac{1}{N} \sum_{i=1}^n x_i f_i\right.} & \left.=\bar{x} \text { or } \sum_{i=1}^n x_i f_i=\mathrm{N} \bar{x}\right] \\ & =\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i x_i^2+\bar{x}^2-2 \bar{x}^2=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i x_i^2-\bar{x}^2\end{aligned}$