Q.If for a distribution $\sum(x-5) = 3$, $\sum(x-5)^2 = 43$ and the total number of item is 18, find the mean and standard deviation.

Variance and Standard Deviation of Data

For a data set these measure spread about the mean. For ungrouped data

$x_1,\dots,x_n$ with mean $\bar x$,

$\sigma^2=\dfrac1n\sum_{i=1}^n (x_i-\bar x)^2=\dfrac{\sum x_i^2}{n}-\bar x^2,$

and the standard deviation is $\sigma=\sqrt{\sigma^2}$. For a frequency distribution with values $x_i$ (class midpoints for grouped data) and frequencies

$f_i$, $N=\sum f_i$,

$\sigma^2=\dfrac{\sum f_i x_i^2}{N}-\Big(\dfrac{\sum f_i x_i}{N}\Big)^2.$

The coefficient of variation $\mathrm{CV}=\dfrac{\sigma}{\bar x}\times100$ compares

relative spread. Variance is unchanged by a shift of origin and scales as

$a^2$ under $x\mapsto ax+b$. For two groups combined, the pooled variance uses the

group means $\bar x_1,\bar x_2$ and the overall mean $\bar x$:

$\sigma^2=\dfrac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}$ with …