Question

Find the standard deviation for the following data:

$\begin{array}{cccccc}{x}_i& 3& 8& 13& 18& 23\\ f_i& 6& 10& 14& 10& 10\end{array}$

Answer 1

We have

Mean, $\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{(18+80+182+180+230)}{50}=\frac{690}{50}=13.8$

This value being in decimal form, the calculation will become tedious.

So, we use the formula, $\sigma =\frac{1}{N}.\sqrt{N.\sum f_i{x}_i^2-(\sum f_i{)}^2}$

Now, we prepare the table given below:

$\begin{array}{ccccc}{x}_i& f_i& f_i{x}_i& x_i^2& f_i{x}_i^2\\ 3& 6& 18& 9& 54\\ 8& 10& 80& 64& 640\\ 13& 14& 182& 169& 2366\\ 18& 10& 180& 324& 3240\\ 23& 10& 230& 529& 5290\\ & 50& 690& & 11590\end{array}$

$\therefore N=\sum f_i=50,\sum f_i{x}_i=690$ and $\sum f_i{x}_i^2=11590$

$\therefore$ standard deviation, $\sigma =\frac{1}{N}.\left\{\sqrt{N.\sum f_i{x}_i^2-(\sum f_i{x}_i{)}^2}\right\}$

$\Rightarrow \sigma =\frac{1}{50}.\sqrt{50\times 11590-(690{)}^2}=\frac{1}{50}\times \sqrt{579500-(690{)}^2}$

$=\frac{1}{50}\sqrt{579500-476100}=\frac{1}{50}\times \sqrt{103400}=\frac{321.5}{50}=6.43$

Hence, $\sigma =6.43$