Question

Find the standard deviation for the following distribution:

x :	4.5	14.5	24.5	34.5	44.5	54.5	64.5
f :	1	5	12	22	17	9	4

Answer 1

x:	4.5	14.5	24.5	34.5	44.5	54.5	64.5
f:	1	5	12	22	17	9	4

Median value of  x  is 34.5.

\[x_i\]	\[f_i\]	\[d_i = x_i - 34 . 5\]	\[u_i = \frac{x_i - 34 . 5}{10}\]	\[f_i u_{i_{}}\]	\[{u_i}^2\]	\[f_i {u_i}^2\]
4.5	1	- 30	- 3	- 3	9	9
14.5	5	- 20	- 2	- 10	4	20
24.5	12	- 10	- 1	- 12	1	12
34.5	22	0	0	0	0	0
44.5	17	10	1	17	1	17
54.5	9	20	2	18	4	36
64.5	4	30	3	12	9	36
	\[N = \sum f_i = 70\]			\[\sum f_i u_i = 22\]		\[\sum f_i u_i^2 = 130\]

\[Var(X) = h^2 \left[ \frac{1}{N} \sum^n_{i = 1} f_i {u_i}^2 - \left( \frac{1}{N} \sum^n_{i = 1} f_i u_i \right)^2 \right]\]

We have

\[N = 70, \sum^n_{i = 1} f_i u_i = 22, \sum^n_{i = 1} f_i {u_i}^2 = 130, h = 10\]

Plugging all the values in the formula of variance:

 

\[\text{ Var} (X) = {10}^2 \left[ \frac{1}{70} \times 130 - \left( \frac{1}{70} \times 22 \right)^2 \right]\]
\[ = 100\left[ \frac{130}{70} - \left( \frac{22}{70} \right)^2 \right]\]
\[ = 100\left[ \frac{13}{7} - \frac{121}{1225} \right]\]
\[ = 100\left[ 1 . 857 - 0 . 0987 \right]\]
\[ = 100\left[ 1 . 7583 \right] \]
\[ = 175 . 83\]

Standard deviation,

\[SD = \sqrt{\text{ Var } (X)}\]

\[SD = \sqrt{Var(X)}\]
\[ = \sqrt{175 . 83}\]
\[ = 13 . 26\]