Question

Find the mean variance and standard deviation using short-cut method

Height in cms	No. of children
70-75	3
75-80	4
80-85	7
85-90	7
90-95	15
95-100	9
100-105	6
105-110	6
110-115	3

Answer 1

Class interval	$f_i$	$x_i$	$u_i=\frac{x_i-A}{h}$	$u_i^2$	$f_i{u}_i$	$f_i{u}_i^2$
70-75	3	72.5	-4	16	-12	48
75-80	4	77.5	-3	9	-12	36
80-85	7	82.5	-2	4	-14	28
85-90	7	87.5	-1	1	-7	7
90-95	15	92.5	0	0	0	0
95-100	9	97.5	1	1	9	9
100-105	6	102.5	2	4	12	24
105-110	6	107.5	3	9	18	54
110-115	3	112.5	4	16	12	48
Total	60				6	254

We have $N=\sum f_i=60$

Width of class $h=5$

Assumed mean $A=92.5\left(say\right)$

Mean, $\overline{x}=A+\frac{{\sum }_{i=1}^9{f}_i{u}_i}{N}\times h$

$=92.5+\frac{6}{60}\times 5=92.5+0.5=93$

$\Rightarrow \overline{x}=93$

Variance ${\sigma }^2=h^2[\frac{{\sum }_{i=1}^9{f}_i{u}_i^2}{N}-{\left(\frac{{\sum }_{i=1}^9{f}_i{u}_i}{N}\right)}^2]$
${\sigma }^2=5^2[\frac{254}{60}-{\left(\frac{6}{60}\right)}^2]$

$=5^2\frac{[60\times 254-6^2]}{{\left(60\right)}^2}$

$=\frac{25}{3600}[15240-36]$

$=\frac{25}{3600}\times 15204$

${\sigma }^2=105.58$

Standard Deviation $=\sqrt{\text{Variance}}$

$\sigma =\sqrt{105.58}$

$\sigma =10.27$