Find the mean, variance and standard deviation using short-cut method
Height in cms
$70-75$
$75-80$
$80-85$
$85-90$
$90-95$
$95-100$
$100-105$
$105-110$
$110-115$
No. of children
$3$
$4$
$7$
$7$
$15$
$9$
$6$
$6$
$3$
Find the mean, variance and standard deviation using short-cut method
| Height in cms | $70-75$ | $75-80$ | $80-85$ | $85-90$ | $90-95$ | $95-100$ | $100-105$ | $105-110$ | $110-115$ |
| No. of children | $3$ | $4$ | $7$ | $7$ | $15$ | $9$ | $6$ | $6$ | $3$ |
| Class Interval | Frequency ${f_i}$ | Mid-point ${f_i}$ | ${y_i} = \frac{{{x_i} - 92.5}}{5}$ | ${y_i}^2$ | ${f_i}{y_i}$ | ${f_i}{y_i}^2$ |
| $70-7$ | $3$ | $72.5$ | $-4$ | $16$ | $-12$ | $48$ |
| $75-80$ | $4$ | $77.5$ | $-3$ | $9$ | $-12$ | $36$ |
| $80-85$ | $7$ | $82.5$ | $-2$ | $4$ | $-14$ | $28$ |
| $85-90$ | $7$ | $87.5$ | $-1$ | $1$ | $-7$ | $7$ |
| $90-95$ | $15$ | $92.5$ | $0$ | $0$ | $0$ | $0$ |
| $95-100$ | $9$ | $97.5$ | $1$ | $1$ | $9$ | $9$ |
| $100-105$ | $6$ | $102.5$ | $2$ | $4$ | $12$ | $24$ |
| $105-110$ | $6$ | $107.5$ | $3$ | $9$ | $18$ | $54$ |
| $110-115$ | $3$ | $112.5$ | $4$ | $16$ | $12$ | $48$ |
| $60$ | $6$ | $254$ |
Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h$
$ = 92.5 + \frac{6}{{60}} \times 5 = 92.5 + 0.5 = 93$
Variance, $\left( {{\sigma ^2}} \right) = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)}^2}} } \right]$
$=\frac{(5)^{2}}{(60)^{2}}\left[60 \times 254-(6)^{2}\right]$
$=\frac{25}{3600}(15204)=105.58$
$\therefore$ Standard deviation $(\sigma)=\sqrt{105.58}=10.27$
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