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 $3$ $4$ $7$ $7$ $15$ $9$ $6$ $6$ $3$

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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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