# 9. 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 students  3 4 7 7 15 9 6 6 3

Let the assumed mean, A = 92.5 and h = 5

 Height in cms Frequency $f_i$ Midpoint $x_i$ $\dpi{100} y_i = \frac{x_i-A}{h}$ $y_i^2$ $f_iy_i$ $f_iy_i^2$ 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 $\sum{f_i}$ =N = 60 $\sum f_iy_i$ = 6 $\sum f_iy_i ^2$ =254

Mean,

$\overline{y} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =92.5 + \frac{6}{60}\times5 = 93$

We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

$\\ \implies \sigma^2 = \frac{1}{(60)^2}\left [60(254) - (6)^2 \right ] \\ = \frac{1}{(60)^2}\left [15240 - 36 \right ] \\ = \frac{15204}{144} = 105.583$

We know,  Standard Deviation = $\sigma = \sqrt{Variance}$

$\therefore \sigma = \sqrt{105.583} = 10.275$

Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275

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