# 6     Find the mean and standard deviation using short-cut method. $\small x_i$ 60 61 62 63 64 65 66 67 68 $\small f_i$ 2 1 12 29 25 12 10 4 5

Let the assumed mean, A = 64 and h = 1

 $x_i$ $f_i$ $\dpi{100} y_i = \frac{x_i-A}{h}$ $y_i^2$ $f_iy_i$ $f_iy_i^2$ 60 2 -4 16 -8 32 61 1 -3 9 -3 9 62 12 -2 4 -24 48 63 29 -1 1 -29 29 64 25 0 0 0 0 65 12 1 1 12 12 66 10 2 4 20 40 67 4 3 9 12 36 68 5 4 16 20 80 $\sum{f_i}$ =100 $\sum f_iy_i$ = 0 $\sum f_iy_i ^2$ =286

$N = \sum_{i=1}^{9}{f_i} = 100 ; \sum_{i=1}^{9}{f_iy_i} = 0$

Mean,

$\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =64 + \frac{0}{100} = 64$

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

$\\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(286) - (0)^2 \right ] \\ = \frac{28600}{10000} = 2.86$

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

$\therefore \sigma = \sqrt{2.86} = 1.691$

Hence, Mean = 64 and Standard Deviation = 1.691

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