# 10.  The diameters of circles (in mm) drawn in a design are given below: Diameters 33-36 37-40 41-44 45-48 49-52 No. of circles 15 17 21 22 25 Calculate the standard deviation and mean diameter of the circles. [ Hint First make the data continuous by making the classes as $32.5-36.5,36.5-40.4,40.5-44.5,44.5-48.5,48.5-52.5$  and then proceed.]

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

 Diameters No. of circles $f_i$ Midpoint $x_i$ $\dpi{100} y_i = \frac{x_i-A}{h}$ $y_i^2$ $f_iy_i$ $f_iy_i^2$ 32.5-36.5 15 34.5 -2 4 -30 60 36.5-40.5 17 38.5 -1 1 -17 17 40.5-44.5 21 42.5 0 0 0 0 44.5-48.5 22 46.5 1 1 22 22 48.5-52.5 25 50.5 2 4 50 100 $\sum{f_i}$ =N = 100 $\sum f_iy_i$ = 25 $\sum f_iy_i ^2$ =199

Mean,

$\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =42.5 + \frac{25}{100}\times4 = 43.5$

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

$\\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(199) - (25)^2 \right ]\times4^2 \\ = \frac{1}{625}\left [19900 - 625 \right ] \\ = \frac{19275}{625} = 30.84$

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

$\therefore \sigma = \sqrt{30.84} = 5.553$

Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553

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