Question

The diameters of circles (in mm) drawn in a design are given below :

$\begin{array}{cccccc}\text{Diameters}& 33-36& 37-40& 41-44& 45-48& 49-52\\ \text{No. of circles}& 15& 17& 21& 22& 25\end{array}$

Calculate the standard deviation and mean diameter of the circles.

Answer 1

$\begin{array}{ccccccc}\text{Diameters}& \text{Classes}& \text{Mid values}& & & & \\ & & x_i& f_i& u=\frac{x-42.5}{4}& fu& f{u}^2\\ 33-36& 32.5-36.5& 34.5& 15& -2& -30& 60\\ 37-40& 36.5-40.5& 38.5& 17& -1& -17& 17\\ 41-44& 40.5-44.5& 42.5& 21& 0& 0& 0\\ 45-48& 44.5-52.5& 46.5& 22& 1& 22& 22\\ 49-52& 48.5-52.5& 50.5& 25& 2& 50& 100\\ & & & 100& & 25& 199\end{array}$

Mean $\left(\overline{x}\right)=A+\frac{\sum fu}{N}\times h=42.5+\frac{25}{100}\times 4=42.5+1=43.5mm$

Standard deviation $\left(\sigma \right)=\sqrt{N\sum f{u}^2-(\sum fu{)}^2}$

= $\frac{4}{100}\sqrt{100\times 199-(25{)}^2}=\frac{1}{25}\sqrt{19275}=\frac{1}{25}\times 138.83=5.55$