Question

Given below are the diameters of circles (in mm) drawn in a design.

$\begin{array}{cccccc}\text{Diameter}& 33-36& 37-40& 41-41& 45-48& 49-52\\ \text{Number of circles}& 15& 17& 21& 22& 25\end{array}$

Calculate the mean diameter of the circles, variance and standard deviation.

Answer 1

Converting the given series into an exclusive series, we prepare the table given below.

$\begin{array}{ccccccc}\text{Class}& \text{Frequency}& \text{Midpoint}& y_i=\frac{(x_i-42.5)}{4}& y_i^2& f_i{y}_i& f_i{y}_i^2\\ 32.5-36.5& 15& 34.5& -2& 4& -30& 60\\ 36.5-40.5& 17& 38.5& -1& 1& -17& 17\\ 40.5-44.5& 21& 42.5=A& 0& 0& 0& 0\\ 44.5-48.5& 22& 46.5& 1& 1& 22& 22\\ 48.5-52.5& 25& 50.5& 2& 4& 50& 100\\ & N=100& & & & 25& 199\end{array}$

$\therefore A=42.5,h=4,N=\sum f_i=100,\sum f_i{y}_i=25$ and $\sum f_1{y}_i^2=199$

$\therefore \overline{x}=(A+\frac{\sum f_i{y}_i}{N}\times h)\Rightarrow \overline{x}=(42.5+\frac{25}{100}\times 4)=43.5$

Thus, mean = 43.5 mm.

Variance, ${\sigma }^2=\frac{h^2}{N^2}.\{N.\sum f_i{y}_i^2-(\sum f_i{y}_i{)}^2\}$

$=\frac{16}{10000}\times \{100\times 199-(25{)}^2\}=\frac{16}{10000}\times (19900-625)$

$=(\frac{16}{10000}\times 19275)=\frac{3084}{100}=30.84$

Standard deviation, $\sigma =\sqrt{30.84}=5.55$

Hence, mean = 43.5 mm, variance = 30.84, and standard deviation = 5.55.