Question

Calculate the mean and median for the following continuous frequency distribution.

$$\begin{array}{cc}\text{Class}& f_i\\ 0-10& 6\\ 10-20& 8\\ 20-30& 20\\ 30-40& 25\\ 40-50& 15\\ 50-60& 10\\ 60-70& 4\end{array}$$

• A. 34.25, 32.4
• B. 34.20, 34
• C. 34.20, 32.1
• D. 34, 32.1

Answer 1

The correct option is B 34.20, 34

$$\begin{array}{ccccc}\text{Class}& f_i& x_i& c{f}_i& x_i{f}_i\\ 0-10& 6& 5& 6& 30\\ 10-20& 8& 15& 14& 120\\ 20-30& 20& 25& 34& 500\\ 30-40& 25& 35& 59& 875\\ 40-50& 15& 45& 74& 675\\ 50-60& 10& 55& 84& 550\\ 60-70& 4& 65& 88& 260\\ & 88& & & 3010\end{array}$$

$\mathrm{To}\mathrm{calculate}\mathrm{mean}:\mathrm{We}\mathrm{first}\mathrm{compute}{\sum }^{}{x}_i{f}_i=3010\left(\mathrm{from}\mathrm{table}\mathrm{provided}\mathrm{above}\right)\mathrm{Mean}=\frac{{\sum }^{}{x}_i{f}_i}{{\sum }^{}{f}_i}=\frac{3010}{88}=34.2\mathrm{To}\mathrm{calculate}\mathrm{median}:N=88,\mathrm{therefore},\frac{N}{2}=44\mathrm{So},\mathrm{the}\mathrm{median}\mathrm{class}\mathrm{is}30-40\left(\mathrm{as}44\mathrm{th}\mathrm{term}\mathrm{lies}\mathrm{in}30-40\mathrm{class}\right)\mathrm{Median}=L+\left(\frac{\frac{N}{2}-C}{f}\right)h=30+\left(\frac{44-34}{25}\right)10=34$