Question

Calculate the mean of the following distribution using step - deviation method.
$\begin{array}{cc}\text{Class Interval}& \text{Frequency}\\ 50-55& 5\\ 55-60& 20\\ 60-65& 10\\ 65-70& 10\\ 70-75& 9\end{array}$

• A. 62.3

Answer 1

The correct option is A 62.3

1. Choose an arbitrary constant 'a'. (also called assumed mean)

2. $x_i$ = $\frac{Upperlimit+Lowerlimit}{2}$

3. Subtract the value of 'a' from $x_i$.

4. The reduced value of $(x_i-a)$ is called the deviation of $x_i$ from 'a'.

5. Divide the deviation by constant 'h' where h is the width of the class interval.

6. $u_i=\frac{x_i-a}{h}$

$\begin{array}{ccccc}\text{Class Interval}& \text{Frequency}\left(f_i\right)& x_i& u_i=\frac{x_i-a}{h}& f_i{u}_i\\ 50-55& 5& 52.5& -2& -10\\ 55-60& 20& 57.5& -1& -20\\ 60-65& 10& 62.5=a& 0& 0\\ 65-70& 10& 67.5& 1& 10\\ 70-75& 9& 72.5& 2& 18\\ & \sum f_i=54& & & \sum f_i{u}_i=-2\end{array}$

$\therefore \text{Mean}=a+h\frac{\sum f_i{u}_i}{\sum f}$
$=62.5+5\times \frac{(-2)}{54}=62.5-\frac{10}{54}=62.3$