Question

Find arithmetic mean of the following frequency distribution by step deviation method.
$\begin{array}{ccccccc}\text{Variate}& 5& 10& 15& 20& 25& 30\\ \text{frequency}& 20& 43& 75& 67& 72& 45\end{array}$

• A. 19.09
• B. 20.09
• C. 12.09
• D. 30.09

Answer 1

The correct option is A 19.09

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 where 'h' is the width of the class interval.

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

Let the assumed mean be a = 20 and h = 5

$\begin{array}{ccccc}\text{Variate}& \text{frequency}& \text{Deviation}& u_i=\frac{(x_i-20)}{5}& f_i{u}_i\\ \left(x_i\right)& \left(f_i\right)& d_i=x_i-20& & \\ 5& 20& -15& -3& -60\\ 10& 43& -10& -2& -86\\ 15& 75& -5& -1& -75\\ 20& 67& 0& 0& 0\\ 25& 72& 5& 1& 72\\ 30& 45& 10& 2& 90\\ & N=\sum f_i=322& & & \sum f_i{u}_i=-59\end{array}$

Here,
$N=322,a=20,h=5$ and $\sum f_i{u}_i=-59$

$\therefore Mean=a+\frac{\sum f_i{u}_i}{\sum f}\times h$
$\Rightarrow Mean=20+\left(\frac{-59}{322}\right)5$
$=20-0.91$
$=19.09$