Question

Find the mean, using assumed-mean method:

$\begin{array}{cccccc}\text{Class}& 100-120& 120-140& 140-160& 160-180& 180-200\\ \text{Frequency}& 10& 20& 30& 15& 5\end{array}$

Answer 1

Lets form a frequency distribution table in assumed-mean method for finding the Mean.

$\begin{array}{ccccc}\text{Classr}& \text{Frequency}\left(f_i\right)& \text{Mid-values}\left(x_i\right)& \text{Deviation}\left(d_i\right)& f_i{d}_i\\ 100-120& 10& 110& -40& -400\\ 120-140& 20& 130& -20& -400\\ 140-160& 30& 150\left(A\right)& 0& 0\\ 160-180& 15& 170& 20& 300\\ 180-200& 5& 190& 40& 200\\ & \sum f_i=80& & & \sum f_i{d}_i=-300\end{array}$

From the table,

Assumed mean $\left(A\right)=150,\sum f_i{d}_i=-300$ and $\sum f_i=80$

Mean in deviation method,

$\overline{x}=A+\frac{\sum f_i{d}_i}{\sum f_i}$

$\Rightarrow \overline{x}=150+\left(\frac{-300}{80}\right)$

$\Rightarrow \overline{x}=150-\left(\frac{15}{4}\right)$

$\Rightarrow \overline{x}=150-3.75$

$\Rightarrow \overline{x}=146.25$

Therefore, Mean in assumed-mean method is $146.25$.