Question

Find the missing frequencies in the following distribution , if the sum of the frequencies is $120$ and the mean is $50$.

$\begin{array}{cccccc}\text{Class}& 0-20& 20-40& 40-60& 60-80& 80-100\\ \text{Frequency}& 17& f_1& 32& f_2& 19\end{array}$

Answer 1

It is given that mean of the data is $50$ and $N=120$

Let the assumed mean $A=50$ and $h=10$.

$\begin{array}{ccccc}\text{Marks}& \text{Mid values}\left(x_i\right)& \text{Frequency}& u_i=\frac{x_i-50}{A}& f_i{u}_i\\ 0-20& 10& 17& -2& -34\\ 20-40& 30& f_1& -1& -f_1\\ 40-60& 50& 32& 0& 0\\ 60-80& 70& f_2& 1& f_2\\ 80-100& 90& 19& 2& 38\\ & & N=68+f_1+f_2& & \sum f_i{u}_i=4-f_1+f_2\end{array}$

From the table, $N=68+f_1+f_2$, $\sum f_i{u}_i=4-f_1+f_2$ and $\overline{x}=50$

$68+f_1+f_2=120$

$\Rightarrow f_1+f_2=120-68$

$\Rightarrow f_1+f_2=52$---(1)

Mean in short-cut method,

$\left(\overline{x}\right)=A+\left(\frac{\sum f_i{u}_i}{N}\right)\times h$

$\Rightarrow 50=50+\left(\frac{4-f_1+f_2}{N}\right)\times h$

$\Rightarrow 50-50=\left(\frac{4-f_1+f_2}{N}\right)\times h$

$\Rightarrow 0=4-f_1+f_2$

$\Rightarrow f_1-f_2=4$---(2)

add (1) and (2)

$f_1+f_2+f_1-f_2=52+4$

$\Rightarrow 2f_1=56$

$\Rightarrow f_1=28$---(3)

From $f_1+f_2=64$

$\Rightarrow 28+f_2=52$

$\Rightarrow f_2=52-28$

$\Rightarrow f_2=24$---(4)