Question

The daily expenditure of $100$ families are given below. Calculate $f_1$and $f_2$ if the mean daily expenditure is $Rs188$.

Expenditure in Rs.	$140-160$	$160-180$	$180-200$	$200-220$	$220-240$
No. of families	$5$	$25$	$f_1$	$f_2$	$5$

Answer 1

Step 1. Make a frequency distribution table for the given data.

Expenditure in Rs.	No. of families $f_i$	Class mark $x_i=\frac{Upperlimit-lowerlimit}{2}$	$f_i{x}_i$
$140-160$	$5$	$150$	$750$
$160-180$	$25$	$170$	$4250$
$180-200$	$f_1$	$190$	$190f_1$
$200-220$	$f_2$	$210$	$210f_2$
$220-240$	$5$	$230$	$1150$
	$\sum f_i=35+f_1+f_2$		$\sum f_i{x}_i=6150+190f_1+210f_2$

Step 2: Find the relation between $f_1$and $f_2$

Sum of frequencies $\sum _{}^{}{f}_i=100$

$$\begin{gathered} \Rightarrow 35+f_1+f_2=100 \\ \Rightarrow f_1+f_2=65 \\ \Rightarrow f_2=65-f_1.................................\left(i\right) \end{gathered}$$

Step 3: Find the value of $f_1$

Also mean of daily expenditure $=Rs188$.

since Mean $=\frac{{\displaystyle \sum _{}^{}}{f}_i{x}_i}{{\displaystyle \sum _{}^{}}{f}_i}$

Therefore we have ,

$$\begin{gathered} \Rightarrow 188=\frac{6150+190f_1+210f_2}{100} \\ \Rightarrow 188x100=6150+190f_1+210f_2 \\ \Rightarrow 18800=6150+190f_1+210f_2 \\ \Rightarrow 12650=190f_1+210f_2 \\ \Rightarrow 12650=190f_1+210(65–f_1)(from(i\left)\right) \\ \Rightarrow 20f_1=1000 \\ \Rightarrow f_1=\frac{1000}{20} \\ \Rightarrow f_1=50 \end{gathered}$$

Step 4: Find the value of $f_2$

Substitute $f_1$ in $\left(i\right)$ to find $f_2$

$f_2=65–50=15$

Hence $f_1=50and{f}_2=15$