Question

The table below shows the daily expenditure on food of 30 households in a locality:
$\begin{array}{cc}\text{Daily expenditure (in Rs.)}& \text{Number of households}\\ 100-150& 6\\ 150-200& 7\\ 200-250& 12\\ 250-300& 3\\ 300-350& 2\end{array}$
Find:

(a) mean of daily expenditure on food.

(b) median of daily expenditure on food.

Answer 1

Let assumed mean A = 225 and h = 50

$\begin{array}{cccccc}\text{Daily expenditure (in Rs.)}& \text{Frequency}& \text{Mid- value(}{x}_i=\frac{l+h}{2}\text{)}& u_i=\frac{x_i-A}{h}& f_i{u}_i& cf\\ 100-150& 6& 125& -2& -12& 6\\ 150-200& 7& 175& -1& -7& 13\\ 200-250& 12& 225& 0& 0& 25\\ 250-300& 3& 275& 1& 3& 28\\ 300-350& 2& 325& 2& 4& 30\\ & & & & \sum \left(f_i{u}_i\right)=-12& \end{array}$

= x¯=A+h(∑fiui/N)

(i) Mean = A + h $\frac{\sum f_i{u}_i}{N}$ = 225 + 50$\times$ $\frac{-12}{30}$

= 225 - 20 = 205

(ii) N = 30,$\frac{N}{2}$ = 15

Cumulative frequency just after 15 is 25

Corresponding frequency interval is 200 - 250

Median class is 200 - 250

Cumulative frequency c just before this class = 13

So I = 200, f = 12, $\frac{N}{2}$ = 15, h = 50, c = 13

Therefore,
Median = l + h$\frac{\frac{N}{2}-c}{f}$
= 200 + 50$\times$$\frac{15-13}{12}$
= 200 + 50 $\times$ $\frac{1}{6}$
= 200 + 8.333
=208.33

Hence, Mean = 205 and Median = 208.33