Question

The following frequency distribution gives the monthly consumption of an electricity of $68$ consumers in a locality. Find the median, mean, and mode of the data and compare them.

Monthly consumption(in units)	No. of customers
$65-85$	$4$
$85-105$	$5$
$105-125$	$13$
$125-145$	$20$
$145-165$	$14$
$165-185$	$8$
$185-205$	$4$

Answer 1

The cumulative frequency can be calculated as follows

Class Interval	Frequency	Cumulative frequency
$65-85$	$4$	$4$
$85-105$	$5$	$9$
$105-125$	$13$	$22$
$125-145$	$20$	$42$
$145-165$	$14$	$56$
$165-185$	$8$	$64$
$185-205$	$4$	$68$
	$n=68$

Step 1: Calculate the median of the given data.

From the table, we get to know that $n=68$ , and hence $\frac{n}{2}=34$

The median class is $125-145$ with cumulative frequency $=42$ Where, $l=125,n=68,C_f=22,f=20,h=20$

Formula of median

Median $=l+\frac{{\displaystyle \frac{n}{2}}-c_f}{f}\times h$

$$\begin{gathered} =25+\left(\frac{(34-22)}{20}\right)\times 20 \\ =125+12 \\ =137 \end{gathered}$$

Therefore, the median is $137.$

Step 2: Determining the mode of the given data.

Modal class $=125-145,f_1=20,f_0=13,f_2=14,\&h=20$

Mode formula:

$Mode=l+\left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]\times h$

$$\begin{gathered} =125+\left(\frac{(20-13)}{(40-13-14)}\right)\times 20 \\ =125+\frac{140}{13} \\ =125+10.77 \\ =135.77 \end{gathered}$$

Therefore, the mode is $135.77$

Step 3: Determination of the mean of the given data.

Class Interval	$f_i$	$x_i$	$d_i=x_i-a$	$u_i=\frac{d_i}{h}$	$f_i{u}_i$
$65-85$	$4$	$75$	$-60$	$-3$	$-12$
$85-105$	$5$	$95$	$-40$	$-2$	$-10$
$105-125$	$13$	$115$	$-20$	$-1$	$-13$
$125-145$	$20$	$135$	$0$	$0$	$0$
$145-165$	$14$	$155$	$20$	$1$	$14$
$165-185$	$8$	$175$	$40$	$2$	$16$
$185-205$	$4$	$195$	$60$	$3$	$12$
Total	$\sum _{}{f}_i=68$				$\sum _{}{f}_i{u}_i=7$

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

$$\begin{gathered} =135+\left(\frac{7}{68}\right)\times 20 \\ =135+2.05 \\ =137.05 \end{gathered}$$

Therefore, the mean of the given data is $137.05$

Hence, in this case, mean, median, and mode are more/less equal in this distribution.