Question

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

Monthly Consumption (in units)	Number of consumers
65 – 85	4
85 – 105	5
105 – 125	13
125 – 145	20
145 – 165	14
165 – 185	8
185 – 205	4

Answer 1

Hint: First we will calculate mean using the formula $\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$, where $\sum {{f_i}{x_i}} $ is the total sum of the product of mid-value of class interval and frequency and $\sum {{f_i}} $ is the sum of frequency. After that calculate the median by the formula $l + \dfrac{{\dfrac{N}{2} - cf}}{f} \times h$. Then use the formula of mode $l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}$ to calculate the mode.

Complete step-by-step answer:
We are given that the monthly consumption is the confidence interval C.I.
Let us assume that ${f_i}$ represents the number of consumers and ${x_i}$ is the mid-value of the interval.
We will now form a table to find the value of the product ${f_i}{x_i}$ for the mean.

C.I.	${x_i}$	${f_i}$	${f_i}{x_i}$
65 – 85	75	4	300
85 – 105	95	5	475
105 – 125	115	13	1495
125 – 145	135	20	2700
145 – 165	155	14	2170
165 – 185	175	8	1400
185 – 205	195	4	780
Total		$\sum {{f_i}} = 68$	$\sum {{f_i}{x_i}} = 9320$

We know that the formula to calculate mean using the formula $\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$, where $\sum {{f_i}{x_i}} $ is the total sum of the product of mid-value of class interval and frequency and $\sum {{f_i}} $ is the sum of frequency.
Substitute the values to find the value of mean in the above formula, we get,
$ \Rightarrow $ Mean $ = \dfrac{{9320}}{{68}}$
Divide the numerator by the denominator,
$ \Rightarrow $ Mean $ = 137.05$
Hence, the mean is 137.05.
We will now find the value of $cf$ from the above table for median and mode.
We know from the above table that the value of $n$ is 68.

C.I.	${x_i}$	${f_i}$	${f_i}{x_i}$	$cf$
65 – 85	75	4	300	4
85 – 105	95	5	475	9
105 – 125	115	13	1495	22
125 – 145	135	20	2700	42
145 – 165	155	14	2170	56
165 – 185	175	8	1400	64
185 – 205	195	4	780	68

The value of $\dfrac{n}{2}$ is,
$ \Rightarrow \dfrac{{68}}{2} = 34$
So, the median class is 125 – 145.
The lowest value of the median class is,
$ \Rightarrow l = 125$
The frequency of the median class is,
$ \Rightarrow f = 20$
The difference of interval is,
$ \Rightarrow h = 22$
The cumulative frequency above the median class is,
$ \Rightarrow cf = 22$
Substitute these values in the median formula,
$ \Rightarrow $ Median $ = 125 + \dfrac{{34 - 22}}{{20}} \times 20$
Simplify the terms,
$ \Rightarrow $ Median $ = 125 + 12$
Add the terms,
$ \Rightarrow $ Median $ = 137$
Hence, the median is 137.
Now the modal class is the class where ${f_i}$ is the highest, thus the modal class from the above table is,
$ \Rightarrow 125 - 145$
Then in the mode class, we have
$ \Rightarrow l = 125$
$ \Rightarrow {f_0} = 13$
$ \Rightarrow {f_1} = 20$
$ \Rightarrow {f_2} = 14$
$h = 20$
Substitute the values in mode formula,
$ \Rightarrow $ Mode $ = 125 + \dfrac{{20 - 13}}{{2\left( {20} \right) - 13 - 14}} \times 20$
Simplify the terms,
$ \Rightarrow $ Mode $ = 125 + \dfrac{7}{{13}} \times 20$
Multiply the numerator and then divide by denominator,
$ \Rightarrow $ Mode $ = 125 + 10.77$
Add the terms,
$ \Rightarrow $ Mode $ = 135.77$
Hence, the mode is 135.77.

Note: In solving these types of questions, students should know the formulae of mean, median, and mode. The question is really simple, students should note down the values from the problem carefully, else the answer can be wrong. The other possibility of a mistake in this problem is while calculating as it has a lot of mathematical calculations.