Question

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality of a locality. Find the median, mean 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 \[{f_i}{x_i}\] and \[\sum {{f_i}} \] is the total sum of \[{f_i}\], formula to calculate median is \[Median = l + \dfrac{{\dfrac{n}{2} - c \cdot f}}{f} \times h\] and formula to calculate mode is \[Mode = l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h\].

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 difference in the confidence intervals divided by 2.
We will now form a table to find the value of the product \[{f_i}{x_i}\] for 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		68	9320

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

We know that the formula to calculate mean is \[\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\] , where \[\sum {{f_i}{x_i}} \] is the total sum of \[{f_i}{x_i}\] and \[\sum {{f_i}} \] is the total sum of \[{f_i}\].
Substituting the values of \[\sum {{f_i}{x_i}} \] and \[\sum {{f_i}} \] to find the value of mean in the above formula, we get
\[
   \Rightarrow {\text{Mean}} = \dfrac{{9320}}{{68}} \\
   \Rightarrow {\text{Mean}} = 137.05 \\
 \]
Hence, the mean is \[137.05\].
We will now find the value of \[c \cdot f\] from the above table for median and mode.
We know from the above table that the value of \[n\] is 68.

C.I.	\[{x_i}\]	\[{f_i}\]	\[{f_i}{x_i}\]	\[c \cdot f\]
65-85	75	4	300	4
85-105	95	5	475	\[4 + 5 = 9\]
105-125	115	13	1495	\[9 + 13 = 22\]
125-145	135	20	2700	\[22 + 20 = 42\]
145-165	155	14	2170	\[42 + 14 = 56\]
165-185	175	8	1400	\[56 + 8 = 64\]
185-205	195	4	780	\[64 + 4 = 68\]
Total		68	9320

Since the value of \[n\] is even, then we will use the formula of \[\dfrac{n}{2}\], we get
\[\dfrac{{68}}{2} = 34\]
Therefore, the median class is \[125 - 145\].
We know that the formula to calculate median is \[Median = l + \dfrac{{\dfrac{n}{2} - c \cdot f}}{f} \times h\].
Then in the above median class, we have \[l = 125\], \[c \cdot f = 22\], \[f = 20\] and \[h = 20\].
Substituting these values in the above formula of median, we get
\[
   \Rightarrow Median = 125 + \dfrac{{34 - 22}}{{20}} \times 20 \\
   \Rightarrow Median = 125 + 12 \\
   \Rightarrow Median = 137 \\
 \]
Now the modal class is the class where \[{f_i}\] is the highest, thus the modal class from the above table is \[125 - 145\].
We know that the formula to calculate mode is \[Mode = l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h\].
Then in the above median class, we have \[l = 125\], \[{f_0} = 13\], \[{f_1} = 20\], \[{f_2} = 14\]and \[h = 20\].
Substituting these values in the above formula of mode, we get
\[
   \Rightarrow Mode = 125 + \dfrac{{20 - 13}}{{2\left( {20} \right) - 13 - 14}} \times 20 \\
   \Rightarrow Mode = 125 + \dfrac{7}{{13}} \times 20 \\
   \Rightarrow Mode = 125 + 10.77 \\
   \Rightarrow Mode = 135.77 \\
 \]
We have seen that the values of mean and median are really similar, but mode is less than both of them.


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 really carefully, else the answer can be wrong. The other possibility of mistake in this problem is while calculating as it has a lot of mathematical calculations.