The following frequency distribution gives the monthly consumption of electricity of \[\mathbf{68}\] consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly Consumption (in units) | Number of Consumers |
\[65-85\] | \[4\] |
\[85-105\] | \[5\] |
\[105-125\] | \[13\] |
\[125-145\] | \[20\] |
\[145-165\] | \[14\] |
$165-185$ | \[8\] |
$185-205$ | \[4\] |
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Ans: The mean can be found as given below:
$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$
Suppose the assured mean $\left( a \right)$ of the data is \[32.5\].
Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:
To find the class mark for each interval, the following relation is used.
Class mark \[{{x}_{i}}=\frac{\text{Upper class limit+Lower class limit}}{2}\]
Class size (\[h\]) of this data is:
\[h=85-65\]
\[h=20\]
[\[{{d}_{i}},\]\[{{u}_{i}},\]and \[{{f}_{i}}{{u}_{i}}\] can be calculated according to step deviation method as follows:
Monthly consumption (in units) | Number of Consumers \[{{f}_{i}}\] | Class Mark \[{{x}_{i}}\] | ${{d}_{i}}={{x}_{i}}-135$ | ${{u}_{i}}=\frac{{{d}_{i}}}{20}$ | \[{{f}_{i}}{{u}_{i}}\] |
\[65-85\] | \[4\] | \[75\] | \[-60\] | \[-3\] | \[-12\] |
\[85-105\] | \[5\] | \[95\] | \[-40\] | \[-2\] | \[-10\] |
\[105-125\] | \[13\] | \[115\] | \[-20\] | \[-1~\] | \[-13\] |
\[125-145\] | \[20\] | \[135\] | \[0~\] | \[0~\] | \[0~\] |
\[145-165\] | \[14\] | \[155\] | \[20\] | \[1\] | \[14\] |
\[165-185\] | \[8\] | \[175\] | \[40\] | \[2\] | \[16\] |
\[185-205\] | \[4\] | \[195\] | \[60\] | \[3\] | \[12\] |
Total | \[68\] | \[7\] |
It can be observed from the above table
\[\sum{{{f}_{i}}=68}\]
$\sum{{{f}_{i}}{{u}_{i}}}=7$
Class size \[\left( h \right)\text{ }=\text{ }20\]
Substituting \[{{u}_{i}},\]and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean
The required mean:
\[\overline{X}=a+\left(\frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}}\right)\times h\]
\[\Rightarrow\overline{X}=135+\left(\frac{7}{68} \right)\times (20)\]
\[\Rightarrow\overline{X}=135+\frac{140}{68}\]
\[\Rightarrow\overline{X}=137.058\]
Hence, the mean of given data is $137.058$.
For Mode
Mode can be calculated as
$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$
Where
\[l=\] Lower limit of modal class
${{\text{f}}_{1}}=$ Frequency of modal class
\[{{f}_{0}}=\] Frequency of class preceding the modal class
\[{{f}_{2}}=\] Frequency of class succeeding the modal class
\[h=\]Class size
From the table, it can be noticed that the maximum class frequency is $20$ ,
Belongs to class interval \[125-145\].
Modal class = \[125-145\]
\[l=125\]
Class size \[\text{h}=20\]
\[{{f}_{1}}=20\]
\[{{f}_{0}}=13\]
\[{{f}_{2}}=14\]
Substituting these values in the formula of mode we get:
\[M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h\]
\[\Rightarrow M=125+\left( \frac{20-13}{2(20)-13-14} \right)\times 20\]
\[\Rightarrow M=125+\frac{7}{13}\times 20\]
\[\Rightarrow M=135.76\]
Hence, the value of mode is $135.76$
For Median
We can calculate the median as given below:
$m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h$
Where
\[l=\] Lower limit of median class
\[h=\] Class size
\[f=\] Frequency of median class
\[cf=\] cumulative frequency of class preceding median class
To find the median of the given data, cumulative frequency is calculated as follows.
Monthly consumption (in units) | Number of Consumers | Cumulative Frequency |
\[65-85\] | \[4\] | \[4\] |
\[85-105\] | \[5\] | \[4+5=9\] |
\[105-125\] | \[13\] | \[9+13=22\] |
\[125-145\] | \[20\] | \[22+20=42\] |
\[145-165\] | \[14\] | \[42+14=56\] |
\[165-185\] | \[8\] | \[56+8=64\] |
\[185-205\] | \[4\] | \[64+4=68\] |
It can be observed from the given table
\[n\text{ }=\text{ }68\]
$\frac{\text{n}}{2}=34$
Cumulative frequency just greater than $\frac{\text{n}}{2}$ is $42$ , belonging to
interval \[125-145\].
Therefore, median class = \[125-145\].
\[l=\text{ }125\]
\[\text{h }=\text{ }20\]
\[f=20\]
\[cf=22\]
Substituting these values in the formula of median we get:
\[m=l+\left( \frac{\frac{n}{2}-cf}{f} \right)\times h\]
\[\Rightarrow m=125+\left( \frac{34-22}{20} \right)\times 20\]
\[\Rightarrow m=125+12\]
\[\Rightarrow m=137\]
Hence, median, mode, mean of the given data is \[137,\text{ }135.76,\] and \[137.05\] respectively.
Mean, mode and median are almost equal in this case.