Question

Find the mean, median, and mode of the following data

Classes	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	5	10	18	30	20	12	5

Answer 1

Hint: We solve this problem in three parts.
(i) For finding the mean we create a table by adding the columns to given data. One column is for midpoint of class, \[{{x}_{i}}\] and other column for product of frequency and midpoint of class \[{{f}_{i}}{{x}_{i}}\] then we find mean using the formula
\[\bar{x}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]
(ii) For finding the median we create a table by adding the columns to given data. One column is for cumulative frequency which means sum of frequencies up to that class interval. Then we find the median class for which class interval \[{{\left( \dfrac{N}{2} \right)}^{th}}\] frequency lies. Then median is given by
\[\text{median}= L+\left( \dfrac{\dfrac{N}{2}-{{f}_{0}}}{f} \right)h\]
\[L\] is a lower interval of the median class
\[N\] is the sum of all frequencies
\[{{f}_{0}}\] is the cumulative frequency of preceding median class
\[f\] is the frequency of the median class
\[h\] is the height of class interval
(iii) For calculating the mode we take the given data and find the modal class where the frequency is highest. Then, we formula of mode as
\[\text{Mode}=L+ \left( \dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)h\]
\[L\] is lower interval of modal class
\[{{f}_{0}}\] is frequency of preceding modal class
\[{{f}_{1}}\] is frequency of modal class
\[{{f}_{2}}\] is frequency of succeeding modal class
\[h\] is height of class interval

Complete step-by-step solution:
(i) Let us solve for the mean.
Let us create a table by adding the columns to given data. One column is for midpoint of class, \[{{x}_{i}}\] and other column for product of frequency and midpoint of class \[{{f}_{i}}{{x}_{i}}\] as follows-

Class interval	Frequency (\[{{f}_{i}}\])	Midpoint of class (\[{{x}_{i}}\])	\[{{f}_{i}}{{x}_{i}}\]
0-10	5	5	25
10-20	10	15	150
20-30	18	25	450
30-40	30	35	1050
40-50	20	45	900
50-60	12	55	660
60-70	5	65	325

Now, we know that mean is calculated by using the formula
\[\bar{x}=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]
By substituting the required values in the formula we get
\[\begin{align}
  & \Rightarrow \bar{x}=\dfrac{25+150+450+1050+900+660+325}{5+10+18+30+20+12+5} \\
 & \Rightarrow \bar{x}=\dfrac{3560}{100} \\
 & \Rightarrow \bar{x}=35.6 \\
\end{align}\]
Therefore, the mean of given data is 35.6.
(ii) Let us solve for median
Let us create a table by adding the columns to given data. One column is for cumulative frequency which means sum of frequencies up to that class interval as follows

Class interval	Frequency (\[{{f}_{i}}\])	Cumulative frequency
0-10	5	5
10-20	10	15
20-30	18	33
30-40	30	63
40-50	20	83
50-60	12	95
60-70	5	100

We know that \[N=\sum{{{f}_{i}}}=100\].
Now, we need to find in which class \[{{\left( \dfrac{N}{2} \right)}^{th}}\] frequency lies.
We know that \[\dfrac{N}{2}=50\].
So, frequency 50 lies in the class of 30.
So, the median class is 30-40.
We know that the formula of median is
\[\text{median}=L+\left( \dfrac{\dfrac{N}{2}-{{f}_{0}}}{f} \right)h\]
\[L\] is lower interval of median class \[\left( L=30 \right)\]
\[N\] is sum of all frequencies \[\left( N=100 \right)\]
\[{{f}_{0}}\] is cumulative frequency of preceding median class \[\left( {{f}_{0}}=33 \right)\]
\[f\] is frequency of median class \[\left( f=30 \right)\]
\[h\] is height of class interval \[\left( h=10 \right)\]
By substituting these values in the formula we get
\[\begin{align}
  & \Rightarrow \text{median}=30+\left( \dfrac{\dfrac{100}{2}-33}{30} \right)\times 10 \\
 & \Rightarrow \text{median}=30+\dfrac{17}{3} \\
 & \Rightarrow \text{median}=35.67 \\
\end{align}\]
Therefore, the median of given data is 35.67.
(iii) Let us solve for mode.
Let us take the given data

Class interval	Frequency (\[{{f}_{i}}\])
0-10	5
10-20	10
20-30	18
30-40	30
40-50	20
50-60	12
60-70	5

Here, we can see that the highest frequency is at 30-40.
So, we can say that class 30-40 is the modal class.
We know that the formula for mode as
\[\text{Mode}=L+\left( \dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)h\]
\[L\] is lower interval of modal class \[\left( L=30 \right)\]
\[{{f}_{0}}\] is frequency of preceding modal class \[\left( {{f}_{0}}=18 \right)\]
\[{{f}_{1}}\] is frequency of modal class \[\left( {{f}_{1}}=30 \right)\]
\[{{f}_{2}}\] is frequency of succeeding modal class \[\left( {{f}_{2}}=20 \right)\]
\[h\] is height of class interval \[\left( h=10 \right)\]
By substituting the values in the formula we get
\[\begin{align}
  & \Rightarrow \text{Mode}=30+\left( \dfrac{30-18}{\left( 2\times 30 \right)-18-20} \right)\times 10 \\
 & \Rightarrow \text{Mode}=30+\left( \dfrac{12}{22} \right)\times 10 \\
 & \Rightarrow \text{Mode}=35.45 \\
\end{align}\]
Therefore, the mode of given data is 35.45.

Note: The only difficulty where students will face is selecting the median and modal classes. After selecting the classes taking the frequencies and cumulative frequencies for substituting in the formula is important. Due to confusion students will take frequencies of median and modal classes instead of preceding and succeeding frequencies. The selection needs to be taken care of.