Solveeit Logo

Question

Question: Find the median of the following data. Class Interval| Frequency ---|--- 0 – 10| 12 10 – 2...

Find the median of the following data.

Class IntervalFrequency
0 – 1012
10 – 2013
20 – 3025
30 – 4020
40 – 5010

(a) 25
(b) 23
(c) 24
(d) 26

Explanation

Solution

Here, we need to find the median of the data. Median is defined as the middle value of a list arranged in ascending order. First, we will find the median class. Then, using the formula for the median of a grouped distribution, we will find the value of the median.
Formula Used: The median of a grouped distribution is given by the formula M=L+N2c.f.f×iM = L + \dfrac{{\dfrac{N}{2} - c.f.}}{f} \times i, where LL is the lower limit of the median class, NN is the total number of terms, ff is the frequency of the median class, c.f.c.f. is the cumulative frequency of the class interval preceding the median class, and ii is the class size of the median class.

Complete step by step solution:
First, we will make another column in the table, showing the cumulative frequencies.
Cumulative frequency of a class interval is the sum of the frequency of the class interval, and all frequencies before the class interval.
For example, the cumulative frequency of the class interval 10 – 20 is the sum of the frequency of the class interval 10 – 20, and 0 – 10, that is 13+12=2513 + 12 = 25.
Adding the column for cumulative frequencies, we get

Class IntervalFrequencyCumulative frequency
ffc.f.c.f.
0 – 101212
10 – 201325
20 – 302550
30 – 402070
40 – 501080

Now, we know that the median class is the class where the (N2)th{\left( {\dfrac{N}{2}} \right)^{th}} term lies, where NN is the sum of all frequencies.
The sum of all frequencies is 80.
Therefore, the median class is the class where the (802)th=40th{\left( {\dfrac{{80}}{2}} \right)^{th}} = {40^{th}} term lies.
From the column of cumulative frequencies, we can see that the first 25 terms lie in the first two class intervals, and first 50 terms lie in the first three class intervals.
Therefore, the 40th term lies in the third class interval, that is 20 – 30.
Thus, the class interval 20 – 30 is the median class.
Finally, we will use the formula for median to find the value of the median.

Class IntervalFrequencyCumulative frequency
0 – 101212
10 – 201325
20 – 302550
30 – 402070
40 – 501080

The median of a grouped distribution is given by the formula M=L+N2c.f.f×iM = L + \dfrac{{\dfrac{N}{2} - c.f.}}{f} \times i, where LL is the lower limit of the median class, NN is the total number of terms, ff is the frequency of the median class, c.f.c.f. is the cumulative frequency of the class interval preceding the median class, and ii is the class size of the median class.
The class size of the median class is =i=3020=10 = i = 30 - 20 = 10.
The total number of terms is N=80N = 80.
The lower limit of the median class 20 – 30 is L=20L = 20.
The frequency of the median class 20 – 30 is f=25f = 25.
The cumulative frequency of the class interval preceding the median class, that is 10 – 20, is c.f.=25c.f. = 25.
Substituting i=10i = 10, N=80N = 80, L=20L = 20, c.f.=25c.f. = 25, and f=25f = 25 in the formula for median, we get
M=20+8022525×10M = 20 + \dfrac{{\dfrac{{80}}{2} - 25}}{{25}} \times 10
Simplifying the expression, we get
M=20+402525×10 M=20+1525×10 M=20+35×10\begin{array}{l} \Rightarrow M = 20 + \dfrac{{40 - 25}}{{25}} \times 10\\\ \Rightarrow M = 20 + \dfrac{{15}}{{25}} \times 10\\\ \Rightarrow M = 20 + \dfrac{3}{5} \times 10\end{array}
Multiplying 35\dfrac{3}{5} and 10, we get
M=20+6=26\Rightarrow M = 20 + 6 = 26
\therefore The median of the given information is 26.

Thus, the correct option is option (d).

Note:
We used the term class size in the solution. Class size is the difference in the upper limit and the lower limit of a class interval. It is also known as class width. The formula to calculate class size of a class interval is the upper limit of the class interval - lower limit of the class interval.