Question

Find the median of the following data.

Age greater than (Years)	Number of Persons
0	230
10	218
20	200
30	165
40	123
50	73
60	28
70	8

(A) 31.5
(B) 41.5
(C) 41.6
(D) 31.6

Answer 1

Hint: We solve this problem by first recognizing that given data about the number of persons is cumulative frequency of greater than typical. Then we find the frequency of every class by subtracting the cumulative frequency of the next class from the cumulative frequency of the considered class. Then we find the cumulative frequency of less than type from the frequencies of the classes found. Then we consider the formula for the median, $Median=l+{\displaystyle \frac{{\displaystyle \frac{N}{2}}-C}{f}}\times h$, and then find the median class by finding the class which has the cumulative frequency just greater than ${\displaystyle \frac{N}{2}}$. Then we substitute the values accordingly and find the required value of the median.

Complete step-by-step solution:
First, let us acknowledge that given data about the number of persons is the cumulative frequency of greater than typical.
Cumulative frequency of greater than type is the sum of frequencies of all the classes succeeding the class considered.
As we need to find the median, let us find the cumulative frequency of each class.
From the table given in the question, we can say that the total number of persons, that is total frequency is 230, as it is a cumulative frequency of greater than type
So, let us find the frequencies and cumulative frequencies of each class and record them in a table.

Class Interval	Cumulative Frequency (greater than)	Frequency	Cumulative Frequency (less than)
0-10	230	230 – 218 =12	12
10-20	218	218 – 200 = 18	30
20-30	200	200 – 165 = 35	65
30-40	165	165 – 123 = 42	107
40-50	123	123 – 73 = 50	157
50-60	73	73 – 28 = 45	202
60-70	28	28 – 8 = 20	222
70+	8	8	230

Now let us consider the formula for the median of the above frequency distribution.
$Median=l+{\displaystyle \frac{{\displaystyle \frac{N}{2}}-C}{f}}\times h$
where $l=$ Lower limit of the median class
$h=$ width of the class interval
$f=$ frequency of the median class
$N=$ Sum of all frequencies
$C=$ Cumulative frequency of the class preceding median class
The median class is the class that has the cumulative frequency just greater than or equal to ${\displaystyle \frac{N}{2}}$.
Here $N=230$
Here ${\displaystyle \frac{N}{2}}={\displaystyle \frac{230}{2}}=115$
The class having the frequency just greater than ${\displaystyle \frac{N}{2}}$ is 40-50.
So, the median class is 40-50.
So, we get the values of $l,h,f,C$ as,
$\begin{array}{rl}& \Rightarrow l=40\\ & \Rightarrow h=10\\ & \Rightarrow f=50\\ & \Rightarrow C=107\end{array}$
So, substituting these values in the above formula of the median, we get,
$$\begin{array}{rl}& \Rightarrow Median=40+\left({\displaystyle \frac{115-107}{50}}\right)\times 10\\ & \Rightarrow Median=40+\left({\displaystyle \frac{8}{50}}\right)\times 10\end{array}$$
Calculating the above value, we get,
$$\begin{array}{rl}& \Rightarrow Median=40+{\displaystyle \frac{80}{50}}\\ & \Rightarrow Median=40+{\displaystyle \frac{8}{5}}\\ & \Rightarrow Median=40+1.6\\ & \Rightarrow Median=41.6\end{array}$$
So, we get the value of median as 41.6. Hence the answer is Option C.

Note: The common mistake one makes while solving this question is one might not recognize that the given data is a cumulative frequency of greater than type and just solve it by taking it as normal frequency and applying the formula.