Question

Calculate the median in the following frequency distribution.

Wages	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
Number of labourers	5	10	15	20	25

(A) 90
(B) 70
(C) 83.75
(D) 85

Answer 1

Hint: In the question, we have to find the median of the given distribution. As we know, median is the central tendency that is used to determine an approximate average. Thus, in the given problem, we will apply the formula of median for continuous distribution, which is equal to $l+\dfrac{\left( \dfrac{N}{2}-cf \right)}{f}\times h$, where $l$ is the lower limit, $h$ is the height, $N$ is the sum of the frequencies and $cf$ is the cumulative frequency. Thus, in this problem, we will find the cumulative frequency and then obtain the median class, where $\dfrac{{{n}^{th}}}{2}$ term lies. After that we will substitute these values in the formula to get the required solution for the problem.

Complete step by step answer:
According to the problem, we have to find the median for the given distribution table. Thus, we will use the median for continuous distribution formulas. First, we will find the midpoint of the given distribution and then find $d$ which is equal to $\dfrac{\text{mid-point }-75}{10}$ , we get,

Wages	Number of labourers	Mid-point	$d=\dfrac{\text{mid-point }-75}{10}$
50 - 60	5	55	-2
60 -70	10	65	-1
70 -80	15	75	0
80 - 90	20	85	1
90 - 100	25	95	2

Since, in the above table, we let that the midpoint for the given distribution is equal to 75, and so we find the value of $d$.
Now, we will find the $fd$, which is the product of frequency, that is the number of labourers and $d$. Thus, we get,

Wages	Number of labourers	Mid-point	$d$	$fd$
50 - 60	5	55	-2	-10
60 -70	10	65	-1	-10
70 -80	15	75	0	0
80 - 90	20	85	1	20
90 - 100	25	95	2	50

So, we will find the cumulative frequency, which is equal to the total of all the observations, thus we get,

Wages	Number of labourers	Mid-point	$d$	$fd$	$cf$
50 - 60	5	55	-2	-10	5
60 -70	10	65	-1	-10	15
70 -80	15	75	0	0	30
80 - 90	20	85	1	20	50
90 - 100	25	95	2	50	75
	75			50

Thus, we will now find the median class, that is $\dfrac{{{n}^{th}}}{2}$term. So,
$\Rightarrow {{\left( \dfrac{75}{2} \right)}^{th}}$ term
$\Rightarrow {{\left( 37.5 \right)}^{th}}$ term
Therefore, we obtain the median class as (70 - 80) wages. Thus,
$l$ = lower limit of median class = 70 ………(1)
$cf$= cumulative frequency of the class preceding the median class = 15 ………(2)
$\dfrac{n}{2}$ = 37.5 ………(3)
$f$ = frequency of median class = 15 ………(4)
$h$ = upper limit - lower limit = 80 - 70 = 10 ………(5)
Now, Median = $l+\dfrac{\left( \dfrac{n}{2}-cf \right)}{f}\times h$
Thus, we will substitute (1), (2), (3), (4) and (5) in the above formula, so we get median as,
$\begin{align}
  & =70+\dfrac{\left( 37.5-15 \right)}{15}\times 10 \\
 & \Rightarrow 70+\dfrac{\left( 22.5 \right)}{15}\times 10 \\
 & \Rightarrow 70+\dfrac{225}{15} \\
 & \Rightarrow 70+15=85 \\
\end{align}$
Thus, the median = 85.
So, the correct answer is “Option D”.

Note: While solving this problem, do mention all the steps properly. Do mention the formula you are using and always remember the $cf$ is the frequency cumulative of the class preceding the median class and not the $cf$ of the median class. Also, remember that since, ${{\left( \dfrac{n}{2} \right)}^{th}}$ term = ${{\left( 37.5 \right)}^{th}}$ term, thus it must lie in (70 - 80) median class because 30 is the $cf$ which is lower than 37.5.