Question

Find the median of the following frequency distribution:

Weekly wages (in Rs.)	60-69	70-79	80-89	90-99	100-109	110-119
No. of Days:	5	15	20	30	20	8

Answer 1

Hint: We start solving the problem by checking whether the given frequency distribution is continuous or not and making it continuous if it is not. We then calculate the total no. of days by adding all the terms in total no. of days. We then find the cumulative no. of days in each interval. We then use the formula ${{m}_{d}}={{L}_{i}}+\left( \dfrac{\dfrac{N}{2}-c{{f}_{i-1}}}{{{f}_{i}}} \right)\times I$ and substitute all the required valued in this to get the required value of median.

Complete step by step answer:
According to the problem, we need to find the median of the given frequency distribution,

Weekly wages (in Rs.)	60-69	70-79	80-89	90-99	100-109	110-119
No. of Days:	5	15	20	30	20	8

We can see that the upper boundary of the previous class interval is not equal to the lower boundary of the next class interval. Also, we can see the difference between those is 1. So, we add each upper boundary with half of this difference i.e., 0.5 and subtract each lower boundary with half of this difference i.e., 0.5 in order to make the given frequency distribution continuous.
So, let us rewrite the given frequency distribution by making addition and subtraction operations as mentioned above.
So, we get

Weekly wages (in Rs.)	59.5-69.5	69.5-79.5	79.5-89.5	89.5-99.5	99.5-109.5	109.5-119.5
No. of Days:	5	15	20	30	20	8

Now, let us find the sum of total no. of days given,

Weekly wages (in Rs.)	No. of Days
59.5-69.5	5
69.5-79.5	15
79.5-89.5	20
89.5-99.5	30
99.5-109.5	20
109.5-119.5	8
	Total days: 98

Let us find the cumulative no. of days for each of the interval of weekly wages (in Rs.),

Weekly wages (in Rs.)	No. of Days	Cumulative no. of days
59.5-69.5	5	5
69.5-79.5	15	20
79.5-89.5	20	40
89.5-99.5	30	70
99.5-109.5	20	90
109.5-119.5	8	98
	Total days: 98

Now, let us find the interval of weekly wages (in Rs.) at which the cumulative frequency is nearer to the \[\dfrac{\left( \text{sum of total no of days }\right)}{2}\] and \[\dfrac{\left( \text{sum of total no of days }\right)}{2}+1\], as the sum of total no. of days is even.
We can see that \[\dfrac{\left( \text{sum of total no of days }\right)}{2}=\dfrac{98}{2}=49\] and \[\dfrac{\left( \text{sum of total no of days }\right)}{2}+1=\dfrac{98}{2}+1=50\]. From the frequency distribution table, we can see that these values lie in the $89.5-99.5$ interval of Weekly wages (in Rs.).
We know that the median of the grouped frequency distribution is defined as ${{m}_{d}}={{L}_{i}}+\left( \dfrac{\dfrac{N}{2}-c{{f}_{i-1}}}{{{f}_{i}}} \right)\times I$.
Where, ${{L}_{i}}$ = Lower class boundary of the interval at which the median lies = 89.5.
N = Sum of the total no. of days = 98.
$c{{f}_{i-1}}$ = Cumulative no. of days that lies before the interval that median lies = 40.
${{f}_{i}}$ = No. of days present in the interval that median lies = 30.
I = Length of the interval = 10.
Now, let us substitute all these values in equation (1) to get the median.
$\Rightarrow {{m}_{d}}=89.5+\left( \dfrac{\dfrac{98}{2}-40}{30} \right)\times 10$.
$\Rightarrow {{m}_{d}}=89.5+\left( \dfrac{49-40}{3} \right)$.
$\Rightarrow {{m}_{d}}=89.5+\left( \dfrac{9}{3} \right)$.
$\Rightarrow {{m}_{d}}=92.5$.

So, we have found the median of the given frequency distribution as Rs.92.5.

Note: We should not directly proceed by taking the lower and upper limits as present in the problem as the wages are not continuous which will lead us to the wrong answer. We should know that the obtained median is an estimate not the accurate or absolute value. We should not confuse the formulas of median while solving this problem. We can also find mean and mode for the given frequency distribution.