The median of 230 observations of the following frequency distribution is 46. Find a and b:

Class	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	12	30	A	65	b	25	18

Here, we have grouped data given in the table. We need to find cumulative frequency of the data, which will predict our answer.

So,

We have added up all the values of the frequency in the second column and have got,

Total = n = 150 + a + b …(i)

But given is, sum of frequencies, ∑f = n = 230 …(ii)

⇒

⇒

Comparing equations (i) and (ii), we get

150 + a + b = 230

⇒ a + b = 230 – 150

⇒ a + b = 80 …(iii)

Also, given that, median = 46

Corresponding to this value of median, we can say that median lies in the class is 40 – 50. [∵ 46 lies between 40 – 50]

⇒ Median class = 40 – 50

Median is given by,

Where,

l = lower limit of the median class = 40

n = Total number of observation (sum of frequencies) = 230

cf = cumulative frequency of the class preceding the median class = 42 + a

f = frequency of the median class = 65

c = class size (class sizes are equal) = 10

Putting the values, l = 40, n/2 = 115, cf = 42 + a, f = 65 and c = 10 in the given formula of median, we get

⇒ [∵ given that, median = 46]

⇒

⇒

⇒

⇒ 730 – 10a = 390

⇒ 10a = 730 – 390

⇒ 10a = 340

⇒

⇒ a = 34

Substituting the value, a = 34 in equation (iii), we get

a + b = 80

⇒ 34 + b = 80

⇒ b = 80 – 34

⇒ b = 46

Thus, a = 34 and b = 46.