The median of the following frequency distribution is 38. Find the value of a and b if the sum of frequences is 400:

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 = 202 + a + b …(i)

But given is, sum of frequencies, ∑f = n = 400 …(ii)

⇒

⇒

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

202 + a + b = 400

⇒ a + b = 400 – 202

⇒ a + b = 198 …(iii)

Also, given that, median = 38

Corresponding to this value of median, we can say that median lies in the class is 30 – 40. [∵ 38 lies between 30 – 40]

⇒ Median class = 30 – 40

Median is given by,

Where,

l = lower limit of the median class = 30

n = Total number of observation (sum of frequencies) = 400

cf = cumulative frequency of the class preceding the median class = 80

f = frequency of the median class = a

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

Putting the values, l = 30, n/2 = 200, cf = 80, f = a and c = 10 in the given formula of median, we get

⇒ [∵ given that, median = 38]

⇒

⇒

⇒

⇒

⇒ a = 150

Substituting the value, a = 150 in equation (iii), we get

a + b = 198

⇒ 150 + b = 198

⇒ b = 198 – 150

⇒ b = 48

Thus, a = 150 and b = 48.