Find the mean, mode and median of the following frequency distribution: [CBSE 2010]

Class interval	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70
Number of batsmen	4	4	7	10	12	8	5

To find mean, we will solve by direct method:

We have got

Σf_i = 50 & Σf_ix_i = 1910

∵ mean is given by

⇒

⇒

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 50

⇒ N/2 = 50/2 = 25

The cumulative frequency just greater than (N/2 = ) 25 is 37, so the corresponding median class is 40 - 50 and accordingly we get C_f = 25(cumulative frequency before the median class).

Now, since median class is 40 - 50.

∴ l = 40, h = 10, f = 37, N/2 = 25 and C_f = 25

Median is given by,

⇒

= 40 + 0

= 40

And we know that,

Mode = 3(Median) – 2(Mean)

= 3(40) – 2(38.2)

= 120 – 76.4

= 43.6

Hence, mean is 38.2, median is 40 and mode is 43.6.