Find the mode, median and mean for the following data: [CBSE 2009]

Marks obtained	25 - 35	35 - 45	45 - 55	55 - 65	65 - 75	75 - 85
Number of students	7	31	33	17	11	1

To find mean, we will solve by direct method:

We have got

Σf_i = 100 & Σf_ix_i = 4970

∵ 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 = 100

⇒ N/2 = 100/2 = 50

The cumulative frequency just greater than (N/2 = ) 50 is 71, so the corresponding median class is 45 - 55 and accordingly we get C_f = 38(cumulative frequency before the median class).

Now, since median class is 45 - 55.

∴ l = 45, h = 10, f = 33, N/2 = 50 and C_f = 38

Median is given by,

⇒

= 45 + 3.64

= 48.64

And we know that,

Mode = 3(Median) – 2(Mean)

= 3(48.64) – 2(49.7)

= 145.92 – 99.4

= 46.52

Hence, mean is 49.7, median is 48.64 and mode is 46.52.