Question

Find the ‘median’ and ‘mode’ of the following data:

Class	100 – 105	105 – 110	110 – 115	115 – 120	120 – 125	125 – 130
Frequency	6	8	10	4	9	3

Answer 1

Class	Frequency	Cumulative frequency
100 – 105	6	6
105 – 110	8 → f0	14
110 – 115	10 → f1	24
115 – 120	4 → f2	28
120 – 125	9	37
125 – 130	3	40

Now, n = 40   ...(Even)

So, `n/2 = 40/2` = 20

This observation lies in the class 110 – 115.

Then lower limit (l) = 110

Cumulative frequency of the class preceding (110 – 115) (c.f.) = 14

The frequency of the median class (110 – 115) (f) = 10

Class size (h) = 115 – 110 = 5

We know that,

Median = `l + ((n/2 - c.f.))/f xx h`

= `110 + ((20 - 14)/10) xx 5`

= `110 + (6/10) xx 5`

= 110 + 3

= 113

So, the median of the data is 113.

Mode: Here, the maximum class frequency = 10

The class corresponding to this frequency = 110 – 115

Modal class = 110 – 115

Lower limit (1) = 110

Class size (h) = 115 – 110 = 5

Frequency (f₁) of the modal class = 10

Frequency (f₀) of the class preceding the modal class = 8

Frequency (f₂) of the class succeeding the modal class = 4

We know that,

Mode = `l + ((f_1 - f_0))/((2f_1 - f_0 - f_2)) xx h`

= `110 + ((10 - 8))/((2 xx 10 - 8 - 4)) xx 5`

= `110 + (2/(20 - 12)) xx 5`

= `110 + (2/8) xx 5`

= 110 + 1.25

= 111.25

So, the mode of the given data is 111.25.