Question

The mean of the following distribution is 53. Find the missing frequency p.

Class Interval:	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
Frequency:	12	15	p	28	13

Hence, find mode of the distribution.

Answer 1

Class Interval	Frequency (fi)	Class Mark (xi)	fixi
0 – 20	12	10	120
20 – 40	15	30	450
40 – 60	p	50	50p
60 – 80	28	70	1960
80 – 100	13	90	1170
Total	68 + p		3700 + 50p

Given Mean = 53.

`53 = (3700 + 50p)/(68 + p)`

53(68 + p) = 3700 + 50p

⇒ 3604 + 53p = 3700 + 50p

3p = 3700 – 3604

3p = 96

⇒ p = 32

Calculating Mode:

Since p = 32, the maximum frequency is 32.

Modal Class: 40 – 60

l = 40, f₁ = 32, f₀ = 15, f₂ = 28, h = 20.

Mode = `40 + ((32 - 15)/(2(32) - 15 - 28)) xx 20`

Mode = `40 + (17/(64 - 43)) xx 20`

= `40 + 17/21 xx 20`

= `40 + 340/21`

= 40 + 16.19

= 56.19

The missing frequency p is 32 and the mode of the distribution is approximately 56.19.