Question

What is the mean and standard deviation of this aggregate data set?

18-23	13
24-30	15
31-40	6
41-50	3
51-60	6
61+	1

Answer 1

First, We need to find out Midpoint:

Midpoint = (Upper class Limit + Lower class limit) / 2

Midpoint = (18+23)/2 = 20.5 (For, first class)

(24+30)/2 = 27 (For, second class)

and so on ...

Frequency Distribution Table:

Class interval	Frequency (fi)	Midpoints (Xi)	Xi*Fi	Fi*Xi2
18-23	13	20.5	266.5	5463.25
24-30	15	27	405	10935
31-40	6	35.5	213	7561.5
41-50	3	45.5	136.5	6210.75
51-60	6	55.5	333	18481.5
61+	1	65.5	65.5	4290.25
Total	44		1419.5	52942.25

Mean:

$\bar{X} = \frac{\sum FiXi}{N} = \frac{1419.5}{44} = 32.26$

Where,

$N = \sum fi$

Standard Deviation:

$SD (S) = \sqrt{\frac{\sum FiXi^2 - \frac{(\sum FiXi)^2}{N}}{N-1}} = \sqrt{\frac{52942.25 - \frac{(1419.5)^2}{44}}{44-1}} = 12.89$