Question

From the data given below state which group is more variable, A or B?

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Group A	9	17	32	33	40	10	9
Group B 1	10	20	30	25	43	15	7

Answer 1

Marks	Group A $f_i$	$mid-point\,x_i$	$y_i=\frac{x_i-45}{10}$	$f_i^2$	$f_iy_i$	$f_iy_1^2$
10-20	9	15	3	9	-27	81
20-30	17	25	2	4	-34	68
30-40	32	35	1	1	-32	32
40-50	33	45	0	0	0	0
50-60	40	55	1	1	40	40
60-70	10	65	2	4	20	40
70-80	9	75	3	9	27	81
150				6	342

Here, h = 10, N = 150, A = 45

Mean, $=A\frac{\sum_{i=1}^7f_ix_i}{n}×h=45+\frac{-6×10}{150}=45-0.4-44.6$

Variance (σ2) = $\frac{h^2}{N^2}(N\sum_{i=1}^7f_iy_i^2-(\sum_{i=1}^7f_iy_i)^2)$

$=\frac{100}{22500}(150×342-(6)^2)$

$\frac{1}{225}(51264)$

$=227.84$

$∴\,Standard\,deviation\\.(σ) =√2227.84=15.09$

The standard deviation of group B is calculated as follows.

Marks	Group B $f_i$	$mid-point\,x_i$	$y_i=\frac{x_i-45}{10}$	$f_i^2$	$f_iy_i$	$f_iy_1^2$
10-20	10	15	-3	9	-30	90
20-30	20	25	-2	4	-40	80
30-40	30	35	-1	1	-30	30
40-50	25	45	0	0	0	0
50-60	43	55	1	1	43	43
60-70	15	65	2	4	30	60
70-80	7	75	3	9	21	63
150				6	366

Mean=$=A\frac{\sum_{i=1}^7f_ix_i}{n}×h=45+\frac{-6×10}{150}=45-0.4-44.6$

Variance (σ2)= $\frac{h^2}{N^2}[N\sum_{i=1}^7f_iy_i^2-(\sum_{i=1}^7f_iy_i)^2]$

$=\frac{100}{22500}(150×366-(6)^2)$

$=\frac{1}{225}[54864]=243.84$

$∴\,Standard\,deviation\\.(σ_2) =√243.84=15.61$

Since the mean of both the groups is same, the group with greater standard deviation will be more variable.

Thus, group B has more variability in the marks.