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.