Question

The following distribution was obtained change of origin and scale of variable X.

di	– 4	– 3	– 2	– 1	0	1	2	3	4
fi	4	8	14	18	20	14	10	6	6

If it is given that mean and variance are 59.5 and 413 respectively, determine actual class intervals.

Answer 1

Here, Mean = `bar"x"` = 59.5 and
Var(X) = `sigma^2` = 413

d = `("x-a")/"h"`, where a is assumed mean and h is class width
We prepare the following table for calculation of the mean and variance of d_i

di	fi	fidi	fidi2
– 4	4	– 16	64
– 3	8	– 24	72
– 2	14	– 28	56
– 1	18	– 18	18
0	20	0	0
1	14	14	14
2	10	20	40
3	6	18	54
4	6	24	96
Total	N = 100	∑fidi = – 10	∑fidi2 = 414

`bar"d" = 1/"N"sum"f"_"i""d"_"i"`

= `(1)/(100) xx (- 10)` = – 0.1

Here, `bar"d"=(bar"x"-"a")/"h"`

∴ – 0.1 = `(59.5-"a")/"h"`
∴ – 0.1 h = 59.5 – a
∴ – 0.1 h + a = 59.5 ...............(i)

Var(D) = `sigma_"d"^2 = (1)/"N" sum"f"_"i""d"_"i"^2 - (bar"d")^2`

= `(1)/(100)(414) - (-0.1)^2`

= 4.14 – 0.01

= 4.13
Now, Var(X) = h² . Var(D)

∴ 413 = h² (4.13)

∴ h² = `(413)/(4.13)`

∴ h² = 100
∴ h = 10
Substitting h = 10 in equation (i), we get
– 0.1 × 10 + a = 59.5
∴ –1 + a = 59.5
∴ 59.5 = a – 1
∴ a = 59.5 + 1
∴ a = 60.5
We prepare the following table to determine actual class-intervals for corresponding values of d_i

di	Mid value xi = di × h + a	Class-interval
– 4	20.5	15.5 – 25.5
– 3	30.5	25.5 – 35.5
– 2	40.5	35.5 – 45.5
– 1	50.5	45.5 – 55.5
0	60.5	55.5 – 65.5
1	70.5	65.5 – 75.5
2	80.5	75.5 – 85.5
3	90.5	85.5 – 95.5
4	100.5	95.5 – 105.5

∴ The actual class intervals are 15.5 – 25.5, 25.5 – 35.5, …., 95.5 – 105.5