Question

Calculate the appropriate measure of dispersion for the following data:

Wages (In ₹)	Less than 35	35 – 40	40 – 45	45 – 50	50 – 55	55 – 60
No. of workers	15	50	85	40	27	33

Answer 1

Since, open-ended classes are given, the appropriate measure of dispersion that we can compute is the quartile deviation.
We construct the less than cumulative frequency table as follows:

Wages (in ₹)	No. of workers (f)	Less than cumulative frequency (c.f.)
Less than 35	15	15
35 – 40	50	65 ← Q1
40 – 45	85	150
45 – 50	40	190 ← Q3
50 – 55	27	217
55 – 60	33	250
Total	N = 250

Here, N = 250

Q₁ class = class containing `("N"/4)^"th"` observation

∴ `"N"/4 = 250/4` = 62.5
Cumulative frequency which is just greater than (or equal to) 62.5 is 65.
∴ Q₁ lies in the class 35 – 40.
∴ L = 35, c.f. = 15, f = 50, h = 5

∴ Q₁ = `"L" + "h"/"f" ("N"/4 - "c.f.")`

= `35 + 5/50(62.5 - 15)`

= `35 + 1/10 xx 47.5`

= 35 + 4.75
∴ Q₁ = 39.75

Q₃ class = class containing `((3"N")/4)^"th"` observation

∴ `(3"N")/4 = (3 xx 250)/4` = 187.5
Cumulative frequency which is just greater than (or equal to) 187.5 is 190.
∴ Q₃ lies in the class 45 – 50.
∴ L = 45, c.f. = 150, f = 40, h = 5

∴ Q₃ = `"L" + "h"/"f" ((3"N")/4 - "c.f.")`

= `45 + 5/40(187.5 - 150)`

= `45 + 1/8 xx 37.5`

= 45 + 4.6875
∴ Q₃ = 49.6875

∴ Q.D. = `("Q"_3 - "Q"_1)/2`

= `(49.6875 - 39.75)/2`

= `(9.9375)/2`

∴ Q.D. = 4.9688