Question

Find the quartile deviation of advertisement expenditure using following frequency distribution of advertising expenditure of 50 companies:

Answer 1

First quartile:

Q₁ class = Class that Includes (n / 4) th observation
= Class that includes (50 / 4)
= 12.5th observation

Referring to column cf Q₁ class = 15 – 30
Now, Q₁ = L + (n / 4)−cf / f × c
Putting L = 15. (n / 4) = 12.5, cf = 11, f = 15 and
c = 15 in the formula,
Q₁ = 15 + 12.5−11 / 15 × 15
= 15 + 1.5 = ₹ 16.5 thousand

Third quartile:

Q₃ class = Class that Includes 3(n / 4)th observation
= Class that Includes 3 (12.5)
= 37.5 th observation

Referring to column cf, Q₃ class = 40 – 60
Now, Q₃ = L + \(\frac{3(\frac{n}{4})−cf}{f}\) × c
Putting. L = 40. 3(n / 4) = 37.5, cf = 36, f = 8 and c = 20 in the formula.
Q₃ = 40 + \(\frac{37.5−36}{8}\) × 20
= 40 + 1.5×20 / 8
= 40 + 30 / 8
= 40 + 3.75 = ₹ 43.75 thousand

Quartile deviation:

Q_d = \(\frac{Q_3−Q_1}{2} = \frac{43.75−16.5}{2}\)
= 27.5 / 2 = 13.625 ≈ ₹ 13.63 thousand