Question

For a distribution, Bowley’s coefficient of skewness is 0.6. The sum of upper and lower quartiles is 100 and median is 38. Find the upper and lower quartiles.

Answer 1

Given, Sk_b = 0.6, Q₃ + Q₁ = 100,
Median = Q₂ = 38

Sk_b = `("Q"_3 + "Q"_1 - 2"Q"_2)/("Q"_3 - "Q"_1)`

∴ 0.6 = `(100 - 2(38))/("Q"_3 - "Q"_1)`

∴ 0.6(Q₃ – Q₁) = 100 – 76 = 24

∴ Q₃ – Q₁ = `24/0.6`

∴ Q₃ – Q₁ = 40             ....(i)
Q₃ + Q₁ = 100             ....(ii) (given)
Adding (i) and (ii), we get
2Q₃ = 140

∴ Q₃ = `140/2` = 70
Substituting the value of Q₃ in (ii), we get
70 + Q₁ = 100
∴ Q₁ = 100 – 70 = 30
∴ upper quartile = 70 and lower quartile = 30.