Advertisement Remove all ads

Advertisement Remove all ads

Sum

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.

Advertisement Remove all ads

#### Solution

Given, Sk_{b} = 0.6, Q_{3} + Q_{1} = 100,

Median = Q_{2} = 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(Q_{3} – Q_{1}) = 100 – 76 = 24

∴ Q_{3} – Q_{1} = `24/0.6`

∴ Q_{3} – Q_{1} = 40 ....(i)

Q_{3} + Q_{1} = 100 ....(ii) (given)

Adding (i) and (ii), we get

2Q_{3} = 140

∴ Q_{3} = `140/2` = 70

Substituting the value of Q_{3} in (ii), we get

70 + Q_{1} = 100

∴ Q_{1} = 100 – 70 = 30

∴ upper quartile = 70 and lower quartile = 30.

Concept: Measures of Skewness - Bowley’s Coefficient of Skewness

Is there an error in this question or solution?

Advertisement Remove all ads

#### APPEARS IN

Advertisement Remove all ads