Question

In a data distribution, the first quartile, the median and the means are 30.8, 48.5 and 42.0

Answer 1

Answer:

$Q_3 = 56.45$ --- The third quartile

$Var = 370.18$ -- Variance

Step-by-step explanation:

Given

$Q_1 = 30.8$ -- First quartile

$Q_2 = 48.5$ --- Median

$\bar x = 42$ --- Mean

$Skp = -0.38$ --- Coefficient of skewness

Solving (9): The third quartile $Q_3$

This is calculated from

$Skp = \frac{Q_1 + Q_3 - 2Q_2}{Q_3 - Q_1}$

So, we have:

$-0.38 = \frac{30.8 + Q_3- 2*48.5}{Q_3 - 30.8}$

Cross Multiply

$-0.38 (Q_3 - 30.8)= 30.8 + Q_3- 2*48.5$

Open bracket

$-0.38Q_3 + 11.704= 30.8 + Q_3- 97.0$

Collect like terms

$-0.38Q_3 -Q_3= 30.8 - 97.0- 11.704$

$-1.38Q_3= -77.904$

Divide both sides -1.38

$Q_3 = \frac{-77.904}{-1.38}$

$Q_3 = 56.45$ --- approximated

Solving (b): The variance

First, calculate the standard deviation from:

$3IQR = 4SD$

$IQR= Q_3 - Q_1$

So:

$3IQR = 4SD$

$3(Q_3 - Q_1) = 4SD$

Make SD the subject

$SD = \frac{3}{4}(Q_3 - Q_1)$

$SD = \frac{3}{4}(56.45 - 30.8)$

$SD = \frac{3}{4}*25.65$

$SD = \frac{3*25.65}{4}$

$SD = \frac{76.95}{4}$

$SD = 19.24$

So, the variance is:

$Var = SD^2$

$Var = 19.24^2$

$Var = 370.18$