Answer:
a) The third quartile Q₃ = 56.45
b) Variance [tex]\mathbf{ \sigma^2 =2633.31}[/tex]
Step-by-step explanation:
Given that :
[tex]Q_1[/tex] Â = Â 30.8
Median [tex]Q_2[/tex] = Â 48.5
Mean  = 42
a) The mean is less than median; thus the expression showing the coefficient of skewness is given by the formula :
[tex]SK = \dfrac{Q_3+Q_1-2Q_2}{Q_3-Q_1}[/tex]
[tex]-0.38 = \dfrac{Q_3+30.8-2(48.5)}{Q_3-30.8}[/tex]
[tex]-0.38Q_3 + 11.704 = Q_3 +30.8 - 97[/tex]
[tex]1.38Q_3 = 77.904[/tex]
Divide both sides by 1.38
[tex]Q_3 = 56.45[/tex]
b) The objective here is to determine the approximate value of the variance;
Using the relation
[tex]SK_p = \dfrac{Mean- (3*Median-2 *Mean) }{\sigma}[/tex]
[tex]-0.38= \dfrac{42- (3 *48.5-2*42) }{\sigma}[/tex]
[tex]-0.38= \dfrac{(-19.5) }{\sigma}[/tex]
[tex]-0.38* \sigma = {(-19.5) }{}[/tex]
[tex]\sigma =\dfrac {(-19.5) }{-0.38 }[/tex]
[tex]\sigma =51.32[/tex]
Variance = Â [tex]\sigma^2 =51.32^2[/tex]
[tex]\mathbf{ \sigma^2 =2633.31}[/tex]
