Answer:
[tex]Q_3 = 56.45[/tex] --- The third quartile
[tex]Var = 370.18[/tex] -- Variance
Step-by-step explanation:
Given
[tex]Q_1 = 30.8[/tex] -- First quartile
[tex]Q_2 = 48.5[/tex] --- Median
[tex]\bar x = 42[/tex] --- Mean
[tex]Skp = -0.38[/tex] --- Coefficient of skewness
Solving (9): The third quartile [tex]Q_3[/tex]
This is calculated from
[tex]Skp = \frac{Q_1 + Q_3 - 2Q_2}{Q_3 - Q_1}[/tex]
So, we have:
[tex]-0.38 = \frac{30.8 + Q_3- 2*48.5}{Q_3 - 30.8}[/tex]
Cross Multiply
[tex]-0.38 (Q_3 - 30.8)= 30.8 + Q_3- 2*48.5[/tex]
Open bracket
[tex]-0.38Q_3 + 11.704= 30.8 + Q_3- 97.0[/tex]
Collect like terms
[tex]-0.38Q_3 -Q_3= 30.8 - 97.0- 11.704[/tex]
[tex]-1.38Q_3= -77.904[/tex]
Divide both sides -1.38
[tex]Q_3 = \frac{-77.904}{-1.38}[/tex]
[tex]Q_3 = 56.45[/tex] --- approximated
Solving (b): The variance
First, calculate the standard deviation from:
[tex]3IQR = 4SD[/tex]
[tex]IQR= Q_3 - Q_1[/tex]
So:
[tex]3IQR = 4SD[/tex]
[tex]3(Q_3 - Q_1) = 4SD[/tex]
Make SD the subject
[tex]SD = \frac{3}{4}(Q_3 - Q_1)[/tex]
[tex]SD = \frac{3}{4}(56.45 - 30.8)[/tex]
[tex]SD = \frac{3}{4}*25.65[/tex]
[tex]SD = \frac{3*25.65}{4}[/tex]
[tex]SD = \frac{76.95}{4}[/tex]
[tex]SD = 19.24[/tex]
So, the variance is:
[tex]Var = SD^2[/tex]
[tex]Var = 19.24^2[/tex]
[tex]Var = 370.18[/tex]
