Question

In a data distribution, the first quartile, the median and the means are 30.8, 48.5 and 42.0 respectively. If the coefficient skewness is −0.38
a) What is the approximate value of the third quartile (Q3 ), correct to 2 decimal places.
b)What is the approximate value of the variance, correct to the nearest whole number

Answer 1

Answer:

a) The third quartile Q₃ = 56.45

b) Variance $\mathbf{ \sigma^2 =2633.31}$

Step-by-step explanation:

Given that :

$Q_1$  =  30.8

Median $Q_2$ =  48.5

Mean  = 42

a) The mean is less than median; thus the expression showing the coefficient of skewness is given by the formula :

$SK = \dfrac{Q_3+Q_1-2Q_2}{Q_3-Q_1}$

$-0.38 = \dfrac{Q_3+30.8-2(48.5)}{Q_3-30.8}$

$-0.38Q_3 + 11.704 = Q_3 +30.8 - 97$

$1.38Q_3 = 77.904$

Divide both sides by 1.38

$Q_3 = 56.45$

b) The objective here is to determine the approximate value of the variance;

Using the relation

$SK_p = \dfrac{Mean- (3*Median-2 *Mean) }{\sigma}$

$-0.38= \dfrac{42- (3 *48.5-2*42) }{\sigma}$

$-0.38= \dfrac{(-19.5) }{\sigma}$

$-0.38* \sigma = {(-19.5) }{}$

$\sigma =\dfrac {(-19.5) }{-0.38 }$

$\sigma =51.32$

Variance =  $\sigma^2 =51.32^2$

$\mathbf{ \sigma^2 =2633.31}$