Question

A random sample of five policyholders is taken from a group of 10 comprising 4men and 6women. Let x represent the number of females in the sample. Calculate the expected value and variance of the number of female policyholders

Answer 1

$$\dbinom{10}{5}=\dfrac{10!}{5!(10-5)!}=\dfrac{10(9)(8)(7)(6)}{120}=252$$

$$P(X=0)=0$$

$$P(X=1)=\dfrac{\dbinom{6}{1}\dbinom{4}{4}}{\dbinom{10}{5}}=\dfrac{6}{252}$$

$$P(X=2)=\dfrac{\dbinom{6}{2}\dbinom{4}{3}}{\dbinom{10}{5}}=\dfrac{15(4)}{252}=\dfrac{60}{252}$$

$$P(X=3)=\dfrac{\dbinom{6}{3}\dbinom{4}{2}}{\dbinom{10}{5}}=\dfrac{20(6)}{252}=\dfrac{120}{252}$$

$$P(X=4)=\dfrac{\dbinom{6}{4}\dbinom{4}{1}}{\dbinom{10}{5}}=\dfrac{15(4)}{252}=\dfrac{60}{252}$$

$$P(X=5)=\dfrac{\dbinom{6}{5}\dbinom{4}{0}}{\dbinom{10}{5}}=\dfrac{6}{252}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline p(x) & 0 & 1/42 & 10/42 & 20/42 & 10/42 & 1/42 \\ \end{array}$$

$$E(X)=0(0)+\dfrac{1}{42}(1)+\dfrac{10}{42}(2)+\dfrac{20}{42}(3)$$

$$+\dfrac{10}{42}(4)+\dfrac{1}{42}(5)=3$$

$$E(X^2)=0(0)+\dfrac{1}{42}(1)^2+\dfrac{10}{42}(2)^2+\dfrac{20}{42}(3)^2$$

$$+\dfrac{10}{42}(4)^2+\dfrac{1}{42}(5)^2=\dfrac{29}{3}$$

$$Var(X)=E(X^2)-[E(X)]^2$$

$$=\dfrac{29}{3}-(3)^2=\dfrac{2}{3}$$