Question

Suppose that n students are selected at random without replacement from a class containing 28 students, of whom 8 are boys and 20 are girls. We assume that 0 < n < 28. Let X denote the number of boys that are obtained. Answer the following questions: a (4 marks) State the distribution of X, with parameters b (1 mark) Write down the possible values of X c (1 mark) Express E(X) in terms of n. d (4 marks) For what sample size n will Var(X) be a maximum'? Show your working. Hint: If you have difficulties to find the answer analytically, cal culate all possible values of Var(X). e (5 marks) Let X be the proportion of boys in the sample. Calcu late the mean and the variance of X for the case when n = 14.

Answer 1

a) The distribution of $\dpi{100} \fn_jvn X$ is hypergeometric distribution given by,

$\dpi{100} \fn_jvn P(X=x)=\frac{\binom{8}{x}\binom{20}{n-x}}{\binom{28}{n}};x=0,1,...,\min(n,8)$

Thus, $\dpi{100} \fn_jvn {\color{Blue} X\sim Hypergeometric(28,8,n)}$

b) The possible values of $\dpi{100} \fn_jvn X$ are $\dpi{100} \fn_jvn {\color{Blue} x=0,1,...,\min(n,8)}$ .

c) The expectation of the hypergeometric distribution is $\dpi{100} \fn_jvn E(X)=\frac{8n}{28}={\color{Blue} \frac{2n}{7}}$

d) The variance is given by,

$\dpi{100} \fn_jvn Var(X)=n\frac{8}{28}\frac{28-8}{28}\frac{28-n}{28-1}\\ Var(X)=\frac{2}{7}\frac{5}{7}\frac{n\left (28-n \right )}{27}\\$

The variance is maximum when $\dpi{100} \fn_jvn n\left (28-n \right )=14^2-\left ( n-14 \right )^2$ is maximum. We see $\dpi{100} \fn_jvn {\color{Blue} n=14}$

e) Here $\dpi{100} \fn_jvn \widetilde{X}=\frac{X}{28}$ . Now the expectation and variance are

$\dpi{100} \fn_jvn E\left (\widetilde{X}\right )=\frac{E\left (X \right )}{28} \\ E\left (\widetilde{X}\right )=\frac{2(n)/7}{28} \\ E\left (\widetilde{X}\right )=\frac{2(14)/7}{28} \\ {\color{Blue} E\left (\widetilde{X}\right )=\frac{1}{7}} \\$

$\dpi{100} \fn_jvn Var\left (\widetilde{X}\right )=\frac{Var\left (X \right )}{28^2} \\ Var\left (\widetilde{X}\right )=\frac{2}{7}\frac{5}{7}\frac{n\left (28-n \right )}{27}\frac{1}{28^2}\\ Var\left (\widetilde{X}\right )=\frac{2}{7}\frac{5}{7}\frac{14\left (28-14 \right )}{27}\frac{1}{28^2}\\ {\color{Blue} Var\left (\widetilde{X}\right )=\frac{5}{2646}}$