Question 3 of 8

Question 3

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (A) The discrete distribution of $Y$ over $y=10,25,40,55,70,85,100$ with $\Pr(Y=y)=0.10,0.20,0.25,0.20,0.10,0.10,0.05$ (sums to 1.00, confirmed). (B) A staff of 18 (11 male, 7 female); a committee of $n=6$ is drawn without replacement.

Find. (A) $E[Y]$, $\mathrm{Var}(Y)$, $\mathrm{sd}(Y)$. (B)(a) $P(F>3)$; (B)(b) $P(M>1)$, where $F,M$ are the numbers of females/males on the committee.

Approach. (A) apply the discrete-distribution moment formulas directly to the table. (B) both parts are sampling-without-replacement from two finite subpopulations — the Hypergeometric distribution.

1. A) Mean, variance, sd of Y. $$E[Y]=\sum y\Pr(Y=y)=10(.10)+25(.20)+40(.25)+55(.20)+70(.10)+85(.10)+100(.05)=\boxed{47.5}.$$ $$\mathrm{Var}(Y)=\sum (y-47.5)^2\Pr(Y=y)=\boxed{596.25},\qquad \mathrm{sd}(Y)=\sqrt{596.25}=\boxed{24.42}.$$
2. B(a) P(committee has more than 3 females). With $N=18$ total, $K=7$ females, $n=6$ drawn, the hypergeometric pmf is $h(f)=\binom{7}{f}\binom{11}{6-f}/\binom{18}{6}$. "More than three" means $f=4,5,6$: $$P(F>3)=h(4)+h(5)+h(6)=0.0839+0.0288+0.0037=\boxed{0.1165}.$$
3. B(b) P(committee has more than 1 male). By symmetry the male count $M=6-F$ is $\text{Hypergeom}(N=18,K=11,n=6)$. $P(M>1)=1-P(M=0)-P(M=1)$: $$P(M=0)=\binom{11}{0}\binom{7}{6}/\binom{18}{6}=0.0004,\qquad P(M=1)=\binom{11}{1}\binom{7}{5}/\binom{18}{6}=0.0124,$$ $$P(M>1)=1-0.0004-0.0124=\boxed{0.9872}.$$