Match height questions to the correct probability model

Let $M\sim N(170,10^2)$ and $W\sim N(160,7.5^2)$ denote men’s and women’s heights in cm. Use one unit consistently for means, standard deviations and thresholds.

The source gives $p=P(M>183)=0.0968005$, $P(M<170)=0.5$ and $P(170<M<183)=1-0.0968005-0.5=0.403199$ (displayed intermediate values). Under independent selection, the required probability is $p(0.5)(0.5-p)\,3!=0.117$ to three significant figures. The factor $3!$ counts every assignment of the three categories to the three selected men; use the unrounded p internally.

Let $X$ count the men taller than 183 cm out of five. The Teacher fills $X\sim B(5,0.0968005494)$ and $P(X=3)=0.00740$ to three significant figures. Equivalently evaluate $\binom53p^3(1-p)^2$ using the unrounded probability.

$S=M_1+M_2+M_3+W_1+W_2\sim N(3(170)+2(160),3(10^2)+2(7.5^2))=N(830,412.5)$. Thus $P(S/5>168)=P(S>840)=0.311$ to three significant figures. The average is over five distinct people, not five times one height.

Under the stated independence assumption, $D=M-W\sim N(170-160,10^2+7.5^2)=N(10,156.25)$. “Not more than 5 cm difference” means $|D|\le5$, so $P(-5\le M-W\le5)=0.230$ to three significant figures. It includes either spouse being taller; normal continuity makes endpoint inclusion immaterial to this probability.

The source assumes that all selected people’s heights are independent, including the husband and wife. This is a modelling assumption, not something implied by marriage or by separate normal marginal distributions. It is needed for the source sum/difference variances and the normal combination model. The Teacher adds a pointer to the examiners’ report for 2009 Paper 2 Question 9(iv); that pointer is retained here without a claim that the report has been inspected.