IB Mathematics SL 4.8 Binomial distribution. Mean and variance-AI SL Paper 2 - Exam Style Questions - New Syllabus

Two equivalent calculations:

\[ P(T>72)=P(T=73)+P(T=74) =\binom{74}{73}0.9^{73}(0.1)^1+\binom{74}{74}0.9^{74}(0.1)^0, \] or \[ P(T>72)=1-P(T\le 72). \]

Using technology (as required): \(\boxed{P(T>72)=0.00379\ \text{(0.00379124…)} }\). [3]

(i) For \(T\sim\mathrm{Bin}(72,0.9)\), \(E[T]=np=72\times 0.9=\boxed{64.8}\). [2]

(ii) If \(n\) tickets are sold, the expected arrivals are \(0.9n\). We require \(0.9n\le 72\Rightarrow n\le \dfrac{72}{0.9}=80\). Thus the maximum is \(\boxed{80}\) tickets. [2]

Ticket revenue: \(74\times \$150=\$11{,}100\). If at most 72 arrive, no compensation. If 73 arrive, one is denied boarding \(\Rightarrow \$300\) paid. If 74 arrive, two are denied \(\Rightarrow \$600\) paid.

Expected income when selling 74 tickets:
\[ \begin{aligned} \mathbb E[I] &=(11100)(0.9962\ldots)+(10800)(0.003380\ldots)+(10500)(0.0004110\ldots)\\ &\approx \boxed{\$11\,099}. \end{aligned} \]

Income when selling 72 tickets: \(72\times \$150=\boxed{\$10\,800}\).

Expected increase (74 instead of 72) \(=\$11\,099-\$10\,800=\boxed{\$299}\) (to the nearest dollar). [8]

Alternative equivalent method: expected compensation for 74 tickets is \(300\times P(T=73)+600\times P(T=74)=300(0.003380\ldots)+600(0.0004110\ldots)=1.26070\ldots\),
so expected income is \(11100-1.26070\ldots\approx \$11\,098.74\) \(\Rightarrow\) expected increase \(\approx \$299\).