Question 3 of 8: Bus Maintenance (Poisson) and a Hardware Shipment (Hypergeometric)

Question 3: Bus Maintenance (Poisson) and a Hardware Shipment (Hypergeometric) (20 marks)

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. 3(A): buses needing maintenance ~ Poisson, average 3/day. 3(B): lot of N=14 units, K=5 substandard, a sample of n=6 sold.

Find. 3(A)(a) P(X>3) in a day; (b) P(4<X<8) in a 2-day period; (c) P(X>100) in a 30-day period. 3(B) the pmf of X = number substandard among the 6 sold, and P(X>3).

Approach. The Poisson rate scales linearly with the time window (2-day → λ=6, 30-day → λ=90); for the 30-day window the normal approximation to the Poisson (CLT, since λ is large) is used. 3(B) is sampling without replacement from a finite lot of two types — the hypergeometric distribution.

1. 3(A)(a) One day, λ=3, P(X>3). $$P(X\gt 3)=1-\sum_{k=0}^{3}\dfrac{e^{-3}3^k}{k!}=1-(0.0498+0.1494+0.2240+0.2240)$$ $$\boxed{P(X\gt 3)=0.3528}$$
2. 3(A)(b) Two-day period, λ=2(3)=6, P(X∈{5,6,7}). $$P(4\lt X\lt 8)=\sum_{k=5}^{7}\dfrac{e^{-6}6^k}{k!}=0.1606+0.1606+0.1377$$ $$\boxed{P(4\lt X\lt 8)=0.4589}$$
3. 3(A)(c) Thirty-day period, λ=30(3)=90, normal approximation. With λ=90 large, $X\approx N(\lambda,\lambda)$: $\mu=90$, $\sigma=\sqrt{90}=9.487$. Applying the continuity correction, $z=\dfrac{100.5-90}{9.487}=1.107$, so $$\boxed{P(X\gt 100)\approx P(Z\gt 1.107)=0.1342}$$
4. 3(B) Hypergeometric distribution and P(X>3). N=14, K=5 substandard, n=6 sampled without replacement, so $$P(X=x)=\dfrac{\dbinom{5}{x}\dbinom{9}{6-x}}{\dbinom{14}{6}}, \qquad x=0,1,\dots,5$$ Evaluating for every x gives the full distribution below (table). $$\boxed{P(X\gt 3)=P(X=4)+P(X=5)=0.0629}$$