Question 3 of 8

Question 3 (20 marks: (a) 5, (b) 5, (c) 5, (d) 5)

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. Collisions on the highway follow a Poisson process with rate λ = 3 per week.

Find. (a) P(<4 in a week). (b) P(5<X<9 in 2 weeks). (c) P(3 in week 1 AND 2 in week 2). (d) P(>210 in a year, via Normal approximation).

Approach. The Poisson rate scales linearly with the length of the observation window ($\lambda_t = \lambda t$); independent weeks multiply; for the year-long window use the Normal approximation to the Poisson (valid since $\lambda$ is large).

1. (a) P(X < 4) with λ = 3. $$P(X<4)=P(X\le 3)=\sum_{k=0}^{3}\frac{e^{-3}3^k}{k!} = \boxed{0.6472}.$$
2. (b) Two-week window, λ = 6. "More than five but fewer than nine" means $6\le X\le 8$: $$P(6\le X\le 8) = P(X\le 8)-P(X\le 5) = 0.8472-0.4457 = \boxed{0.4016}.$$
3. (c) Two independent weeks, each λ = 3. $$P(X_1=3\ \text{and}\ X_2=2) = P(X_1=3)\cdot P(X_2=2) = \left(\frac{e^{-3}3^3}{3!}\right)\left(\frac{e^{-3}3^2}{2!}\right) = (0.2240)(0.2240) = \boxed{0.0502}.$$
4. (d) One-year window, λ = 3(52) = 156. With λ this large, approximate $X\sim N(156,156)$, $\sigma=\sqrt{156}=12.49$. With the continuity correction, "more than 210" → $X\ge 210.5$: $$Z=\frac{210.5-156}{12.49}=4.36, \qquad P(X>210)\approx 1-\Phi(4.36) = \boxed{6.4\times10^{-6}}.$$