Question 3 of 8

Question 3 (15 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. (A) Poisson rate $\lambda_0=1.5$ stoppages / 500 h, constant over time (a Poisson process). (B) $p=0.002$ (fired for gross negligence), $n=1{,}000$ employees.

Find. (A)(a) $P(1<X<5)$ in 500 h; (A)(b) $P(X\le3)$ in 1,000 h; (A)(c) $P(X_1{=}1\text{ in first }500\text{h AND }X_2{=}2\text{ in next }500\text{h})$; (B) $P(X>2)$.

Approach. Rescale the Poisson rate to the interval length asked for ($\lambda=1.5$ for 500 h, $\lambda=3.0$ for 1,000 h); part (c) uses the independent-increments property of a Poisson process to multiply two separate 500-hour probabilities; (B) uses the Poisson approximation to the binomial since $n$ is large and $p$ is small.

1. (A)(a) Poisson(1.5), P(1<X<5). $P(X{=}2)+P(X{=}3)+P(X{=}4)=e^{-1.5}\!\left(\dfrac{1.5^2}{2!}+\dfrac{1.5^3}{3!}+\dfrac{1.5^4}{4!}\right)=0.2510+0.1255+0.0471=\boxed{0.4236}$.
2. (A)(b) Poisson(3.0) for the doubled interval, P(X≤3). Doubling the operating time doubles the mean: $\lambda=2(1.5)=3.0$. $P(X\le3)=e^{-3}\left(1+3+\dfrac{9}{2}+\dfrac{9}{2}\right)=0.0498+0.1494+0.2240+0.2240=\boxed{0.6472}$.
3. (A)(c) Independent increments of a Poisson process. Non-overlapping time intervals of a Poisson process generate independent counts, so $P(X_1{=}1\text{ in the first 500h AND }X_2{=}2\text{ in the next 500h})=P(X_1{=}1)\cdot P(X_2{=}2)$, each with $\lambda=1.5$: $P(X_1{=}1)=e^{-1.5}(1.5)=0.3347$, $P(X_2{=}2)=e^{-1.5}\dfrac{1.5^2}{2}=0.2510$. Product $=0.3347(0.2510)=\boxed{0.0840}$. The multiplication (rather than addition or a single-interval Poisson with $\lambda=4.5$) is justified precisely because the two 500-hour windows do not overlap, so their stoppage counts are statistically independent Poisson variables that can be joined with the product rule.
4. (B) Poisson approximation to the binomial, n=1,000, p=0.002. $\lambda=np=1{,}000(0.002)=2.0$ (large $n$, small $p$, $np$ moderate — the classic Poisson-approximation regime). $P(X>2)=1-P(X\le2)=1-e^{-2}\!\left(1+2+2\right)=1-0.6767=\boxed{0.3233}$.