04-BS-2, May 2016, Q3 | Nivaar Exam Prep

Question 3 (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. (A) Poisson rate $\lambda_0=3$ calls/hour, constant over the 8:00–20:00 span. (B) $p=0.001$ (expensive claim), $n=2{,}500$ clients.

Find. (A)(a) $P(X>3)$ over 1 hour; (A)(b) $P(1<X<5)$ over 30 minutes; (A)(c) $P(X>40)$ over the full 12-hour span; (B) $P(X<3)$.

Approach. Rescale the Poisson rate to the length of each sub-interval ($\lambda=\lambda_0\times$hours), use the exact Poisson pmf/cdf for the short windows, switch to the normal approximation (with continuity correction) for the large 12-hour count, and use the Poisson approximation to the binomial for the rare-event insurance-claim question.

(A)(a) 1-hour window (10:00–11:00), λ=3. $P(X>3)=1-P(X\le3)=1-\sum_{k=0}^{3}\dfrac{e^{-3}3^k}{k!}=1-0.6472=\boxed{0.3528}$.
(A)(b) 30-minute window (14:30–15:00), λ=3(0.5)=1.5. "More than one but fewer than five" $=P(2\le X\le4)=P(X\le4)-P(X\le1)=0.9814-0.5578=\boxed{0.4236}$.
(A)(c) 12-hour span, λ=3(12)=36 — normal approximation. With $\mu=\sigma^2=36$, $\sigma=6$, and a continuity correction, $P(X>40)=P\!\left(Z>\dfrac{40.5-36}{6}\right)=P(Z>0.75)=1-\Phi(0.75)=\boxed{0.2266}$.
(B) Poisson approximation to the binomial (rare event, large n). $\lambda=np=2{,}500(0.001)=2.5$. $P(X<3)=P(X\le2)=\sum_{k=0}^{2}\dfrac{e^{-2.5}2.5^k}{k!}=\boxed{0.5438}$.

Question 3 — final results
Part	Quantity	Value
(A)(a)	P(X > 3), 1 hr, λ=3	0.3528
(A)(b)	P(1 < X < 5), 30 min, λ=1.5	0.4236
(A)(c)	P(X > 40), 12 hr, λ=36 (normal approx.)	0.2266
(B)	P(X < 3), n=2,500, p=0.001 (Poisson approx.)	0.5438

Topic: Poisson process rescaling, and the normal/Poisson approximations to discrete counting distributions.

Key relations: $P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}$ with $\lambda=\lambda_0\cdot t$; normal approximation to Poisson valid for large $\lambda$ ($Z\approx(x\pm0.5-\lambda)/\sqrt\lambda$); Poisson approximation to binomial valid for large $n$, small $p$, moderate $np$.

Why this works: the Poisson process has an additive rate over disjoint time intervals, so rescaling $\lambda$ to the window length is exact; for large $\lambda$ the Poisson pmf is well approximated by a normal curve with matching mean and variance (CLT); for a rare event repeated many times, the binomial pmf converges to a Poisson pmf with $\lambda=np$ because each trial contributes a vanishingly small probability.

Common pitfall: forgetting to rescale λ to the actual window length before applying the Poisson formula, and omitting the continuity correction in the normal approximation.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 5–6 (Poisson process, approximations).

Question 3 of 8

Question 3 (20 marks)