Poisson Process — Telephone Calls: 04-BS-2, December 2018, Q3

Question 3: Poisson Process — Telephone Calls (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. Calls follow a Poisson process with rate λ = 3.6 calls/hour.

Find. (a) P(X>3) in one hour. (b) P(5<X<9) in a two-hour period. (c) P(X>160) in a 50-hour week. (d) P(3 calls in hour 1 AND 5 calls in hour 2).

Approach. Rescale λ to the interval length required (hour → 2 hours → 50 hours), then use the Poisson pmf directly for (a)/(b)/(d) and the normal approximation (Poisson mean = variance, large λ) for (c). Disjoint time intervals of a Poisson process are independent, which justifies multiplying probabilities in (d).

(a) One hour, λ=3.6. $$P(X\gt 3)=1-\sum_{k=0}^{3}\dfrac{e^{-3.6}3.6^{k}}{k!}=1-0.5152=0.4848$$ $$\boxed{P(X\gt 3)=0.4848}$$
(b) Two-hour period, λ=2(3.6)=7.2. “More than five but fewer than nine” means $X\in\{6,7,8\}$: $$P(6\le X\le8)=\dfrac{e^{-7.2}7.2^{6}}{6!}+\dfrac{e^{-7.2}7.2^{7}}{7!}+\dfrac{e^{-7.2}7.2^{8}}{8!}=0.1445+0.1486+0.1337$$ $$\boxed{P(5\lt X\lt 9)=0.4268}$$
(c) One week (50 hours), normal approximation. $\lambda_{\text{week}}=50(3.6)=180$, and for a Poisson variable $\mu=\sigma^2=180$ so $\sigma=13.416$. With a continuity correction, $$z=\dfrac{160.5-180}{13.416}=-1.453$$ $$P(X\gt 160)\approx P(Z\gt -1.453)=0.5+0.4270=0.9270$$ $$\boxed{P(X\gt 160)\approx0.9270}$$
(d) Two consecutive, independent hours. Because calls in disjoint hours are independent increments of the same Poisson process, the joint probability is the product of the two one-hour Poisson probabilities: $$P(X_1=3,\,X_2=5)=P(X_1=3)\cdot P(X_2=5)=\dfrac{e^{-3.6}3.6^{3}}{3!}\cdot\dfrac{e^{-3.6}3.6^{5}}{5!}=0.2125\times0.1377$$ $$\boxed{P(X_1=3,\,X_2=5)=0.02925}$$ This is NOT a conditional/Markov calculation — the Poisson process has no memory between disjoint intervals, so the two counts are simply multiplied as independent events.

Question 3 — final results
Part	Quantity	Result
(a)	P(X>3), λ=3.6	0.4848
(b)	P(5<X<9), λ=7.2	0.4268
(c)	P(X>160), λ=180 (normal approx.)	0.9270
(d)	P(3 then 5 in consecutive hours)	0.02925

Topic: Poisson process — rate scaling, exact pmf, and the normal approximation.

Key relations: $P(X=k)=\dfrac{e^{-\lambda}\lambda^{k}}{k!}$; over an interval of length $t$ (in the same units as the rate), $\lambda_t=\lambda t$; for large $\lambda$, $X\approx N(\lambda,\lambda)$.

Why this works: The Poisson process has stationary, independent increments, so the count over any interval scales linearly with its length and counts over disjoint intervals multiply as independent probabilities; the normal approximation follows from the CLT as $\lambda$ grows large enough that the discrete pmf is well approximated by a continuous bell curve.

Common pitfall: Forgetting to rescale λ to the new interval length before applying the pmf (using λ=3.6 for a 2-hour or 50-hour window); omitting the continuity correction in the normal approximation.

Source: Walpole et al., Probability & Statistics for Engineers and Scientists, Ch. 5 (Poisson Distribution).

Question 3 of 8: Poisson Process — Telephone Calls

Question 3: Poisson Process — Telephone Calls (20 marks)