Poisson Calls at an Internet Service Provider: 04-BS-2, December 2013, Q2

Question 2: Poisson Calls at an Internet Service Provider (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 arrive as a Poisson process at rate $\lambda=3.6$ calls/hour.

Find. (a) $P(X>3)$ for one hour; (b) $P(6\lt X\lt10)$ for a 2-hour period; (c) $P(X>160)$ over a 50-hour week (Normal approximation); (d) $P(X_1=3\text{ and }X_2=5)$ for two independent consecutive hours.

Approach. Rescale $\lambda$ to the length of the interval in question, then use the Poisson pmf directly for (a),(b),(d) and the Normal approximation to Poisson (with continuity correction) for the large mean in (c).

(a) One-hour tail. $X\sim Poisson(\lambda=3.6)$. $P(X>3)=1-P(X\le3)=1-\sum_{k=0}^{3}\dfrac{e^{-3.6}3.6^k}{k!}=1-0.5152=\boxed{0.4848}$.
(b) Two-hour window. Over 2 hours, $\lambda_2=2(3.6)=7.2$. $P(6\lt X\lt10)=P(X=7)+P(X=8)+P(X=9)$ $$=\dfrac{e^{-7.2}7.2^{7}}{7!}+\dfrac{e^{-7.2}7.2^{8}}{8!}+\dfrac{e^{-7.2}7.2^{9}}{9!}=0.1490+0.1341+0.1073=\boxed{0.3893}$$
(c) Weekly count via Normal approximation. Over 50 hours, $\lambda_{50}=50(3.6)=180$, so $E(X)=Var(X)=180$ and $sd(X)=\sqrt{180}=13.416$. With the half-unit continuity correction for "more than 160" (i.e. $X\ge161$), $$z=\dfrac{160.5-180}{13.416}=-1.453,\qquad P(X>160)\approx 1-\Phi(-1.453)=\Phi(1.453)=\boxed{0.9269}$$
(d) Two independent hours. Calls in disjoint hours of a Poisson process are independent, so $$P(X_1=3,\ X_2=5)=P(X_1=3)\cdot P(X_2=5)=\dfrac{e^{-3.6}3.6^{3}}{3!}\cdot\dfrac{e^{-3.6}3.6^{5}}{5!}=(0.2125)(0.1377)=\boxed{0.02925}$$ The joint probability is simply the product of the two marginal Poisson probabilities because the numbers of calls in non-overlapping time intervals of a Poisson process are mutually independent (the defining property of the process) — there is no additional adjustment for "the following hour" beyond that independence.

Final results — Question 2
Part	Result
(a) $P(X>3)$, 1 hr	0.4848
(b) $P(6\lt X\lt10)$, 2 hr	0.3893
(c) $P(X>160)$, week	0.9269
(d) $P(X_1=3,X_2=5)$	0.02925

Topic: Poisson process — rescaling, exact pmf, and Normal approximation.

Key relations: $P(X=k)=\dfrac{e^{-\lambda t}(\lambda t)^k}{k!}$; for large $\lambda t$, $X\approx N(\lambda t,\lambda t)$ with continuity correction; independence of counts on disjoint intervals.

Why this works: A Poisson process has independent, stationary increments, so the count over any interval of length $t$ is Poisson$(\lambda t)$, and by the CLT a Poisson variable with large mean is well approximated by a Normal with matching mean and variance.

Common pitfall: Forgetting to rescale $\lambda$ to the interval length (using 3.6 instead of 7.2 or 180), and omitting the 0.5 continuity correction when approximating a discrete count by a continuous Normal.

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 5 (Poisson process) & Ch. 6 (Normal approximation).

Question 2 of 8: Poisson Calls at an Internet Service Provider

Question 2: Poisson Calls at an Internet Service Provider (20 marks)