The Poisson distribution — practice questions

(a) The interval is 10 minutes, so we use $\lambda = 2.5$. Let $X$ be the number of emails in 10 minutes. $X \sim Po(2.5)$. We need $P(X=4)$. $P(X=4) = \frac{e^{-2.5} \times 2.5^4}{4!} = \frac{0.08208... \times 39.0625}{24} = 0.13360...$ $P(X=4) \approx 0.134$ (3 s.f.)

(b) The interval has changed to 20 minutes, which is double the original. We must adjust $\lambda$. New mean, $\lambda' = 2.5 \times 2 = 5$. Let $Y$ be the number of emails in 20 minutes. $Y \sim Po(5)$. We need the probability of 'at least 2 emails', which is $P(Y \ge 2)$. It's easier to calculate the complement: $P(Y \ge 2) = 1 - P(Y < 2) = 1 - P(Y \le 1)$. $P(Y \le 1) = P(Y=0) + P(Y=1)$. $P(Y=0) = \frac{e^{-5} 5^0}{0!} = e^{-5} \approx 0.006738$ $P(Y=1) = \frac{e^{-5} 5^1}{1!} = 5e^{-5} \approx 0.033690$ $P(Y \le 1) \approx 0.006738 + 0.033690 = 0.040428$ $P(Y \ge 2) = 1 - 0.040428 = 0.959572...$ $P(Y \ge 2) \approx 0.960$ (3 s.f.) (Using a calculator's Poisson CD function for $P(Y \le 1)$ with $\lambda=5$ is more direct).

Let $X$ be the number of faulty bulbs in a box. The exact distribution is Binomial. $X \sim B(400, 0.005)$. Here, $n=400$ is large and $p=0.005$ is small. This justifies using a Poisson approximation. We set the Poisson parameter $\lambda$ equal to the binomial mean $np$. $\lambda = np = 400 \times 0.005 = 2$. So we can approximate with $Y \sim Po(2)$. We need to find the probability of exactly 3 faulty bulbs, which is $P(Y=3)$. $P(Y=3) = \frac{e^{-2} \times 2^3}{3!} = \frac{e^{-2} \times 8}{6}$ $P(Y=3) = \frac{0.1353... \times 8}{6} = 0.18044...$ Probability $\approx 0.180$ (3 s.f.).