AHL 4.17—Poisson distribution

Customers arrive independently at a constant average rate with no fixed upper limit on arrivals in the interval.

Independence, constant rate, and no fixed number of trials all point to a Poisson model with $\lambda =5$.

The number of arrivals in 15 minutes is $\text{Poisson}(5)$.

Binomial requires a fixed number of trials $n$ with success probability $p$; there is no such fixed trial count here, so both binomial options are invalid.

Normal is continuous and used for measured quantities, not for counting discrete rare events in an interval.

Only Poisson(5) fits counts of independent events at a constant rate.

For $X\sim \text{Poisson}(6)$, $P(X=k)=\frac{e^{-6}{6}^k}{k!}$.

$P(X=4)=\frac{e^{-6}\cdot 6^4}{4!}=\frac{e^{-6}\cdot 1296}{24}$.

$e^{-6}\approx 0.002479$, so $P(X=4)\approx \frac{0.002479\times 1296}{24}\approx 0.1339$.

$P(X=4)\approx 0.1339$.

0.2851 corresponds to using k=5 instead of k=4 in the calculation.

0.6288 is close to $P(X\le 4)$, the cumulative probability, not the exact PMF value.

0.0446 results from mistakenly using $3!$ instead of $4!$ in the denominator.

Only 0.1339 correctly applies the PMF with k=4.