| Exam Board | SPS |
|---|---|
| Module | SPS ASFM (SPS ASFM) |
| Year | 2020 |
| Session | May |
| Marks | 6 |
| Topic | Poisson distribution |
| Type | Consecutive non-overlapping periods |
| Difficulty | Standard +0.3 This is a straightforward Poisson distribution problem requiring scaling the rate parameter (120/hour to 20/10-minutes), computing P(X≥28) using normal approximation or tables, then recognizing that the sum of two independent Poisson variables is Poisson. The calculations are routine and the assumption (independence) is standard. Slightly above average difficulty due to the multi-part structure and need to handle parameter scaling, but no novel insight required. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02m Poisson: mean = variance = lambda |
On any day, the number of orders received in one randomly chosen hour by an online supplier can be modelled by the distribution Po(120).
\begin{enumerate}[label=(\alph*)]
\item Find the probability that at least 28 orders are received in a randomly chosen 10-minute period. [2]
\item Find the probability that in a randomly chosen 10-minute period on one day and a randomly chosen 10-minute period on the next day a total of at least 56 orders are received. [3]
\item State a necessary assumption for the validity of your calculation in part (b). [1]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS ASFM 2020 Q10 [6]}}