| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Finding minimum stock level for demand |
| Difficulty | Moderate -0.3 This is a straightforward application of Poisson distribution with standard calculations: finding P(X=5) for λ=6.8, cumulative probabilities P(X≤4) and P(X≤5) for λ=3.4, and finding minimum n where P(X≤n)≥0.9 using tables. All parts are routine textbook exercises requiring only direct use of formulas/tables with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda = 6.8\) | B1 | |
| \(e^{-6.8} \times \frac{6.8^5}{5!}\) | M1 | any \(\lambda\) |
| \(= 0.135\) (3 sf) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{-3.4}(1 + 3.4 + \frac{3.4^2}{2} + \frac{3.4^3}{3!} + \frac{3.4^4}{4!})\) | M1 | any \(\lambda\), allow one end-error |
| \(= 0.744\) (3 sf) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.744 + e^{-3.4} \times \frac{3.4^5}{5!}\) | M1 | or complete method, any \(\lambda\), allow one end-error |
| \(= 0.87(0)\) (3 sf) or \(0.871\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X \leq 6) = 0.870 + e^{-3.4} \times \frac{3.4^6}{6!}\) | M1 | or complete method, any \(\lambda\); fully correct un-simplified expression or better |
| \(= 0.94\) | A1 | |
| Need 6 hair driers | A1 [3] | dep M1A1 with numerical justification (\(0.94\) or better) |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda = 6.8$ | B1 | |
| $e^{-6.8} \times \frac{6.8^5}{5!}$ | M1 | any $\lambda$ |
| $= 0.135$ (3 sf) | A1 [3] | |
### Part (ii)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{-3.4}(1 + 3.4 + \frac{3.4^2}{2} + \frac{3.4^3}{3!} + \frac{3.4^4}{4!})$ | M1 | any $\lambda$, allow one end-error |
| $= 0.744$ (3 sf) | A1 [2] | |
### Part (ii)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.744 + e^{-3.4} \times \frac{3.4^5}{5!}$ | M1 | or complete method, any $\lambda$, allow one end-error |
| $= 0.87(0)$ (3 sf) or $0.871$ | A1 [2] | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X \leq 6) = 0.870 + e^{-3.4} \times \frac{3.4^6}{6!}$ | M1 | or complete method, any $\lambda$; fully correct un-simplified expression or better |
| $= 0.94$ | A1 | |
| Need 6 hair driers | A1 [3] | dep M1A1 with numerical justification ($0.94$ or better) |6 At a certain shop the demand for hair dryers has a Poisson distribution with mean 3.4 per week.\\
(i) Find the probability that, in a randomly chosen two-week period, the demand is for exactly 5 hair dryers.\\
(ii) At the beginning of a week the shop has a certain number of hair dryers for sale. Find the probability that the shop has enough hair dryers to satisfy the demand for the week if
\begin{enumerate}[label=(\alph*)]
\item they have 4 hair dryers in the shop,
\item they have 5 hair dryers in the shop.\\
(iii) Find the smallest number of hair dryers that the shop needs to have at the beginning of a week so that the probability of being able to satisfy the demand that week is at least 0.9 .
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2016 Q6 [10]}}