| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Finding minimum stock level for demand |
| Difficulty | Standard +0.3 This is a standard S2 Poisson question with routine applications: (a) direct probability calculation with scaled parameter, (b) inverse cumulative probability lookup requiring tables/calculator, (c) normal approximation hypothesis test. All techniques are textbook exercises with no novel insight required, though part (b) requires careful interpretation of 'running out' and part (c) needs correct setup of one-tailed test. Slightly above average due to multi-step nature and need for approximation justification, but well within expected S2 scope. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim Po(7)\) | B1 | State or use Po(7) |
| \(P(X > 10) = 1 - P(X \leq 10)\) | M1 | State or use \(1 - P(X \leq 10)\) |
| \(= 1 - 0.9015\) | ||
| \(= 0.0985\) | A1 | awrt 0.0985 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X > d) < 0.05\) or \(P(X \geq d) < 0.05\) | M1 | Using or writing \(P(X > d) < 0.05\) or \(P(X < d) > 0.95\) (condone \(\geq\) instead of \(>\) and \(\leq\) instead of \(<\)) |
| \(P(X \leq 11) = 0.9467\), \(P(X < 12) = 0.9467\) | A1 | \(P(X \leq 12)/P(X < 13) =\) awrt \(0.973\) or \(P(X \leq 11)/P(X < 12) =\) awrt \(0.947\) |
| \(P(X \leq 12) = 0.9730\), \(P(X < 13) = 0.9730\) | A1 | |
| Least number of games \(= 12\) | NB answer of 12/13 on its own with no working gains M1A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \lambda = 1\ (\mu = 28)\), \(H_1: \lambda > 1\ (\mu > 28)\) | B1 | Both hypotheses correct using \(\lambda\) or \(\mu\), and 1 or 28 |
| \(Y \sim Po(28)\) approximated by \(N(28, 28)\) | B1 | Writing or using normal approximation with correct mean and variance |
| \(P(Y \geq 36) = P\left(Z \geq \frac{35.5 - 28}{\sqrt{28}}\right)\) | M1 | Use of continuity correction 35.5 or 36.5 or \(x \pm 0.5\) |
| \(1.6449 = \frac{x - 0.5 - 28}{\sqrt{28}}\) | M1 | Standardising using mean and sd |
| \(= P(Z \geq 1.42)\) | ||
| \(= 0.0778\) or \(1.42 < 1.6449\) | A1 | awrt 0.0778 or 0.9222 or \(1.42 <\) awrt \(1.65/1.64\); CR \(X \geq 37.2\) |
| \(0.0778 > 0.05\) so do not reject \(H_0\) | M1 | Correct conclusion for probability |
| There is no evidence that the average rate of sales per day has increased | A1cso | Correct contextual conclusion; need words "rate/average number", "sales" and "increased" |
# Question 3:
## Part 3(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim Po(7)$ | B1 | State or use Po(7) |
| $P(X > 10) = 1 - P(X \leq 10)$ | M1 | State or use $1 - P(X \leq 10)$ |
| $= 1 - 0.9015$ | | |
| $= 0.0985$ | A1 | awrt 0.0985 |
## Part 3(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X > d) < 0.05$ **or** $P(X \geq d) < 0.05$ | M1 | Using or writing $P(X > d) < 0.05$ or $P(X < d) > 0.95$ (condone $\geq$ instead of $>$ and $\leq$ instead of $<$) |
| $P(X \leq 11) = 0.9467$, $P(X < 12) = 0.9467$ | A1 | $P(X \leq 12)/P(X < 13) =$ awrt $0.973$ or $P(X \leq 11)/P(X < 12) =$ awrt $0.947$ |
| $P(X \leq 12) = 0.9730$, $P(X < 13) = 0.9730$ | A1 | |
| Least number of games $= 12$ | | NB answer of 12/13 on its own with no working gains M1A1A1 |
## Part 3(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \lambda = 1\ (\mu = 28)$, $H_1: \lambda > 1\ (\mu > 28)$ | B1 | Both hypotheses correct using $\lambda$ or $\mu$, and 1 or 28 |
| $Y \sim Po(28)$ approximated by $N(28, 28)$ | B1 | Writing or using normal approximation with correct mean and variance |
| $P(Y \geq 36) = P\left(Z \geq \frac{35.5 - 28}{\sqrt{28}}\right)$ | M1 | Use of continuity correction 35.5 or 36.5 or $x \pm 0.5$ |
| $1.6449 = \frac{x - 0.5 - 28}{\sqrt{28}}$ | M1 | Standardising using mean and sd |
| $= P(Z \geq 1.42)$ | | |
| $= 0.0778$ or $1.42 < 1.6449$ | A1 | awrt 0.0778 or 0.9222 or $1.42 <$ awrt $1.65/1.64$; CR $X \geq 37.2$ |
| $0.0778 > 0.05$ so do not reject $H_0$ | M1 | Correct conclusion for probability |
| There is no evidence that the average **rate** of **sales** per day has **increased** | A1cso | Correct contextual conclusion; need words "rate/average number", "sales" and "increased" |
---\begin{enumerate}
\item An online shop sells a computer game at an average rate of 1 per day.\\
(a) Find the probability that the shop sells more than 10 games in a 7 day period.
\end{enumerate}
Once every 7 days the shop has games delivered before it opens.\\
(b) Find the least number of games the shop should have in stock immediately after a delivery so that the probability of running out of the game before the next delivery is less than 0.05
In an attempt to increase sales of the computer game, the price is reduced for six months. A random sample of 28 days is taken from these six months. In the sample of 28 days, 36 computer games are sold.\\
(c) Using a suitable approximation and a $5 \%$ level of significance, test whether or not the average rate of sales per day has increased during these six months. State your hypotheses clearly.\\
\hfill \mbox{\textit{Edexcel S2 2013 Q3 [13]}}