| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson hypothesis test |
| Difficulty | Moderate -0.3 This is a standard S2 Poisson distribution question covering routine applications: model identification, probability calculations using tables/formula, and a one-tailed hypothesis test. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average but still requiring correct application of multiple techniques. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| Answer | Marks |
|---|---|
| Poisson with \(\lambda = 4\) | B1 |
| Answer | Marks |
|---|---|
| e.g. more people shopping \(\therefore\) probably sell more so \(\lambda\) higher | B1 |
| Answer | Marks |
|---|---|
| (i) let \(X\) = no. of sales per hour \(\therefore X \sim \text{Po}(4)\) | M1 A1 |
| \(P(X > 4) = 1 - P(X \le 4) = 1 - 0.6288 = 0.3712\) | |
| (ii) let \(Y\) = no. of sales per half-hour \(\therefore Y \sim \text{Po}(2)\) | M1 |
| \(P(Y = 0) = 0.1353\) | A1 |
| (iii) \((0.3712)^3 = 0.0511\) (3sf) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \lambda = 4\); \(H_1: \lambda > 4\) | B1 | |
| \(P(X \ge 7) = 1 - P(X \le 6) = 1 - 0.8893 = 0.1107\) | M1 A1 | |
| more than 5% \(\therefore\) not significant, insufficient evidence of increase | A1 | (12) |
**Part (a)**
Poisson with $\lambda = 4$ | B1 |
**Part (b)**
e.g. more people shopping $\therefore$ probably sell more so $\lambda$ higher | B1 |
**Part (c)**
(i) let $X$ = no. of sales per hour $\therefore X \sim \text{Po}(4)$ | M1 A1 |
$P(X > 4) = 1 - P(X \le 4) = 1 - 0.6288 = 0.3712$ |
(ii) let $Y$ = no. of sales per half-hour $\therefore Y \sim \text{Po}(2)$ | M1 |
$P(Y = 0) = 0.1353$ | A1 |
(iii) $(0.3712)^3 = 0.0511$ (3sf) | M1 A1 |
**Part (d)**
$H_0: \lambda = 4$; $H_1: \lambda > 4$ | B1 |
$P(X \ge 7) = 1 - P(X \le 6) = 1 - 0.8893 = 0.1107$ | M1 A1 |
more than 5% $\therefore$ not significant, insufficient evidence of increase | A1 | (12)
---A shoe shop sells on average 4 pairs of shoes per hour on a weekday morning.
\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable distribution for modelling the number of sales made per hour on a weekday morning and state the value of any parameters needed. [1 mark]
\item Explain why this model might have to be modified for modelling the number of sales made per hour on a Saturday morning. [1 mark]
\item Find the probability that on a weekday morning the shop sells
\begin{enumerate}[label=(\roman*)]
\item more than 4 pairs in a one-hour period,
\item no pairs in a half-hour period,
\item more than 4 pairs during each hour from 9 am until noon. [6 marks]
\end{enumerate}
\end{enumerate}
The area manager visits the shop on a weekday morning, the day after an advert appears in a local paper. In a one-hour period the shop sells 7 pairs of shoes, leading the manager to believe that the advert has increased the shop's sales.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Stating your hypotheses clearly, test at the 5\% level of significance whether or not there is evidence of an increase in sales following the appearance of the advert. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [12]}}