| Exam Board | OCR |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2017 |
| Session | December |
| Marks | 7 |
| Topic | Poisson distribution |
| Type | Mean-variance comparison for Poisson validation |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution with standard bookwork questions: (i) requires calculator use for P(32≤X≤50) with λ=40, (ii) is direct formula writing with rate adjustment, (iii) tests understanding that variance should equal mean in Poisson (here 91.55≠35.14), and (iv) requires recognizing that constant rate assumption likely fails across different days. All parts are routine recall and interpretation with no problem-solving or novel insight required, 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 probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim \text{Po}(40)\) | M1 | |
| \(P(\leq 50) - P(\leq 31)\) | M1 | Allow M1 for 0.832 |
| \(= 0.94737\ldots - 0.08552\ldots = 0.862\) | A1 | Answer, awrt 0.862 |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{-\lambda} \frac{(2t)^5}{5!}\) | M1 | \(e^{-\lambda} \frac{\lambda^5}{5!}\), any \(\lambda\) as a function of \(t\) |
| A1 | Completely correct, oe |
| Answer | Marks |
|---|---|
| Would expect mean and variance to be roughly equal | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Not truly random as regular shoppers may always come on the same day of the week | B1 | Allow "Likely to have different means on weekdays and days at the weekend", or "not independent" with reason |
## (i)
$X \sim \text{Po}(40)$ | M1 |
$P(\leq 50) - P(\leq 31)$ | M1 | Allow M1 for 0.832
$= 0.94737\ldots - 0.08552\ldots = 0.862$ | A1 | Answer, awrt 0.862
## (ii)
$e^{-\lambda} \frac{(2t)^5}{5!}$ | M1 | $e^{-\lambda} \frac{\lambda^5}{5!}$, any $\lambda$ as a function of $t$
| A1 | Completely correct, oe
## (iii)
Would expect mean and variance to be roughly equal | B1 |
## (iv)
Not truly random as regular shoppers may always come on the same day of the week | B1 | Allow "Likely to have different means on weekdays and days at the weekend", or "not independent" with reason
---3 Over a long period Jenny counts the number of trolleys used at her local supermarket between 10 am and 10.20 am each day. She finds that the mean number of trolleys used between these times on a weekday is 40.00. You should assume that the use of trolleys occurs randomly, independently of one another, and at a constant average rate.\\
(i) Calculate the probability that, on a randomly chosen weekday, the number of trolleys used between these times is between 32 and 50 inclusive.\\
(ii) Write down an expression for the probability that, on a randomly chosen weekday, exactly 5 trolleys are used during a time period of $t$ minutes between 10 am and 10.20 am.
Jenny carries out this process for seven consecutive days. She finds that the mean number of trolleys used between 10 am and 10.20 am is 35.14 and the variance is 91.55 .\\
(iii) Explain why this suggests that the distribution of the number of trolleys used between these times on these seven consecutive days is not well modelled by a Poisson distribution.\\
(iv) Give a reason why it might not be appropriate to apply the Poisson model to the total number of trolleys used between these times on seven consecutive days.
\hfill \mbox{\textit{OCR FS1 AS 2017 Q3 [7]}}