| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Poisson to the Normal distribution |
| Type | Sum of independent Poisson variables |
| Difficulty | Standard +0.3 This is a straightforward application of standard S3 content: recognizing that sum of independent Poissons is Poisson, applying normal approximation with continuity correction, and computing mean/variance of linear combinations. The only slight challenge is part (iii) requiring conceptual understanding that X-Y cannot be Poisson (since it can be negative), but this is a standard bookwork point. Overall slightly easier than average A-level as it's mostly routine application of learned techniques. |
| Spec | 5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02n Sum of Poisson variables: is Poisson5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use Poisson distribution | M1 | Po(5.5) or Po(55) seen |
| With \(\mu=55\) | B1 | |
| \(\sigma^2=55\) | A1 | |
| \(\frac{(39.5-55)}{\sqrt{55}}\) | A1 | Standardising, with, without or wrong cc |
| \(-2.09\) | A1 | |
| art 0.982 | A1 | 6 |
| (ii) \(E(X-Y)=37\) | B1 | ft \(\mu\) above |
| \(\text{Var}(X-Y)=55+18\) | M1 | |
| \(=73\) | A1 | 3 |
| (iii) EITHER: Expectation not equal to variance | ||
| OR: \(X-Y\) could be negative | ||
| OR: Difference of two Poisson variables could have a negative expectation | M1 | Any one |
| So \(X-Y\) does not have a Poisson distn | A1 | 2 |
(i) Use Poisson distribution | M1 | Po(5.5) or Po(55) seen
With $\mu=55$ | B1 |
$\sigma^2=55$ | A1 |
$\frac{(39.5-55)}{\sqrt{55}}$ | A1 | Standardising, with, without or wrong cc
$-2.09$ | A1 |
art 0.982 | A1 | 6
(ii) $E(X-Y)=37$ | B1 | ft $\mu$ above
$\text{Var}(X-Y)=55+18$ | M1 |
$=73$ | A1 | 3 | ft $\mu$ above
(iii) EITHER: Expectation not equal to variance | | |
OR: $X-Y$ could be negative | | |
OR: Difference of two Poisson variables could have a negative expectation | M1 | Any one
So $X-Y$ does not have a Poisson distn | A1 | 2
---4 A multi-storey car park has two entrances and one exit. During a morning period the numbers of cars using the two entrances are independent Poisson variables with means 2.3 and 3.2 per minute. The number leaving is an independent Poisson variable with mean 1.8 per minute. For a randomly chosen 10-minute period the total number of cars that enter and the number of cars that leave are denoted by the random variables $X$ and $Y$ respectively.\\
(i) Use a suitable approximation to calculate $\mathrm { P } ( X \geqslant 40 )$.\\
(ii) Calculate $\mathrm { E } ( X - Y )$ and $\operatorname { Var } ( X - Y )$.\\
(iii) State, giving a reason, whether $X - Y$ has a Poisson distribution.
\hfill \mbox{\textit{OCR S3 2006 Q4 [11]}}