| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Mean-variance comparison for Poisson validation |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard Poisson distribution concepts: calculating means, recognizing Poisson properties (mean ≈ variance), basic probability calculations, and sum of independent Poisson variables. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average for A-level. |
| 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 = lambda5.02n Sum of Poisson variables: is Poisson5.06b Fit prescribed distribution: chi-squared test |
| Number of repairs | 0 | 1 | 2 | 3 | \(>3\) |
| Frequency | 53 | 20 | 6 | 1 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Mean \(= \frac{\sum xf}{n} = \frac{0 + 20 + 12 + 3}{80} = \frac{35}{80} = 0.4375\) | B1 for mean; NB answer given | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| So Poisson distribution may be appropriate, since mean is close to variance | B1 for variance; E1 dep on squaring s | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Or: This probability compares reasonably well with the relative frequency 0.25 | M1 for probability calc.; M0 for tables unless interpolated (0.2813); A1; B1 for expectation of 22.6 or r.f. of 0.25; E1 for comparison | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Using tables: \(P(X \geq 12) = 1 - P(X \leq 11) = 1 - 0.9997 = 0.0003\) | B1 for mean (SOI); M1 for using tables to find \(1 - P(X \leq 11)\); A1 FT | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| This is probably because the mean number of free repairs in the launderette will be much higher since the machines will get much more use than usual. | E1 for 'at least 12'; E1 for 'very low'; E1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X = 3) = e^{-0.5875} \frac{0.5875^3}{3!} = 0.0188\) (3 s.f.) | B1 for mean (SOI); M1; A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Each needs just 1}) = 0.282 \times 0.129 = 0.036\) | B1 for 0.129 (SOI); B1FT for 0.036 | 2 |
## (i)
Mean $= \frac{\sum xf}{n} = \frac{0 + 20 + 12 + 3}{80} = \frac{35}{80} = 0.4375$ | B1 for mean; NB answer given | 1
## (ii)
Variance = $0.6907^2 = 0.4771$
So Poisson distribution may be appropriate, since mean is close to variance | B1 for variance; E1 dep on squaring s | 2
## (iii)
$P(X = 1) = e^{-0.4375} \frac{0.4375^1}{1!} = 0.282$ (3 s.f.)
Either: Thus the expected number of 1's is 22.6 which is reasonably close to the observed value of 20.
Or: This probability compares reasonably well with the relative frequency 0.25 | M1 for probability calc.; M0 for tables unless interpolated (0.2813); A1; B1 for expectation of 22.6 or r.f. of 0.25; E1 for comparison | 4
## (iv)
$\lambda = 8 \times 0.4375 = 3.5$
Using tables: $P(X \geq 12) = 1 - P(X \leq 11) = 1 - 0.9997 = 0.0003$ | B1 for mean (SOI); M1 for using tables to find $1 - P(X \leq 11)$; A1 FT | 3
## (v)
The probability of at least 12 free repairs is very low, so the model is not appropriate.
This is probably because the mean number of free repairs in the launderette will be much higher since the machines will get much more use than usual. | E1 for 'at least 12'; E1 for 'very low'; E1 | 3
## (vi)
**(A)** $\lambda = 0.4375 + 0.15 = 0.5875$
$P(X = 3) = e^{-0.5875} \frac{0.5875^3}{3!} = 0.0188$ (3 s.f.) | B1 for mean (SOI); M1; A1 | 3
**(B)** $P(\text{Drier needs 1}) = e^{-0.15} \frac{0.15^1}{1!} = 0.129$
$P(\text{Each needs just 1}) = 0.282 \times 0.129 = 0.036$ | B1 for 0.129 (SOI); B1FT for 0.036 | 2
---An electrical retailer gives customers extended guarantees on washing machines. Under this guarantee all repairs in the first 3 years are free. The retailer records the numbers of free repairs made to 80 machines.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Number of repairs & 0 & 1 & 2 & 3 & $>3$ \\
\hline
Frequency & 53 & 20 & 6 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Show that the sample mean is 0.4375. [1]
\item The sample standard deviation $s$ is 0.6907. Explain why this supports a suggestion that a Poisson distribution may be a suitable model for the distribution of the number of free repairs required by a randomly chosen washing machine. [2]
\end{enumerate}
The random variable $X$ denotes the number of free repairs required by a randomly chosen washing machine. For the remainder of this question you should assume that $X$ may be modelled by a Poisson distribution with mean 0.4375.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{2}
\item Find P$(X = 1)$. Comment on your answer in relation to the data in the table. [4]
\item The manager decides to monitor 8 washing machines sold on one day. Find the probability that there are at least 12 free repairs in total on these 8 machines. You may assume that the 8 machines form an independent random sample. [3]
\item A launderette with 8 washing machines has needed 12 free repairs. Why does your answer to part (iv) suggest that the Poisson model with mean 0.4375 is unlikely to be a suitable model for free repairs on the machines in the launderette? Give a reason why the model may not be appropriate for the launderette. [3]
\end{enumerate}
The retailer also sells tumble driers with the same guarantee. The number of free repairs on a tumble drier in three years can be modelled by a Poisson distribution with mean 0.15. A customer buys a tumble drier and a washing machine.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{5}
\item Assuming that free repairs are required independently, find the probability that
\begin{enumerate}[label=(\Alph*)]
\item the two appliances need a total of 3 free repairs between them,
\item each appliance needs exactly one free repair.
\end{enumerate}
[5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2007 Q3 [18]}}