| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Standard +0.3 This is a straightforward multi-part Poisson question requiring standard calculations: basic probability computations with given λ, scaling the rate parameter for different volumes, using binomial distribution for repeated trials, and applying normal approximation for large λ. All techniques are routine S2 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02m Poisson: mean = variance = lambda |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=2) = e^{-0.37}\frac{0.37^2}{2!} = 0.0473\) | M1, A1 (2 s.f.) | — |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X>2) = 1 - \left(e^{-0.37}\frac{0.37^2}{2!} + e^{-0.37}\frac{0.37^1}{1!} + e^{-0.37}\frac{0.37^0}{0!}\right)\) | M1 for \(P(X=1)\) and \(P(X=0)\), M1 for complete method | — |
| \(= 1-(0.0473+0.2556+0.6907) = 0.0064\) | A1 NB Answer given | 5 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{at most one day more than 2}) = \binom{30}{1}\times 0.9936^{29}\times 0.0064 + 0.9936^{30}\) | M1 for coefficient, M1 for \(0.9936^{29}\times 0.0064\) | |
| \(= 0.1594 + 0.8248 = 0.9842\) | M1 for \(0.993630\), A1 CAO (min 2sf) | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda = 0.37\times 10 = 3.7\) | B1 for mean (SOI) | |
| \(P(X>8) = 1-0.9863\) | M1 for probability | |
| \(= 0.0137\) | A1 CAO | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Mean no. per \(1000\text{ml} = 200\times 0.37 = 74\) | B1 for Normal approx. with correct parameters (SOI) | |
| \(X \sim N(74, 74)\) | — | |
| \(P(X>90) = P\left(Z > \frac{90.5-74}{\sqrt{74}}\right)\) | B1 for continuity correction | |
| \(= P(Z>1.918) = 1-\Phi(1.918)\) | M1 for probability using correct tail | |
| \(= 1-0.9724 = 0.0276\) | A1 CAO (min 2 s.f.), (but FT wrong or omitted CC) | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{questionable}) = 0.0064\times 0.0137\times 0.0276\) | M1 | |
| \(= 2.42\times 10^{-6}\) | A1 CAO | 2 marks total |
# Question 2:
## Part (i)(A)
$P(X=2) = e^{-0.37}\frac{0.37^2}{2!} = 0.0473$ | M1, A1 (2 s.f.) | —
## Part (i)(B)
$P(X>2) = 1 - \left(e^{-0.37}\frac{0.37^2}{2!} + e^{-0.37}\frac{0.37^1}{1!} + e^{-0.37}\frac{0.37^0}{0!}\right)$ | M1 for $P(X=1)$ and $P(X=0)$, M1 for complete method | —
$= 1-(0.0473+0.2556+0.6907) = 0.0064$ | A1 **NB Answer given** | 5 marks total
## Part (ii)
$P(\text{at most one day more than 2}) = \binom{30}{1}\times 0.9936^{29}\times 0.0064 + 0.9936^{30}$ | M1 for coefficient, M1 for $0.9936^{29}\times 0.0064$ |
$= 0.1594 + 0.8248 = 0.9842$ | M1 for $0.993630$, A1 CAO (min 2sf) | 4 marks total
## Part (iii)
$\lambda = 0.37\times 10 = 3.7$ | B1 for mean (SOI) |
$P(X>8) = 1-0.9863$ | M1 for probability |
$= 0.0137$ | A1 CAO | 3 marks total
## Part (iv)
Mean no. per $1000\text{ml} = 200\times 0.37 = 74$ | B1 for Normal approx. with correct parameters (SOI) |
$X \sim N(74, 74)$ | — |
$P(X>90) = P\left(Z > \frac{90.5-74}{\sqrt{74}}\right)$ | B1 for continuity correction |
$= P(Z>1.918) = 1-\Phi(1.918)$ | M1 for probability using correct tail |
$= 1-0.9724 = 0.0276$ | A1 CAO (min 2 s.f.), (but FT wrong or omitted CC) | 4 marks total
## Part (v)
$P(\text{questionable}) = 0.0064\times 0.0137\times 0.0276$ | M1 |
$= 2.42\times 10^{-6}$ | A1 CAO | 2 marks total
---2 A public water supply contains bacteria. Each day an analyst checks the water quality by counting the number of bacteria in a random sample of 5 ml of water.
Throughout this question, you should assume that the bacteria occur randomly at a mean rate of 0.37 bacteria per 5 ml of water.
\begin{enumerate}[label=(\roman*)]
\item Use a Poisson distribution to\\
(A) find the probability that a 5 ml sample contains exactly 2 bacteria,\\
(B) show that the probability that a 5 ml sample contains more than 2 bacteria is 0.0064 .
\item The month of September has 30 days. Find the probability that during September there is at most one day when a 5 ml sample contains more than 2 bacteria.
The daily 5 ml sample is the first stage of the quality control process. The remainder of the process is as follows.
\begin{itemize}
\item If the 5 ml sample contains more than 2 bacteria, then a 50 ml sample is taken.
\item If this 50 ml sample contains more than 8 bacteria, then a sample of 1000 ml is taken.
\item If this 1000 ml sample contains more than 90 bacteria, then the supply is declared to be 'questionable'.
\item Find the probability that a random sample of 50 ml contains more than 8 bacteria.
\item Use a suitable approximating distribution to find the probability that a random sample of 1000 ml contains more than 90 bacteria.
\item Find the probability that the supply is declared to be questionable.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2008 Q2 [18]}}