| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | January |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Frequency distribution and Poisson fit |
| Difficulty | Standard +0.3 This is a standard S2 Poisson distribution question with routine calculations: verifying mean/variance, justifying Poisson model (mean ≈ variance), basic probability calculations, and a normal approximation to binomial. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04b Linear combinations: of normal distributions |
| Number of sultanas | 0 | 1 | 2 | 3 | 4 | 5 | \(> 5\) |
| Frequency | 8 | 15 | 12 | 9 | 4 | 2 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Mean} = \frac{\Sigma xf}{n} = \frac{0+15+24+27+16+10}{50} = \frac{92}{50} = 1.84\) | B1 | |
| \(\text{Variance} = \frac{1}{n-1}\left(\Sigma fx^2 - n\bar{x}^2\right) = \frac{1}{49}\left(258 - 50 \times 1.84^2\right) = 1.81\) (to 2 d.p.) | M1, A1 | Use of MSD gets M1 A0. Standard deviation gets M0 A0 unless "Variance \(= 1.81\)" is seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Because the mean is close to the variance | B1 | Must compare mean and variance as found in part (i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (A) \(P(\text{No sultanas}) = e^{-1.84}\frac{1.84^0}{0!} = 0.159\) (3 s.f.) | M1, A1 | \([1.8 \text{ leads to } 0.1653]\) |
| (B) \(P(\text{At least two sultanas}) = 1 - e^{-1.84}\frac{1.84^0}{0!} - e^{-1.84}\frac{1.84^1}{1!}\) | M1 for \(P(1)\), M1 for \(1-[P(0)+P(1)]\) | Or attempt \(P(2)+P(3)+P(4)+\ldots+P(8)\). Use of \(\lambda=1.8\) loses both accuracy marks \([1.8 \text{ leads to } 1-0.4296=0.5372]\) |
| \(= 1 - 0.159 - 0.292 = 0.549\) | A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda = 5 \times 1.84 = 9.2\) | B1 (SOI) | Any \(\lambda\) |
| Using tables: \(P(X \geq 10) = 1 - P(X \leq 9)\) | M1 | |
| \(= 1 - 0.5611 = 0.4389\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(2 \text{ out of 6 contain at least ten sultanas})\) | M1 for \(p^2 \times q^4\) | \(p + q = 1\) |
| \(= \binom{6}{2} \times 0.4389^2 \times 0.5611^4 = 0.2864\) | M1 dep for coefficient, A1 | Coefficient of 15 as part of binomial calculation. ft if \(p\) rounded from part (iv) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use Normal approximation with \(\mu = np = 60 \times 0.4389 = 26.334\) | B1 | Allow 26.3 |
| \(\sigma^2 = npq = 60 \times 0.4389 \times 0.5611 = 14.776\) | B1 | Allow 14.8... |
| \(P(X > 30) = P\!\left(Z > \frac{30.5 - 26.334}{\sqrt{14.776}}\right)\) | B1 for correct continuity correction | |
| \(= P(Z > 1.0838) = 1 - \Phi(1.0838)\) | M1 for probability using correct tail, ft their \(\mu\) & \(\sigma^2\) | giving \(P(Z > 1.091..) = 0.137\) 3sf |
| \(= 1 - 0.8608 = 0.1392\) | A1 cao | Do not ft wrong or omitted CC |
# Question 2 (part i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Mean} = \frac{\Sigma xf}{n} = \frac{0+15+24+27+16+10}{50} = \frac{92}{50} = 1.84$ | B1 | |
| $\text{Variance} = \frac{1}{n-1}\left(\Sigma fx^2 - n\bar{x}^2\right) = \frac{1}{49}\left(258 - 50 \times 1.84^2\right) = 1.81$ (to 2 d.p.) | M1, A1 | Use of MSD gets M1 A0. Standard deviation gets M0 A0 unless "Variance $= 1.81$" is seen |
**Total: 3 marks**
# Question 2 (part ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Because the mean is close to the variance | B1 | Must compare mean and variance as found in part (i) |
**Total: 1 mark**
# Question 2 (part iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (A) $P(\text{No sultanas}) = e^{-1.84}\frac{1.84^0}{0!} = 0.159$ (3 s.f.) | M1, A1 | $[1.8 \text{ leads to } 0.1653]$ |
| (B) $P(\text{At least two sultanas}) = 1 - e^{-1.84}\frac{1.84^0}{0!} - e^{-1.84}\frac{1.84^1}{1!}$ | M1 for $P(1)$, M1 for $1-[P(0)+P(1)]$ | Or attempt $P(2)+P(3)+P(4)+\ldots+P(8)$. Use of $\lambda=1.8$ loses both accuracy marks $[1.8 \text{ leads to } 1-0.4296=0.5372]$ |
| $= 1 - 0.159 - 0.292 = 0.549$ | A1 cao | |
**Total: 5 marks**
# Question 2 (part iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = 5 \times 1.84 = 9.2$ | B1 (SOI) | Any $\lambda$ |
| Using tables: $P(X \geq 10) = 1 - P(X \leq 9)$ | M1 | |
| $= 1 - 0.5611 = 0.4389$ | A1 | |
**Total: 3 marks**
# Question 2 (part v):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(2 \text{ out of 6 contain at least ten sultanas})$ | M1 for $p^2 \times q^4$ | $p + q = 1$ |
| $= \binom{6}{2} \times 0.4389^2 \times 0.5611^4 = 0.2864$ | M1 dep for coefficient, A1 | Coefficient of 15 as part of binomial calculation. ft if $p$ rounded from part (iv) |
**Total: 3 marks**
# Question 2 (part vi):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use Normal approximation with $\mu = np = 60 \times 0.4389 = 26.334$ | B1 | Allow 26.3 |
| $\sigma^2 = npq = 60 \times 0.4389 \times 0.5611 = 14.776$ | B1 | Allow 14.8... |
| $P(X > 30) = P\!\left(Z > \frac{30.5 - 26.334}{\sqrt{14.776}}\right)$ | B1 for correct continuity correction | |
| $= P(Z > 1.0838) = 1 - \Phi(1.0838)$ | M1 for probability using correct tail, ft their $\mu$ & $\sigma^2$ | giving $P(Z > 1.091..) = 0.137$ 3sf |
| $= 1 - 0.8608 = 0.1392$ | A1 cao | Do not ft wrong or omitted CC |
**Total: 5 marks**
---2 A student is investigating the numbers of sultanas in a particular brand of biscuit. The data in the table show the numbers of sultanas in a random sample of 50 of these biscuits.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Number of sultanas & 0 & 1 & 2 & 3 & 4 & 5 & $> 5$ \\
\hline
Frequency & 8 & 15 & 12 & 9 & 4 & 2 & 0 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Show that the sample mean is 1.84 and calculate the sample variance.
\item Explain why these results support a suggestion that a Poisson distribution may be a suitable model for the distribution of the numbers of sultanas in this brand of biscuit.
For the remainder of the question you should assume that a Poisson distribution with mean 1.84 is a suitable model for the distribution of the numbers of sultanas in these biscuits.
\item Find the probability of\\
(A) no sultanas in a biscuit,\\
(B) at least two sultanas in a biscuit.
\item Show that the probability that there are at least 10 sultanas in total in a packet containing 5 biscuits is 0.4389 .
\item Six packets each containing 5 biscuits are selected at random. Find the probability that exactly 2 of the six packets contain at least 10 sultanas.
\item Sixty packets each containing 5 biscuits are selected at random. Use a suitable approximating distribution to find the probability that more than half of the sixty packets contain at least 10 sultanas.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2011 Q2 [20]}}