| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Justify Poisson approximation only |
| Difficulty | Moderate -0.3 This is a standard textbook question on Poisson approximation to binomial with routine calculations throughout. Parts (i)-(iii) require only stating conditions and basic Poisson probability, (iv)-(v) apply binomial distribution mechanically, and (vi)-(vii) involve straightforward mean/variance calculations from a frequency table with basic interpretation. No novel insight or complex problem-solving required, making it slightly easier than average. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02i 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 four-leaf clovers | 0 | 1 | 2 | \(> 2\) |
| Number of samples | 11 | 7 | 2 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Binomial(5000, 0.0001) | B1 | For binomial |
| B1 dep | For parameters |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(n\) is large and \(p\) is small | B1, B1 | Allow appropriate numerical ranges |
| \(\lambda = 5000 \times 0.0001 = 0.5\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X \geq 1) = 1 - e^{-0.5} \cdot \frac{0.5^0}{0!} = 1 - 0.6065 = 0.3935\) | M1 | For correct calculation or correct use of tables |
| \(= 0.3935\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{9 of 20 contain at least one}) = \binom{20}{9} \times 0.3935^9 \times 0.6065^{11}\) | M1 | For coefficient |
| M1 | For \(p^9 \times (1-p)^{11}\), \(p\) from part (iii) | |
| \(= 0.1552\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Expected number \(= 20 \times 0.3935 = 7.87\) | M1 A1 FT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= \frac{\Sigma xf}{n} = \frac{7+4}{20} = \frac{11}{20} = 0.55\) | B1 | For mean |
| Variance \(= \frac{1}{n-1}\left(\Sigma fx^2 - n\bar{x}^2\right)\) | M1 | For calculation |
| \(= \frac{1}{19}\left(15 - 20 \times 0.55^2\right) = 0.471\) | A1 CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Yes, since the mean is close to the variance | B1 | |
| And also as the expected frequency for 'at least one' i.e. 7.87 is close to the observed frequency of 9 | E1 | For sensible comparison |
| Observed frequency \(= 7 + 2 = 9\) | B1 |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Binomial(5000, 0.0001) | B1 | For binomial |
| | B1 dep | For parameters |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $n$ is large and $p$ is small | B1, B1 | Allow appropriate numerical ranges |
| $\lambda = 5000 \times 0.0001 = 0.5$ | B1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X \geq 1) = 1 - e^{-0.5} \cdot \frac{0.5^0}{0!} = 1 - 0.6065 = 0.3935$ | M1 | For correct calculation or correct use of tables |
| $= 0.3935$ | A1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{9 of 20 contain at least one}) = \binom{20}{9} \times 0.3935^9 \times 0.6065^{11}$ | M1 | For coefficient |
| | M1 | For $p^9 \times (1-p)^{11}$, $p$ from part (iii) |
| $= 0.1552$ | A1 | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expected number $= 20 \times 0.3935 = 7.87$ | M1 A1 FT | |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= \frac{\Sigma xf}{n} = \frac{7+4}{20} = \frac{11}{20} = 0.55$ | B1 | For mean |
| Variance $= \frac{1}{n-1}\left(\Sigma fx^2 - n\bar{x}^2\right)$ | M1 | For calculation |
| $= \frac{1}{19}\left(15 - 20 \times 0.55^2\right) = 0.471$ | A1 CAO | |
## Part (vii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Yes, since the mean is close to the variance | B1 | |
| And also as the expected frequency for 'at least one' i.e. 7.87 is close to the observed frequency of 9 | E1 | For sensible comparison |
| Observed frequency $= 7 + 2 = 9$ | B1 | |
---2 Clover stems usually have three leaves. Occasionally a clover stem has four leaves. This is considered by some to be lucky and is known as a four-leaf clover. On average 1 in 10000 clover stems is a four-leaf clover. You may assume that four-leaf clovers occur randomly and independently.
A random sample of 5000 clover stems is selected.\\
(i) State the exact distribution of $X$, the number of four-leaf clovers in the sample.\\
(ii) Explain why $X$ may be approximated by a Poisson distribution. Write down the mean of this Poisson distribution.\\
(iii) Use this Poisson distribution to find the probability that the sample contains at least one four-leaf clover.\\
(iv) Find the probability that in 20 samples, each of 5000 clover stems, there are exactly 9 samples which contain at least one four-leaf clover.\\
(v) Find the expected number of these 20 samples which contain at least one four-leaf clover.
The table shows the numbers of four-leaf clovers in these 20 samples.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Number of four-leaf clovers & 0 & 1 & 2 & $> 2$ \\
\hline
Number of samples & 11 & 7 & 2 & 0 \\
\hline
\end{tabular}
\end{center}
(vi) Calculate the mean and variance of the data in the table.\\
(vii) Briefly comment on whether your answers to parts (v) and (vi) support the use of the Poisson approximating distribution in part (iii).
\hfill \mbox{\textit{OCR MEI S2 2009 Q2 [18]}}