| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | State Poisson approximation with justification |
| Difficulty | Standard +0.3 This is a standard S2 question testing routine application of Poisson approximation to binomial (parts i-iii) and Normal approximation (parts v-vi). All parts follow textbook procedures: calculating binomial probabilities, stating standard conditions for Poisson approximation (n large, p small, np moderate), and applying Normal approximation with continuity correction. No novel problem-solving or insight required beyond recalling standard criteria and formulas. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial5.02i Poisson distribution: random events model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{At least one has red hair}) = 1 - 0.97^{10}\) | M1 | M1 for \(1 - 0.97^{10}\) |
| \(= 0.263\) | A1 | Allow 0.26 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (Because \(X\) is binomially distributed,) \(n\) is large and \(p\) is small. | E1 | Allow "sample is large" for \(n\) is large |
| E1 | Allow "\(np < 10\)" or "mean \(\approx\) variance" for "\(p\) is small". Do not allow "the probability is small" | |
| Mean \(= 1.8\) | B1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=2) = e^{-1.8}\frac{1.8^2}{2!} = 0.2678\) | M1 | For calculation for \(P(X=2)\) |
| \(\text{OR} = 0.7306 - 0.4628 = 0.2678\) | A1 | FT their mean. Allow answer to 3sf. |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X > 2) = 1 - P(X \leq 2) = 1 - 0.7306\) | M1 | \(1 - P(X \leq 2)\) used. e.g. \(1 - P(X \leq 2) = 1 - 0.4628\) gets M0. CAO |
| \(= 0.2694\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The mean (\(np = 1.8\)) is too small | E1* | For "mean is too small" or "mean \(< 10\)" |
| It is not appropriate to use a Normal approximation | E1dep* | For "not appropriate". Do not allow "\(p\) is too small". |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Binomial\((5000, 0.03)\) | B1* | For binomial, or B( , ) |
| B1dep* | For parameters | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 5000 \times 0.03 = 150\) | B1 | For mean (soi) |
| Variance \(= 5000 \times 0.03 \times 0.97 = 145.5\) | B1 | For variance (soi) |
| Using Normal approx. to the binomial, \(X \sim N(150, 145.5)\) | ||
| \(P(X \geq 160) = P\!\left(Z \geq \frac{159.5 - 150}{\sqrt{145.5}}\right)\) | B1 | For continuity correction. |
| \(= P(Z > 0.7876) = 1 - \Phi(0.7876) = 1 - 0.7846\) | M1 | For probability using correct tail and structure (condone omission of/incorrect c.c.) |
| \(= 0.215\) (to 3 sig.fig.) | A1 | CAO. (Do not FT wrong or omitted CC). Allow 0.2155. Do not allow 0.216 |
| [5] |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{At least one has red hair}) = 1 - 0.97^{10}$ | M1 | M1 for $1 - 0.97^{10}$ |
| $= 0.263$ | A1 | Allow 0.26 |
| | **[2]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| (Because $X$ is binomially distributed,) $n$ is large and $p$ is small. | E1 | Allow "sample is large" for $n$ is large |
| | E1 | Allow "$np < 10$" or "mean $\approx$ variance" for "$p$ is small". Do not allow "the probability is small" |
| Mean $= 1.8$ | B1 | |
| | **[3]** | |
## Part (iii)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=2) = e^{-1.8}\frac{1.8^2}{2!} = 0.2678$ | M1 | For calculation for $P(X=2)$ |
| $\text{OR} = 0.7306 - 0.4628 = 0.2678$ | A1 | FT their mean. Allow answer to 3sf. |
| | **[2]** | |
## Part (iii)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X > 2) = 1 - P(X \leq 2) = 1 - 0.7306$ | M1 | $1 - P(X \leq 2)$ used. e.g. $1 - P(X \leq 2) = 1 - 0.4628$ gets M0. CAO |
| $= 0.2694$ | A1 | |
| | **[2]** | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| The mean ($np = 1.8$) is too small | E1* | For "mean is too small" or "mean $< 10$" |
| It is not appropriate to use a Normal approximation | E1dep* | For "not appropriate". Do not allow "$p$ is too small". |
| | **[2]** | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Binomial$(5000, 0.03)$ | B1* | For binomial, or B( , ) |
| | B1dep* | For parameters |
| | **[2]** | |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 5000 \times 0.03 = 150$ | B1 | For mean (soi) |
| Variance $= 5000 \times 0.03 \times 0.97 = 145.5$ | B1 | For variance (soi) |
| Using Normal approx. to the binomial, $X \sim N(150, 145.5)$ | | |
| $P(X \geq 160) = P\!\left(Z \geq \frac{159.5 - 150}{\sqrt{145.5}}\right)$ | B1 | For continuity correction. |
| $= P(Z > 0.7876) = 1 - \Phi(0.7876) = 1 - 0.7846$ | M1 | For probability using correct tail and structure (condone omission of/incorrect c.c.) |
| $= 0.215$ (to 3 sig.fig.) | A1 | CAO. (Do not FT wrong or omitted CC). Allow 0.2155. Do not allow 0.216 |
| | **[5]** | |
---2 Suppose that 3\% of the population of a large city have red hair.
\begin{enumerate}[label=(\roman*)]
\item A random sample of 10 people from the city is selected. Find the probability that there is at least one person with red hair in this sample.
A random sample of 60 people from the city is selected. The random variable $X$ represents the number of people in this sample who have red hair.
\item Explain why the distribution of $X$ may be approximated by a Poisson distribution. Write down the mean of this Poisson distribution.
\item Hence find\\
(A) $\mathrm { P } ( X = 2 )$,\\
(B) $\mathrm { P } ( X > 2 )$.
\item Discuss whether or not it would be appropriate to model $X$ using a Normal approximating distribution.
A random sample of 5000 people from the city is selected.
\item State the exact distribution of the number of people with red hair in the sample.
\item Use a suitable Normal approximating distribution to find the probability that there are at least 160 people with red hair in the sample.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2013 Q2 [18]}}