| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 19 |
| 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 S2 question testing routine application of Poisson approximation to binomial. Part (ii) requires stating the standard conditions (n large, p small, np moderate) which is textbook recall. The calculations are straightforward with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02i 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 distributions5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Exactly one}) = \dbinom{25}{1}\times0.03^1\times0.97^{24} = 0.361\) | M1 | Binomial calculation with correct structure; allow \(25\times p\times(1-p)^{24}\) |
| A1 | Allow 0.3611 and 0.36 www; A0 for 0.3612 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(n\) is large | B1 | \(n\) large or sample is large; or \(n > 30\) |
| \(p\) is small | B1 | \(p\) is small, or \(np \approx np(1-p)\); B0 for "probability" is small unless "probability" is correctly defined; or \(np < 10\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 250\times0.03 = 7.5\) | B1 | For mean (SOI) |
| \(P(\text{exactly }10) = e^{-7.5}\dfrac{7.5^{10}}{10!}\) or from tables \(= 0.8622 - 0.7764 = 0.0858\) | M1 | For Poisson probability calculation; or using \(P(X\leq10)-P(X\leq9)\) with Poisson tables |
| A1 | Allow 0.08583 or 0.086www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{At least }10) = 1 - P(X\leq9) = 1 - 0.7764 = 0.2236\) | M1 | For using \(1 - P(X\leq9)\) |
| A1 | CAO; allow 0.224 www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= 2000\times0.03 = 60\); Variance \(= 2000\times0.03\times0.97 = 58.2\) | B1, B1 | Normal approximation used; for parameters (SOI); award full credit for use of Normal approx to Poisson \(N(60,60)\) |
| \(X \sim N(60,\ 58.2)\) | ||
| \(P(X\geq50) = P\!\left(Z \geq \dfrac{49.5-60}{\sqrt{58.2}}\right)\) | B1 | For correct continuity correction |
| \(= P(Z > -1.376) = \Phi(1.376)\) | M1 | For probability using correct structure; \(N(60,60)\) leads to \(P(Z > -1.356) = 0.9125\) (or 0.913) |
| \(= 0.9157\) (allow 0.9156 and 0.916) | A1 | CAO (do not FT wrong or omitted CC); allow 0.9124 (or 0.912) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using Poisson approximation, mean \(\lambda = np\ (=0.002n)\) or \(\lambda = n\times0.002\ (=0.002n)\) | B1 | For evidence of using \(np\) from binomial distribution to give \(\lambda = 0.002n\); need to see use of \(n\times p\) or obtaining \(0.002n\) from \(B(n,0.002)\) |
| \(P(\text{at most one fake coin}) = P(\text{zero or one fake coins})\) \(= e^{-\lambda}\dfrac{\lambda^0}{0!} + e^{-\lambda}\dfrac{\lambda^1}{1!} = e^{-\lambda} + \lambda e^{-\lambda}\) AG | B1 | Evidence of using \(P(X=0)+P(X=1)\) with \(\lambda = 0.002n\); NB ANSWER GIVEN |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 - \dfrac{\lambda^2}{2} = 0.995\) | M1 | For equation in \(\lambda\) or equivalent equation in \(n\) |
| \(\lambda^2 = 0.01\) so \(\lambda = 0.1\) | A1 | For \(\lambda\) SOI or for \(n^2 = 2500\) |
| \(n = 50\) | A1 | CAO |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Exactly one}) = \dbinom{25}{1}\times0.03^1\times0.97^{24} = 0.361$ | M1 | Binomial calculation with correct structure; allow $25\times p\times(1-p)^{24}$ |
| | A1 | Allow 0.3611 and 0.36 www; A0 for 0.3612 |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $n$ is large | B1 | $n$ large or sample is large; or $n > 30$ |
| $p$ is small | B1 | $p$ is small, or $np \approx np(1-p)$; B0 for "probability" is small unless "probability" is correctly defined; or $np < 10$ |
## Part (iii)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 250\times0.03 = 7.5$ | B1 | For mean (SOI) |
| $P(\text{exactly }10) = e^{-7.5}\dfrac{7.5^{10}}{10!}$ or from tables $= 0.8622 - 0.7764 = 0.0858$ | M1 | For Poisson probability calculation; or using $P(X\leq10)-P(X\leq9)$ with Poisson tables |
| | A1 | Allow 0.08583 or 0.086www |
## Part (iii)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{At least }10) = 1 - P(X\leq9) = 1 - 0.7764 = 0.2236$ | M1 | For using $1 - P(X\leq9)$ |
| | A1 | CAO; allow 0.224 www |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= 2000\times0.03 = 60$; Variance $= 2000\times0.03\times0.97 = 58.2$ | B1, B1 | Normal approximation used; for parameters (SOI); award full credit for use of Normal approx to Poisson $N(60,60)$ |
| $X \sim N(60,\ 58.2)$ | | |
| $P(X\geq50) = P\!\left(Z \geq \dfrac{49.5-60}{\sqrt{58.2}}\right)$ | B1 | For correct continuity correction |
| $= P(Z > -1.376) = \Phi(1.376)$ | M1 | For probability using correct structure; $N(60,60)$ leads to $P(Z > -1.356) = 0.9125$ (or 0.913) |
| $= 0.9157$ (allow 0.9156 and 0.916) | A1 | CAO (do not FT wrong or omitted CC); allow 0.9124 (or 0.912) |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using Poisson approximation, mean $\lambda = np\ (=0.002n)$ or $\lambda = n\times0.002\ (=0.002n)$ | B1 | For evidence of using $np$ from binomial distribution to give $\lambda = 0.002n$; need to **see use of $n\times p$** or obtaining $0.002n$ from $B(n,0.002)$ |
| $P(\text{at most one fake coin}) = P(\text{zero or one fake coins})$ $= e^{-\lambda}\dfrac{\lambda^0}{0!} + e^{-\lambda}\dfrac{\lambda^1}{1!} = e^{-\lambda} + \lambda e^{-\lambda}$ **AG** | B1 | Evidence of using $P(X=0)+P(X=1)$ with $\lambda = 0.002n$; **NB ANSWER GIVEN** |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 - \dfrac{\lambda^2}{2} = 0.995$ | M1 | For equation in $\lambda$ or equivalent equation in $n$ |
| $\lambda^2 = 0.01$ so $\lambda = 0.1$ | A1 | For $\lambda$ SOI or for $n^2 = 2500$ |
| $n = 50$ | A1 | CAO |
---2 It was stated in 2012 that $3 \%$ of $\pounds 1$ coins were fakes. Throughout this question, you should assume that this is still the case.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that, in a random selection of $25 \pounds 1$ coins, there is exactly one fake coin.
A random sample of $250 \pounds 1$ coins is selected.
\item Explain why a Poisson distribution is an appropriate approximating distribution for the number of fake coins in the sample.
\item Use a Poisson distribution to find the probability that, in this sample, there are\\
(A) exactly 10 fake coins,\\
(B) at least 10 fake coins.
\item Use a suitable approximating distribution to find the probability that there are at least 50 fake coins in a sample of 2000 coins.
It is known that $0.2 \%$ of another type of coin are fakes.
\item A random sample of size $n$ of these coins is taken. Using a Poisson approximating distribution, show that the probability of at most one fake coin in the sample is equal to $\mathrm { e } ^ { - \lambda } + \lambda \mathrm { e } ^ { - \lambda }$, where $\lambda = 0.002 n$.
\item Use the approximation $\mathrm { e } ^ { - \lambda } + \lambda \mathrm { e } ^ { - \lambda } \approx 1 - \frac { \lambda ^ { 2 } } { 2 }$ for small values of $\lambda$ to estimate the value of $n$ for which the probability in part ( $\mathbf { v }$ ) is equal to 0.995 .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2015 Q2 [19]}}