| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | State conditions for Poisson approximation |
| Difficulty | Moderate -0.5 This question tests standard recall of Poisson approximation conditions and routine application of the approximation with given parameters. Part (ii) requires stating textbook conditions, while computational parts involve straightforward probability calculations using given distributions. The normal approximation in (iv)(B) is also standard procedure for large n, making this easier than average despite multiple parts. |
| Spec | 2.04b 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 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\binom{10}{1} \times 0.02^1 \times 0.98^9 = 0.1667\) | M1 for coefficient, M1 for probabilities, A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(n\) is large and \(p\) is small | B1, B1; allow appropriate numerical ranges | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda = 150 \times 0.02 = 3\); \(P(X=0) = e^{-3}\frac{3^0}{0!} = 0.0498\) or from tables \(= 0.0498\) | B1 for mean, M1 for calculation or use of tables, A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Expected number \(= 3\); \(P(X>3) = 1 - P(X \leq 3) = 1 - 0.6472 = 0.3528\) | B1 expected no \(= 3\), M1, A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Binomial\((2000, 0.02)\) | B1 for binomial, B1 for parameters | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Use Normal approx with \(\mu = np = 2000 \times 0.02 = 40\), \(\sigma^2 = npq = 2000 \times 0.02 \times 0.98 = 39.2\); \(P(X \leq 50) = P\left(Z \leq \frac{50.5-40}{\sqrt{39.2}}\right) = P(Z \leq 1.677) = \Phi(1.677) = 0.9532\) | B1, B1, B1 for continuity correction, M1 for probability using correct tail, A1 CAO | [5] |
# Question 2:
## Part (i)
| $\binom{10}{1} \times 0.02^1 \times 0.98^9 = 0.1667$ | M1 for coefficient, M1 for probabilities, A1 | [3] |
## Part (ii)
| $n$ is large and $p$ is small | B1, B1; allow appropriate numerical ranges | [2] |
## Part (iii)(A)
| $\lambda = 150 \times 0.02 = 3$; $P(X=0) = e^{-3}\frac{3^0}{0!} = 0.0498$ or from tables $= 0.0498$ | B1 for mean, M1 for calculation or use of tables, A1 | [3] |
## Part (iii)(B)
| Expected number $= 3$; $P(X>3) = 1 - P(X \leq 3) = 1 - 0.6472 = 0.3528$ | B1 expected no $= 3$, M1, A1 | [3] |
## Part (iv)(A)
| Binomial$(2000, 0.02)$ | B1 for binomial, B1 for parameters | [2] |
## Part (iv)(B)
| Use Normal approx with $\mu = np = 2000 \times 0.02 = 40$, $\sigma^2 = npq = 2000 \times 0.02 \times 0.98 = 39.2$; $P(X \leq 50) = P\left(Z \leq \frac{50.5-40}{\sqrt{39.2}}\right) = P(Z \leq 1.677) = \Phi(1.677) = 0.9532$ | B1, B1, B1 for continuity correction, M1 for probability using correct tail, A1 CAO | [5] |
---2 On average 2\% of a particular model of laptop computer are faulty. Faults occur independently and randomly.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that exactly 1 of a batch of 10 laptops is faulty.
\item State the conditions under which the use of a Poisson distribution is appropriate as an approximation to a binomial distribution.
\item A school buys a batch of 150 of these laptops. Use a Poisson approximating distribution to find the probability that\\
(A) there are no faulty laptops in the batch,\\
(B) there are more than the expected number of faulty laptops in the batch.
\item A large company buys a batch of 2000 of these laptops for its staff.\\
(A) State the exact distribution of the number of faulty laptops in this batch.\\
(B) Use a suitable approximating distribution to find the probability that there are at most 50 faulty laptops in this batch.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2010 Q2 [18]}}