| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Poisson to the Normal distribution |
| Type | Multiple approximations in one question |
| Difficulty | Standard +0.3 This is a straightforward application of standard approximation techniques taught in S2. Part (a) is routine Poisson-to-Normal approximation with continuity correction. Part (b)(i) requires recalling the condition np small for Poisson approximation. Part (b)(ii) uses a clever but standard trick (considering failures instead) that appears in textbooks. All steps are procedural with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| N(24, 24) | B1 | Normal, mean 24 stated or implied |
| Variance or SD equal to mean | B1 | |
| \(1-\Phi\left(\frac{30.5-24}{\sqrt{24}}\right) = 1-\Phi(1.327)\) | M1 | Standardise 30 with \(\lambda\) and \(\sqrt{\lambda}\), allow cc or \(\sqrt{}\) errors |
| \(= 0.0923\) | A1 | .131 or .1103; 30.5 and \(\sqrt{\lambda}\) correct |
| A1 5 | Answer in range [0.092, 0.0925] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(p\) or \(np\ [=196]\) is too large | B1 1 | Correct reason, no wrong reason |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Consider \((200-E)\) | M1 | Consider complement |
| \((200-E) \sim \text{Po}(4)\) | M1 | \(\text{Po}(200\times0.02)\) |
| \(P(\geq 6)\ [= 1-0.7851]\) | M1 | Poisson tables used, correct tail, e.g. 0.3712 or 0.1107 |
| \(= 0.2149\) | A1 4 | Answer a.r.t. 0.215 only |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| N(24, 24) | B1 | Normal, mean 24 stated or implied |
| Variance or SD equal to mean | B1 | |
| $1-\Phi\left(\frac{30.5-24}{\sqrt{24}}\right) = 1-\Phi(1.327)$ | M1 | Standardise 30 with $\lambda$ and $\sqrt{\lambda}$, allow cc or $\sqrt{}$ errors |
| $= 0.0923$ | A1 | .131 or .1103; 30.5 and $\sqrt{\lambda}$ correct |
| | A1 **5** | Answer in range [0.092, 0.0925] |
## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $p$ or $np\ [=196]$ is too large | B1 **1** | Correct reason, no wrong reason |
## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Consider $(200-E)$ | M1 | Consider complement |
| $(200-E) \sim \text{Po}(4)$ | M1 | $\text{Po}(200\times0.02)$ |
| $P(\geq 6)\ [= 1-0.7851]$ | M1 | Poisson tables used, correct tail, e.g. 0.3712 or 0.1107 |
| $= 0.2149$ | A1 **4** | Answer a.r.t. 0.215 only |
---6
\begin{enumerate}[label=(\alph*)]
\item The random variable $D$ has the distribution $\operatorname { Po } ( 24 )$. Use a suitable approximation to find $P ( D > 30 )$.
\item An experiment consists of 200 trials. For each trial, the probability that the result is a success is 0.98 , independent of all other trials. The total number of successes is denoted by $E$.
\begin{enumerate}[label=(\roman*)]
\item Explain why the distribution of $E$ cannot be well approximated by a Poisson distribution.
\item By considering the number of failures, use an appropriate Poisson approximation to find $\mathrm { P } ( E \leqslant 194 )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR S2 2010 Q6 [10]}}