| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Binomial to the Poisson distribution |
| Type | Find parameter from Poisson probabilities |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on standard Poisson distribution techniques. Part (i) is routine approximation with direct calculation, part (ii) requires simple algebraic manipulation of the Poisson formula, and part (iii) involves setting up and solving a basic inequality. All parts are textbook exercises requiring recall and standard procedures rather than problem-solving or insight. |
| Spec | 2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Po(3) | B1 | SOI |
| \(e^{-3}\left(\frac{3^3}{3!} + \frac{3^4}{4!} + \frac{3^5}{5!}\right)\) | M1 | Allow one or two extra terms (2 or 6 or both) |
| \(0.493\) (3 sf) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| A correct equation from \(P(0) = P(2)\) \(\left(\text{leading to } 1 = \frac{\lambda^2}{2}\right)\) | M1 | |
| \(\lambda = \sqrt{2}\) or \(1.41\) (3 sf) | A1 | CWO |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct inequality \(\left(\text{leading to } \frac{5.2^n}{n!} < \frac{5.2^{n+1}}{(n+1)!}\right)\) | B1 | |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(n + 1 < 5.2\) or \(1 < \frac{5.2}{n+1}\) | M1 | Simplify to a stage without exponentials, powers or factorials |
| Largest \(n\) is 4 | A1 | |
| Total: 2 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Po(3) | B1 | SOI |
| $e^{-3}\left(\frac{3^3}{3!} + \frac{3^4}{4!} + \frac{3^5}{5!}\right)$ | M1 | Allow one or two extra terms (2 or 6 or both) |
| $0.493$ (3 sf) | A1 | |
| **Total: 3** | | |
---
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| A correct equation from $P(0) = P(2)$ $\left(\text{leading to } 1 = \frac{\lambda^2}{2}\right)$ | M1 | |
| $\lambda = \sqrt{2}$ or $1.41$ (3 sf) | A1 | CWO |
| **Total: 2** | | |
---
## Question 5(iii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct inequality $\left(\text{leading to } \frac{5.2^n}{n!} < \frac{5.2^{n+1}}{(n+1)!}\right)$ | B1 | |
| **Total: 1** | | |
---
## Question 5(iii)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $n + 1 < 5.2$ or $1 < \frac{5.2}{n+1}$ | M1 | Simplify to a stage without exponentials, powers or factorials |
| Largest $n$ is 4 | A1 | |
| **Total: 2** | | |
---5\\
\begin{enumerate}[label=(\roman*)]
\item The random variable $X$ has the distribution $\mathrm { B } ( 300,0.01 )$. Use a Poisson approximation to find $\mathrm { P } ( 2 < X < 6 )$.\\
\item The random variable $Y$ has the distribution $\mathrm { Po } ( \lambda )$, and $\mathrm { P } ( Y = 0 ) = \mathrm { P } ( Y = 2 )$. Find $\lambda$.\\
\item The random variable $Z$ has the distribution $\mathrm { Po } ( 5.2 )$ and it is given that $\mathrm { P } ( Z = n ) < \mathrm { P } ( Z = n + 1 )$.
\begin{enumerate}[label=(\alph*)]
\item Write down an inequality in $n$.
\item Hence or otherwise find the largest possible value of $n$.
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2019 Q5 [8]}}