| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Proving Poisson properties from first principles |
| Difficulty | Challenging +1.2 This question requires proving E(X)=λ from first principles using series manipulation, deriving variance using given information, and working with mode inequalities. While it involves multiple parts and some algebraic manipulation of the Poisson pmf, these are standard S3/Further Maths Statistics techniques. Part (b)(i) requires inequality manipulation but follows a predictable pattern, and part (c) is a routine normal approximation. The proof in (a)(i) is the most demanding element but is a well-known derivation that students at this level would have practiced. |
| Spec | 2.04d Normal approximation to binomial5.02a Discrete probability distributions: general5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = \sum_{x=0}^{\infty} x \cdot \frac{e^{-\lambda}\lambda^x}{x!}\) | M1 | Correct formula for expectation with Poisson probabilities |
| \(= \sum_{x=1}^{\infty} \frac{e^{-\lambda}\lambda^x}{(x-1)!}\) | M1 | Cancelling \(x\) with \(x!\) and removing \(x=0\) term |
| \(= \lambda e^{-\lambda} \sum_{x=1}^{\infty} \frac{\lambda^{x-1}}{(x-1)!} = \lambda e^{-\lambda} \cdot e^{\lambda} = \lambda\) | A1 | Recognising sum \(= e^{\lambda}\), completing proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X(X-1)) = E(X^2) - E(X)\), so \(E(X^2) = \lambda^2 + \lambda\) | M1 | Using \(\text{Var}(X) = E(X^2) - [E(X)]^2\) |
| \(\text{Var}(X) = \lambda^2 + \lambda - \lambda^2 = \lambda\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=m) \geqslant P(X=m-1)\) gives \(\frac{e^{-\lambda}\lambda^m}{m!} \geqslant \frac{e^{-\lambda}\lambda^{m-1}}{(m-1)!}\) | M1 | Setting up inequality using Poisson formula |
| Simplifying: \(\lambda \geqslant m\), i.e. \(m \leqslant \lambda\) | A1 | Correct deduction |
| \(P(X=m) \geqslant P(X=m+1)\) gives \(\frac{\lambda^m}{m!} \geqslant \frac{\lambda^{m+1}}{(m+1)!}\), so \(m+1 \geqslant \lambda\), i.e. \(m \geqslant \lambda - 1\) | A1 | Both inequalities giving \(\lambda - 1 \leqslant m \leqslant \lambda\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\lambda = 4.9\), so \(\lambda - 1 = 3.9 \leqslant m \leqslant 4.9\), hence \(m = 4\) | B1 | Identifying \(m = 4\) |
| \(P(X=4) = \frac{e^{-4.9}(4.9)^4}{4!}\) | M1 | Correct calculation |
| \(= 0.1728\) (awrt) | A1 | Accept 0.173 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Standard deviation \(= 15.5\), so \(\text{Var}(Y) = 15.5^2 = 240.25\); for Poisson \(\lambda = 240.25\) | B1 | Identifying \(\lambda = 240.25\) |
| Since \(\lambda\) is large, approximate \(Y \sim N(240.25, 240.25)\) | M1 | Normal approximation stated |
| Mode \(d\): since \(\lambda\) not integer, \(d = 240\) (as \(\lambda - 1 < d \leqslant \lambda\) gives \(d = 240\)) | B1 | Correct value of \(d = 240\) |
| \(P(Y \geqslant 240) = P\left(Z \geqslant \frac{239.5 - 240.25}{\sqrt{240.25}}\right)\) with continuity correction | M1 | Correct continuity correction and standardising |
| \(= P(Z \geqslant -0.0484) = \Phi(0.0484) \approx 0.519\) | A1 | Correct final answer (accept answers rounding to 0.519) |
# Question 7:
## Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = \sum_{x=0}^{\infty} x \cdot \frac{e^{-\lambda}\lambda^x}{x!}$ | M1 | Correct formula for expectation with Poisson probabilities |
| $= \sum_{x=1}^{\infty} \frac{e^{-\lambda}\lambda^x}{(x-1)!}$ | M1 | Cancelling $x$ with $x!$ and removing $x=0$ term |
| $= \lambda e^{-\lambda} \sum_{x=1}^{\infty} \frac{\lambda^{x-1}}{(x-1)!} = \lambda e^{-\lambda} \cdot e^{\lambda} = \lambda$ | A1 | Recognising sum $= e^{\lambda}$, completing proof |
## Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X(X-1)) = E(X^2) - E(X)$, so $E(X^2) = \lambda^2 + \lambda$ | M1 | Using $\text{Var}(X) = E(X^2) - [E(X)]^2$ |
| $\text{Var}(X) = \lambda^2 + \lambda - \lambda^2 = \lambda$ | A1 | Correct answer |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=m) \geqslant P(X=m-1)$ gives $\frac{e^{-\lambda}\lambda^m}{m!} \geqslant \frac{e^{-\lambda}\lambda^{m-1}}{(m-1)!}$ | M1 | Setting up inequality using Poisson formula |
| Simplifying: $\lambda \geqslant m$, i.e. $m \leqslant \lambda$ | A1 | Correct deduction |
| $P(X=m) \geqslant P(X=m+1)$ gives $\frac{\lambda^m}{m!} \geqslant \frac{\lambda^{m+1}}{(m+1)!}$, so $m+1 \geqslant \lambda$, i.e. $m \geqslant \lambda - 1$ | A1 | Both inequalities giving $\lambda - 1 \leqslant m \leqslant \lambda$ |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = 4.9$, so $\lambda - 1 = 3.9 \leqslant m \leqslant 4.9$, hence $m = 4$ | B1 | Identifying $m = 4$ |
| $P(X=4) = \frac{e^{-4.9}(4.9)^4}{4!}$ | M1 | Correct calculation |
| $= 0.1728$ (awrt) | A1 | Accept 0.173 |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Standard deviation $= 15.5$, so $\text{Var}(Y) = 15.5^2 = 240.25$; for Poisson $\lambda = 240.25$ | B1 | Identifying $\lambda = 240.25$ |
| Since $\lambda$ is large, approximate $Y \sim N(240.25, 240.25)$ | M1 | Normal approximation stated |
| Mode $d$: since $\lambda$ not integer, $d = 240$ (as $\lambda - 1 < d \leqslant \lambda$ gives $d = 240$) | B1 | Correct value of $d = 240$ |
| $P(Y \geqslant 240) = P\left(Z \geqslant \frac{239.5 - 240.25}{\sqrt{240.25}}\right)$ with continuity correction | M1 | Correct continuity correction and standardising |
| $= P(Z \geqslant -0.0484) = \Phi(0.0484) \approx 0.519$ | A1 | Correct final answer (accept answers rounding to 0.519) |7 The random variable $X$ has a Poisson distribution with parameter $\lambda$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Prove, from first principles, that $\mathrm { E } ( X ) = \lambda$.
\item Hence, given that $\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }$, find, in terms of $\lambda$, an expression for $\operatorname { Var } ( X )$.
\end{enumerate}\item The mode, $m$, of $X$ is such that
$$\mathrm { P } ( X = m ) \geqslant \mathrm { P } ( X = m - 1 ) \quad \text { and } \quad \mathrm { P } ( X = m ) \geqslant \mathrm { P } ( X = m + 1 )$$
\begin{enumerate}[label=(\roman*)]
\item Show that $\lambda - 1 \leqslant m \leqslant \lambda$.
\item Given that $\lambda = 4.9$, determine $\mathrm { P } ( X = m )$.
\end{enumerate}\item The random variable $Y$ has a Poisson distribution with mode $d$ and standard deviation 15.5.
Use a distributional approximation to estimate $\mathrm { P } ( Y \geqslant d )$.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{b855b5b3-097e-4894-aaec-d77f515949b0-19_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2010 Q7 [15]}}