| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Proving Poisson properties from first principles |
| Difficulty | Standard +0.8 This question requires proving fundamental Poisson properties from first principles using infinite series manipulation (part a(i)), then applying algebraic reasoning to derive variance (a(ii)). Part (b) involves linear transformations and understanding distributional properties. While the techniques are A-level appropriate, proving from first principles with infinite series and recognizing why Z isn't Poisson (despite matching mean-variance property) requires deeper understanding than routine application, placing it moderately above average difficulty. |
| Spec | 5.02a Discrete probability distributions: general5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(E(X(X-1)) = \sum_{x=0}^{\infty} x(x-1) \cdot \frac{e^{-\lambda}\lambda^x}{x!}\) | M1 | Attempt to use definition of expectation with Poisson probabilities |
| \(= \sum_{x=2}^{\infty} \frac{e^{-\lambda}\lambda^x}{(x-2)!}\) | M1 | Cancelling \(x(x-1)\) with \(x!\) to get \((x-2)!\), lower limit becomes 2 |
| \(= \lambda^2 e^{-\lambda} \sum_{x=2}^{\infty} \frac{\lambda^{x-2}}{(x-2)!} = \lambda^2 e^{-\lambda} \cdot e^{\lambda} = \lambda^2\) | A1 | Recognising sum equals \(e^{\lambda}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\text{Var}(X) = E(X^2) - [E(X)]^2\) and \(E(X^2) = E(X(X-1)) + E(X) = \lambda^2 + \lambda\) | M1 | Using \(E(X^2) = E(X(X-1)) + E(X)\) |
| \(\text{Var}(X) = \lambda^2 + \lambda - \lambda^2 = \lambda = E(X)\) | A1 | Correct completion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(E(Z) = 4E(Y) + 30 = 4(2.5) + 30 = 40\) | B1 | Correct \(E(Z) = 40\) |
| \(\text{Var}(Z) = 16\,\text{Var}(Y) = 16 \times 2.5 = 40\) | M1 | Using \(\text{Var}(aY+b) = a^2\text{Var}(Y)\) with \(\text{Var}(Y) = 2.5\) |
| \(\text{Var}(Z) = 40 = E(Z)\) | A1 | Correct conclusion shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(Z\) cannot take the value 0 (or values less than 30), whereas a Poisson distribution can take the value 0 / \(Z\) does not take integer values starting from 0 | B1 | Any valid reason e.g. \(Z\) takes values \(30, 34, 38, \ldots\) not \(0,1,2,\ldots\) |
# Question 7:
## Part (a)(i)
Prove that $E(X(X-1)) = \lambda^2$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $E(X(X-1)) = \sum_{x=0}^{\infty} x(x-1) \cdot \frac{e^{-\lambda}\lambda^x}{x!}$ | M1 | Attempt to use definition of expectation with Poisson probabilities |
| $= \sum_{x=2}^{\infty} \frac{e^{-\lambda}\lambda^x}{(x-2)!}$ | M1 | Cancelling $x(x-1)$ with $x!$ to get $(x-2)!$, lower limit becomes 2 |
| $= \lambda^2 e^{-\lambda} \sum_{x=2}^{\infty} \frac{\lambda^{x-2}}{(x-2)!} = \lambda^2 e^{-\lambda} \cdot e^{\lambda} = \lambda^2$ | A1 | Recognising sum equals $e^{\lambda}$ |
## Part (a)(ii)
Deduce that $\text{Var}(X) = E(X)$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{Var}(X) = E(X^2) - [E(X)]^2$ and $E(X^2) = E(X(X-1)) + E(X) = \lambda^2 + \lambda$ | M1 | Using $E(X^2) = E(X(X-1)) + E(X)$ |
| $\text{Var}(X) = \lambda^2 + \lambda - \lambda^2 = \lambda = E(X)$ | A1 | Correct completion |
## Part (b)(i)
Show that $\text{Var}(Z) = E(Z)$ where $Z = 4Y + 30$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $E(Z) = 4E(Y) + 30 = 4(2.5) + 30 = 40$ | B1 | Correct $E(Z) = 40$ |
| $\text{Var}(Z) = 16\,\text{Var}(Y) = 16 \times 2.5 = 40$ | M1 | Using $\text{Var}(aY+b) = a^2\text{Var}(Y)$ with $\text{Var}(Y) = 2.5$ |
| $\text{Var}(Z) = 40 = E(Z)$ | A1 | Correct conclusion shown |
## Part (b)(ii)
Give a reason why $Z$ is **not** Poisson
| Working/Answer | Mark | Guidance |
|---|---|---|
| $Z$ cannot take the value 0 (or values less than 30), whereas a Poisson distribution can take the value 0 / $Z$ does not take integer values starting from 0 | B1 | Any valid reason e.g. $Z$ takes values $30, 34, 38, \ldots$ not $0,1,2,\ldots$ |7
\begin{enumerate}[label=(\alph*)]
\item The random variable $X$ has a Poisson distribution with $\mathrm { E } ( X ) = \lambda$.
\begin{enumerate}[label=(\roman*)]
\item Prove, from first principles, that $\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }$.
\item Hence deduce that $\operatorname { Var } ( X ) = \mathrm { E } ( X )$.
\end{enumerate}\item The random variable $Y$ has a Poisson distribution with $\mathrm { E } ( Y ) = 2.5$.
Given that $Z = 4 Y + 30$ :
\begin{enumerate}[label=(\roman*)]
\item show that $\operatorname { Var } ( Z ) = \mathrm { E } ( Z )$;
\item give a reason why the distribution of $Z$ is not Poisson.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{fa3bf9d6-f064-4214-acff-d8b88c33a81e-19_2486_1714_221_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{fa3bf9d6-f064-4214-acff-d8b88c33a81e-20_2486_1714_221_153}
\end{center}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2011 Q7 [9]}}