| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Proving Poisson properties from first principles |
| Difficulty | Standard +0.8 This is a substantial multi-part S3 question requiring proof from first principles (manipulating infinite series with Poisson pmf), deducing variance formula, applying properties of linear combinations of Poisson variables, understanding when distributions remain Poisson, and using Normal approximation to Poisson. The proof in part (a)(i) requires careful algebraic manipulation of series and is non-trivial for A-level, though the remaining parts are more standard applications. Overall harder than typical S3 questions but not exceptionally difficult. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02m Poisson: mean = variance = lambda5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a)(i) | ||
| \(E(X(X-1)) = \sum_{x=0}^{\infty} x(x-1) \times \frac{e^{-\lambda}\lambda^x}{x!} = \) | M1 | \(\sum x(x-1) \times P(X = x)\) used; Ignore limits until A1 |
| \(\sum_{x=2}^{\infty} \frac{e^{-\lambda}\lambda^x}{(x-2)!} = \) | M1 | \(\frac{x(x-1)}{x!} = \frac{1}{(x-2)!}\) used |
| \(\lambda^2 e^{-\lambda} \sum_{x=2}^{\infty} \frac{\lambda^{x-2}}{(x-2)!} = \) | M1 | Factor of \(\lambda^2 e^{-\lambda}\) used |
| \((\lambda^2 e^{-\lambda}) \times (e^{\lambda}) = \lambda^2\) | A1 | Fully correct derivation; AG; 4 marks |
| Part (a)(ii) | ||
| \(\text{Var}(X) = E(X(X-1)) + E(X) - (\text{E}(X))^2\) | M1 | Used |
| \(= \lambda^2 + \lambda - \lambda^2 = \lambda\) | A1 | Fully correct derivation; AG; 2 marks |
| Part (b)(i) | ||
| \(E(D) = 5 - 2 = 3\); \(\text{Var}(D) = 5 + 2 = 7\) | B1; B1 | CAO; 2 marks |
| Part (b)(ii) | ||
| \(E(F) = (2 \times 5) + 10 = 20\); \(\text{Var}(F) = 2^2 \times 5 = 20\) | B1; M1; A1 | CAO; \(2^2\text{V}(X_1) + 0\) CAO; 3 marks |
| Part (b)(iii) | ||
| \(D\): Mean \(\neq\) Variance | B1 | Negative values possible |
| \(F\): Values \(< 10\) impossible; Odd values impossible | B1 | \(2X_1 = X_1 + X_1\) is not sum of independent Po variables; 2 marks |
| Part (c) | ||
| \(B \sim \text{Po}(5)\); \(C \sim \text{Po}(2)\) | B1 | CAO |
| \(T = 24 \times (5 + 2) \sim \text{Po}(168)\) | B1 | CAO |
| \(T \sim \text{approx } N(168, 168)\) | M1 | Normal with \(\mu = \sigma^2\) |
| \(P(T_{100} > 175) = P(T_S > 175.5)\) | B1 | 175.5 |
| \(= P\left(Z > \frac{175.5 - 168}{\sqrt{168}}\right) = P(Z > 0.58) = \) | M1 | Standardising 174.5, 175 or 175.5 with \(\mu = \sigma^2\) |
| \(1 - P(Z < 0.58) = 0.28(0) \text{ to } 0.283\) | m1; A1 | Area change AWFW; 6 marks |
| **Part (a)(i)** |
|---|
| $E(X(X-1)) = \sum_{x=0}^{\infty} x(x-1) \times \frac{e^{-\lambda}\lambda^x}{x!} = $ | M1 | $\sum x(x-1) \times P(X = x)$ used; Ignore limits until A1 |
| $\sum_{x=2}^{\infty} \frac{e^{-\lambda}\lambda^x}{(x-2)!} = $ | M1 | $\frac{x(x-1)}{x!} = \frac{1}{(x-2)!}$ used |
| $\lambda^2 e^{-\lambda} \sum_{x=2}^{\infty} \frac{\lambda^{x-2}}{(x-2)!} = $ | M1 | Factor of $\lambda^2 e^{-\lambda}$ used |
| $(\lambda^2 e^{-\lambda}) \times (e^{\lambda}) = \lambda^2$ | A1 | Fully correct derivation; AG; 4 marks |
| **Part (a)(ii)** |
|---|
| $\text{Var}(X) = E(X(X-1)) + E(X) - (\text{E}(X))^2$ | M1 | Used |
| $= \lambda^2 + \lambda - \lambda^2 = \lambda$ | A1 | Fully correct derivation; AG; 2 marks |
| **Part (b)(i)** |
|---|
| $E(D) = 5 - 2 = 3$; $\text{Var}(D) = 5 + 2 = 7$ | B1; B1 | CAO; 2 marks |
| **Part (b)(ii)** |
|---|
| $E(F) = (2 \times 5) + 10 = 20$; $\text{Var}(F) = 2^2 \times 5 = 20$ | B1; M1; A1 | CAO; $2^2\text{V}(X_1) + 0$ CAO; 3 marks |
| **Part (b)(iii)** |
|---|
| $D$: Mean $\neq$ Variance | B1 | Negative values possible |
| $F$: Values $< 10$ impossible; Odd values impossible | B1 | $2X_1 = X_1 + X_1$ is not sum of independent Po variables; 2 marks |
| **Part (c)** |
|---|
| $B \sim \text{Po}(5)$; $C \sim \text{Po}(2)$ | B1 | CAO |
| $T = 24 \times (5 + 2) \sim \text{Po}(168)$ | B1 | CAO |
| $T \sim \text{approx } N(168, 168)$ | M1 | Normal with $\mu = \sigma^2$ |
| $P(T_{100} > 175) = P(T_S > 175.5)$ | B1 | 175.5 |
| $= P\left(Z > \frac{175.5 - 168}{\sqrt{168}}\right) = P(Z > 0.58) = $ | M1 | Standardising 174.5, 175 or 175.5 with $\mu = \sigma^2$ |
| $1 - P(Z < 0.58) = 0.28(0) \text{ to } 0.283$ | m1; A1 | Area change AWFW; 6 marks |
**Total: 19 marks**
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## TOTAL: 75 marks7
\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 $\operatorname { Var } ( X ) = \lambda$.
\end{enumerate}\item The independent Poisson random variables $X _ { 1 }$ and $X _ { 2 }$ are such that $\mathrm { E } \left( X _ { 1 } \right) = 5$ and $\mathrm { E } \left( X _ { 2 } \right) = 2$.
The random variables $D$ and $F$ are defined by
$$D = X _ { 1 } - X _ { 2 } \quad \text { and } \quad F = 2 X _ { 1 } + 10$$
\begin{enumerate}[label=(\roman*)]
\item Determine the mean and the variance of $D$.
\item Determine the mean and the variance of $F$.
\item For each of the variables $D$ and $F$, give a reason why the distribution is not Poisson.
\end{enumerate}\item The daily number of black printer cartridges sold by a shop may be modelled by a Poisson distribution with a mean of 5 .
Independently, the daily number of colour printer cartridges sold by the same shop may be modelled by a Poisson distribution with a mean of 2.
Use a distributional approximation to estimate the probability that the total number of black and colour printer cartridges sold by the shop during a 4 -week period ( 24 days) exceeds 175.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2008 Q7 [19]}}