| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| 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 standard Poisson properties from first principles using summation manipulation and the exponential series. While it involves algebraic manipulation of infinite series and understanding of variance decomposition (Var(X) = E(X²) - [E(X)]²), these are well-rehearsed proofs that follow standard textbook approaches. The hint for part (b) guides students through the key step. More challenging than routine application but less demanding than questions requiring novel insight or complex multi-stage reasoning. |
| Spec | 5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(E(X) = \sum x \times P(X=x)\) | M1 | Use of |
| \(= \sum_{x=0}^{\infty} x \times \frac{e^{-\lambda}\lambda^x}{x!} = \lambda \times \sum_{x=1}^{\infty} \frac{e^{-\lambda}\lambda^{x-1}}{(x-1)!}\) | M1 | Factor of \(\lambda\), cancelling of \(x\), (ignore change in limits) |
| \(= \lambda \times \sum P(X=x) = \lambda \times 1 = \lambda\) | M1 | AG; must be clear |
| \(G(t) = e^{\lambda t - \lambda}\) or \(M(t) = e^{\lambda e^t - \lambda}\) | (B1) | Either CAO |
| Alternative: \(E(X) = \frac{dG(t)}{dt}\bigg\ | _1\) or \(\frac{dM(t)}{dt}\bigg\ | _0\) |
| \(\left[\lambda e^{\lambda t - \lambda}\right]_1\) or \(\left[\lambda e^t e^{\lambda e^t - \lambda}\right]_0 = \lambda\) | (A1) | AG; correct derivation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(E(X(X-1)) = \sum_{x=0}^{\infty} x(x-1) \times \frac{e^{-\lambda}\lambda^x}{x!}\) | M1 | Use of |
| \(= \lambda^2 \times \sum_{x=2}^{\infty} \frac{e^{-\lambda}\lambda^{x-2}}{(x-2)!}\) | M1 | Factor of \(\lambda^2\), cancelling of \(x(x-1)\), (ignore change in limits) |
| \(= \lambda^2 \times \sum P(X=x) = \lambda^2 \times 1 = \lambda^2\) | M1 | AG; must justify |
| \(\text{Var}(X) = E(X^2) - (E(X))^2 = E(X(X-1)) + E(X) - (E(X))^2\) | M1 | |
| \(= \lambda^2 + \lambda - \lambda^2 = \lambda\) | A1 | AG; must be clear |
| Alternative: \(\text{Var}(X) = \frac{d^2G(t)}{d^2t}\bigg\ | _1 + \lambda - \lambda^2\) or \(\frac{d^2M(t)}{d^2t}\bigg\ | _0 - \lambda^2\) |
| \(= \left[\lambda^2 e^{\lambda t-\lambda}\right]_1 + \lambda - \lambda^2 = \lambda\) | (A2) | AG; correct derivation |
| or \(= \left[\lambda e^t e^{\lambda e^t - \lambda} + \lambda^2 e^{2t} e^{\lambda e^t - \lambda}\right]_0 - \lambda^2 = \lambda\) | (A1) | AG; correct derivation |
# Question 6:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $E(X) = \sum x \times P(X=x)$ | M1 | Use of |
| $= \sum_{x=0}^{\infty} x \times \frac{e^{-\lambda}\lambda^x}{x!} = \lambda \times \sum_{x=1}^{\infty} \frac{e^{-\lambda}\lambda^{x-1}}{(x-1)!}$ | M1 | Factor of $\lambda$, cancelling of $x$, (ignore change in limits) |
| $= \lambda \times \sum P(X=x) = \lambda \times 1 = \lambda$ | M1 | AG; must be clear |
| $G(t) = e^{\lambda t - \lambda}$ or $M(t) = e^{\lambda e^t - \lambda}$ | (B1) | Either CAO |
| **Alternative:** $E(X) = \frac{dG(t)}{dt}\bigg\|_1$ or $\frac{dM(t)}{dt}\bigg\|_0$ | (M1) | Use of either |
| $\left[\lambda e^{\lambda t - \lambda}\right]_1$ or $\left[\lambda e^t e^{\lambda e^t - \lambda}\right]_0 = \lambda$ | (A1) | AG; correct derivation |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $E(X(X-1)) = \sum_{x=0}^{\infty} x(x-1) \times \frac{e^{-\lambda}\lambda^x}{x!}$ | M1 | Use of |
| $= \lambda^2 \times \sum_{x=2}^{\infty} \frac{e^{-\lambda}\lambda^{x-2}}{(x-2)!}$ | M1 | Factor of $\lambda^2$, cancelling of $x(x-1)$, (ignore change in limits) |
| $= \lambda^2 \times \sum P(X=x) = \lambda^2 \times 1 = \lambda^2$ | M1 | AG; must justify |
| $\text{Var}(X) = E(X^2) - (E(X))^2 = E(X(X-1)) + E(X) - (E(X))^2$ | M1 | |
| $= \lambda^2 + \lambda - \lambda^2 = \lambda$ | A1 | AG; must be clear |
| **Alternative:** $\text{Var}(X) = \frac{d^2G(t)}{d^2t}\bigg\|_1 + \lambda - \lambda^2$ or $\frac{d^2M(t)}{d^2t}\bigg\|_0 - \lambda^2$ | (M2) | Use of either |
| $= \left[\lambda^2 e^{\lambda t-\lambda}\right]_1 + \lambda - \lambda^2 = \lambda$ | (A2) | AG; correct derivation |
| or $= \left[\lambda e^t e^{\lambda e^t - \lambda} + \lambda^2 e^{2t} e^{\lambda e^t - \lambda}\right]_0 - \lambda^2 = \lambda$ | (A1) | AG; correct derivation |
---6 The random variable $X$ has a Poisson distribution with parameter $\lambda$.
\begin{enumerate}[label=(\alph*)]
\item Prove that $\mathrm { E } ( X ) = \lambda$.
\item By first proving that $\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }$, or otherwise, prove that $\operatorname { Var } ( X ) = \lambda$.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2006 Q6 [8]}}