| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Topic | Poisson distribution |
| Type | Proving Poisson properties from first principles |
| Difficulty | Challenging +1.2 Part (i) requires deriving the MGF for Poisson distribution and proving the sum property - standard Further Maths content requiring calculus and series manipulation but following well-established methods. Part (ii) applies these results to a straightforward context with routine probability calculations. While technically demanding for A-level, these are textbook exercises in Further Maths statistics rather than requiring novel insight. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
**Question 4(i)(a), 4(i)(b), 4(ii)(a), 4(ii)(b), 4(ii)(c)**
**(i)(a)**
$M(t) = \sum \dfrac{\lambda^r}{r!} e^{-\lambda} \cdot e^{tr} = e^{-\lambda} \sum \dfrac{(\lambda e^t)^r}{r!}$ — M1 (Allow via expansion)
$= e^{-\lambda} \cdot e^{\lambda e^t} = e^{\lambda(e^t - 1)}$ — M1A1 (Needs to recognise series for M1; PGF quoted: M0M0A0)
**[3]**
**(i)(b)**
$M_Z(t) = M_X(t) \cdot M_Y(t) = e^{\mu(e^t-1)} \cdot e^{\nu(e^t-1)}$ — M1 (Multiply two MGFs)
$= e^{(\mu+\nu)(e^t-1)} \Rightarrow Z \sim \mathrm{Po}(\mu + \nu)$ — A1 (Needs one intermediate step)
**[2]**
**(ii)(a)**
From tables $2 + k = 3.1 \Rightarrow k = 1.101 = \mathbf{1.10}$ (3sf) — M1A1
Or: $e^{-(2+k)} = 0.045 \Rightarrow 2 + k = \ln 22.22 \Rightarrow k = 1.101 = 1.10$ (3sf)
**[2]**
**(ii)(b)**
$\mathrm{P}(3) = e^{-1.1} \times \dfrac{1.1^3}{3!} = \mathbf{0.0740}$ — M1, A1
Or $\mathrm{P}(\leq 3) - \mathrm{P}(\leq 2) = 0.9743 - 0.9004$; Answer in range $[0.0738, 0.0740]$; $\lambda = k$ needed for M1
**[2]**
**(ii)(c)**
Using mean of 3.1: $\mathrm{P}(\leq 5) - \mathrm{P}(\leq 1) = 0.9057 - 0.1847 = 0.7210 = \mathbf{0.721}$ (3 sf) — M1, A1 (Or from series $\pm 1$ term)
**[2]**4 (i) (a) Derive the moment generating function for a Poisson distribution with mean $\lambda$.\\
(b) The independent random variables $X$ and $Y$ are such that $X \sim \operatorname { Po } ( \mu )$ and $Y \sim \operatorname { Po } ( v )$. Use moment generating functions to show that $( X + Y ) \sim \operatorname { Po } ( \mu + v )$.\\
(ii) The number of goals scored per match by Camford Academicals FC may be modelled by a Poisson distribution with mean 2. The number of goals scored against Camford during a match may be modelled by an independent Poisson distribution with mean $k$. The probability that no goals are scored, by either side, in a match involving Camford is 0.045 . Find\\
(a) the value of $k$,\\
(b) the probability that exactly 3 goals are scored against Camford in a match,\\
(c) the probability that the total number of goals scored, in a match involving Camford, is between 2 and 5 inclusive.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2015 Q4 [11]}}