| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Topic | Poisson distribution |
| Type | Conditional probability with Poisson |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on Poisson distribution requiring standard techniques: (i) conditional probability using P(A|B) = P(A∩B)/P(B) with table lookup, (ii) algebraic manipulation of the Poisson formula to derive a given equation, and (iii) simple iterative numerical method. All parts are routine applications with no novel insight required, making it slightly easier than average. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks |
|---|---|
| (i) \(P(X > 6 | X > 3) = \frac{(1 - 0.7622)}{(1 - 0.2650)}\) M1A1 |
(i) $P(X > 6 | X > 3) = \frac{(1 - 0.7622)}{(1 - 0.2650)}$ **M1A1**
$= 0.324$ **A1** [3]
(ii) $P(\leqslant 1) = e^{-\lambda}(1 - \lambda) = \frac{1}{2}$ **M1A1**
$\Rightarrow e^{\lambda} = 2(1 + \lambda) \Rightarrow \lambda = \ln\{2(1 + \lambda)\}$ (AG) **A1** [3]
(iii) $\lambda_0 = 1.6$ or $1.7$ **B1**
$\lambda_{r+1} = \ln\{2(1 + \lambda_r)\}$ **M1**
$\lambda = 1.678$ (3dp) (Award B2 for correct answer with no working.) **A1** [3]3 (i) Given that $X \sim \operatorname { Po } ( 5 )$, find $\mathrm { P } ( X > 6 \mid X > 3 )$.\\
(ii) Given that $Y \sim \operatorname { Po } ( \lambda )$ and $\mathrm { P } ( Y \leqslant 1 ) = \frac { 1 } { 2 }$, show that $\lambda$ satisfies the equation $\lambda = \ln \{ 2 ( 1 + \lambda ) \}$.\\
(iii) Starting with a suitable approximation from the table of cumulative Poisson probabilities, use iteration to find $\lambda$ correct to 3 decimal places.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q3 [9]}}