| Date | May 2015 | Marks available | 4 | Reference code | 15M.2.hl.TZ1.7 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
The random variable \(X\) follows a Poisson distribution with mean \(m \ne 0\).
Given that \(2{\text{P}}(X = 4) = {\text{P}}(X = 5)\), show that \(m = 10\).
Given that \(X \le 11\), find the probability that \(X = 6\).
Markscheme
\(2\frac{{{{\text{e}}^{ - m}}{m^4}}}{{4!}} = \frac{{{{\text{e}}^{ - m}}{m^5}}}{{5!}}\) M1A1
\(\frac{2}{{4!}} = \frac{m}{{5!}}\;\;\;\)or other simplification M1
Note: accept a labelled graph showing clearly the solution to the equation. Do not accept simple verification that \(m = 10\) is a solution.
\( \Rightarrow m = 10\) AG
[3 marks]
\({\text{P}}(X = 6|X \le 11) = \frac{{{\text{P}}(X = 6)}}{{{\text{P}}(X \le 11)}}\) (M1) (A1)
\( = \frac{{0.063055 \ldots }}{{0.696776 \ldots }}\) (A1)
\( = 0.0905\) A1
[4 marks]
Total [7 marks]
Examiners report
Most candidates successfully finished part (a) with two fundamental errors occurring regularly. Either e was granted powers of 4 and 5 or an attempt to show that the value of \(m\) was 10 was made by evaluation.
Part (b) was challenging for many candidates that showed that the idea of conditional probability was poorly understood. There were many incorrect solutions where often candidates only found \({\text{P}}(X = 6)\).
