| Date | May 2012 | Marks available | 2 | Reference code | 12M.2.hl.TZ2.5 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 5 | Adapted from | N/A |
Question
The random variable X has the distribution \({\text{Po}}(m)\) .
Given that \({\text{P}}(X = 5) = {\text{P}}(X = 3) + {\text{P}}(X = 4)\), find
the value of m ;
P (X > 2) .
Markscheme
\({\text{P}}(X = 5) = {\text{P}}(X = 3) + {\text{P}}(X = 4)\)
\(\frac{{{{\text{e}}^{ - m}}{m^5}}}{{5!}} = \frac{{{{\text{e}}^{ - m}}{m^3}}}{{3!}} + \frac{{{{\text{e}}^{ - m}}{m^4}}}{{4!}}\) M1(A1)
\({m^2} - 5m - 20 = 0\)
\( \Rightarrow m = \frac{{5 + \sqrt {105} }}{2} = (7.62)\) A1
[3 marks]
\({\text{P}}(X > 2) = 1 - {\text{P}}(X \leqslant 2)\) (M1)
\( = 1 - 0.018...\)
\( = 0.982\) A1
[2 marks]
Examiners report
Again this proved to be a successful question for many candidates with a good proportion of wholly correct answers seen. It was good to see students making good use of the calculator.
Again this proved to be a successful question for many candidates with a good proportion of wholly correct answers seen. It was good to see students making good use of the calculator.
