the value of m ;

P (X > 2) .

${\text{P}}(X = 5) = {\text{P}}(X = 3) + {\text{P}}(X = 4)$

$\frac{{{{\text{e}}^{ - m}}{m^5}}}{{5!}} = \frac{{{{\text{e}}^{ - m}}{m^3}}}{{3!}} + \frac{{{{\text{e}}^{ - m}}{m^4}}}{{4!}}$     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]

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.