The random variable \(X\) has a Poisson distribution with mean \(\mu \).

Given that \({\text{P}}(X = 2) + {\text{P}}(X = 3) = {\text{P}}(X = 5)\),

(a)     find the value of \(\mu \);

(b)     find the probability that X lies within one standard deviation of the mean.

(a)     \(\frac{{{\mu ^2}{{\text{e}}^{ - \mu }}}}{{2!}} + \frac{{{\mu ^3}{{\text{e}}^{ - \mu }}}}{{3!}} = \frac{{{\mu ^5}{{\text{e}}^{ - \mu }}}}{{5!}}\)     (M1)

\(\frac{{{\mu ^2}}}{2} + \frac{{{\mu ^3}}}{6} - \frac{{{\mu ^5}}}{{120}} = 0\)

\(\mu  = 5.55\)     A1

[2 marks]

(b)     \(\sigma  = \sqrt {5.55 \ldots }  = 2.35598 \ldots \)     (M1)

\({\text{P}}(3.19 \leqslant X \leqslant 7.9)\)

\({\text{P}}(4 \leqslant X \leqslant 7)\)

\( = 0.607\)     A1

[2 marks]

Total [4 marks]