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]