(a)     Show that the probability that a randomly selected bottle filled by Machine A contains more than $1000$ ml of mineral water is $0.212$.

(b)     A random sample of $5$ bottles is taken from Machine A. Find the probability that exactly $3$ of them each contain more than $1000$ ml of mineral water.

(c)     Find the minimum number of bottles that would need to be sampled to ensure that the probability of getting at least one bottle filled by Machine A containing more than $1000$ ml of mineral water, is greater than $0.99$.

(d)     It has been found that for Machine B the probability of a bottle containing less than $996$ ml of mineral water is $0.1151$. The probability of a bottle containing more than $1000$ ml is $0.3446$. Find the mean and standard deviation for the volume of mineral water contained in bottles filled by Machine B.

(e)     The company that makes the mineral water receives, on average, m phone calls every $10$ minutes. The number of phone calls, $X$ , follows a Poisson distribution such that ${\text{P}}(X = 2) = {\text{P}}(X = 3) + {\text{P}}(X = 4)$ .

  (i)     Find the value of $m$ .

  (ii)     Find the probability that the company receives more than two telephone calls in a randomly selected $10$ minute period.

(a)     $X \sim {\text{N}}$($998$, ${2.5^2}$ )     M1

${\text{P}}(X > 1000) = 0.212$     AG

[1 mark]

(b)     $X \sim {\text{B}}$($5$, $0.2119...$)

evidence of binomial     (M1)

${\text{P}}(X = 3) = \left( {\begin{array}{*{20}{c}}   5 \\   3 \end{array}} \right){\left( {0.2119...} \right)^3}{\left( {0.7881...} \right)^2} = 0.0591$   (accept $0.0592$)     (M1)A1

[3 marks]

(c)     ${\text{P}}(X \geqslant 1) = 1 - {\text{P}}(X = 0)$     (M1)

$1 - {\left( {0.7881...} \right)^n} > 0.99$

${\left( {0.7881...} \right)^n} < 0.01$     A1

Note: Award A1 for line 2 or line 3 or equivalent.

$n > 19.3$     (A1)
minimum number of bottles required is $20$     A1N2

[4 marks]

(d)     $\frac{{996 - \mu }}{\sigma } =  - 1.1998$   (accept $1.2$)     M1A1

$\frac{{1000 - \mu }}{\sigma } = 0.3999$   (accept $0.4$)     M1A1

$\mu  = 999$(ml), $\sigma  = 2.50$(ml)     A1A1

[6 marks]

(e)     (i)     $\frac{{{{\text{e}}^{ - m}}{m^2}}}{{2!}} = \frac{{{{\text{e}}^{ - m}}{m^3}}}{{3!}} + \frac{{{{\text{e}}^{ - m}}{m^4}}}{{4!}}$     M1A1

$\frac{{{m^2}}}{2} = \frac{{{m^3}}}{6} + \frac{{{m^4}}}{{24}}$

$12{m^2} - 4{m^3} - {m^4} = 0$     (A1)

$m = - 6$, $0$, $2$

$\Rightarrow m = 2$     A1N2

(ii)     ${\text{P}}(X > 2) = 1 - {\text{P}}(X \leqslant 2)$     (M1)

$= 1 - {\text{P}}(X = 0) - {\text{P}}(X = 1) - {\text{P}}(X = 2)$

$= 1 - {{\text{e}}^{ - 2}} - 2{{\text{e}}^{ - 2}} - \frac{{{2^2}{{\text{e}}^{ - 2}}}}{{2!}}$

$= 0.323$     A1

[6 marks]

Total [20 marks]

This was the best done of the section B questions, with the majority of candidates making the correct choice of probability distribution for each part. The main sources of errors: (b) missing out the binomial coefficient in the calculation; (c) failure to rearrange 'at least one bottle' in terms of the probability of obtaining no bottles; (d) using $1.2$ rather than $- 1.2$ in the inverse Normal or not performing an inverse Normal at all; (e)(ii) misinterpreting 'more than two'.