(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'.