(a)     A box of biscuits is considered to be underweight if it weighs less than 228 grams. It is known that the weights of these boxes of biscuits are normally distributed with a mean of 231 grams and a standard deviation of 1.5 grams.

What is the probability that a box is underweight?

(b)     The manufacturer decides that the probability of a box being underweight should be reduced to 0.002.

(i)     Bill’s suggestion is to increase the mean and leave the standard deviation unchanged. Find the value of the new mean.

(ii)     Sarah’s suggestion is to reduce the standard deviation and leave the mean unchanged. Find the value of the new standard deviation.

(c)     After the probability of a box being underweight has been reduced to 0.002, a group of customers buys 100 boxes of biscuits. Find the probability that at least two of the boxes are underweight.

There are six boys and five girls in a school tennis club. A team of two boys and two girls will be selected to represent the school in a tennis competition.

(a)     In how many different ways can the team be selected?

(b)     Tim is the youngest boy in the club and Anna is the youngest girl. In how many different ways can the team be selected if it must include both of them?

(c)     What is the probability that the team includes both Tim and Anna?

(d)     Fred is the oldest boy in the club. Given that Fred is selected for the team, what is the probability that the team includes Tim or Anna, but not both?

(a)     \(X \sim {\text{N(231, 1.}}{{\text{5}}^2})\)

\({\text{P}}(X < 228) = 0.0228\)     (M1)A1

Note: Accept 0.0227.

[2 marks]

(b)     (i)     \(X \sim {\text{N(}}\mu {\text{, 1.}}{{\text{5}}^2})\)

\({\text{P}}(X < 228) = 0.002\)

\(\frac{{228 - \mu }}{{1.5}} = - 2.878…\)     M1A1

\(\mu = 232{\text{ grams}}\)     A1     N3

(ii)     \(X \sim {\text{N(231, }}{\sigma ^2})\)

\(\frac{{228 - 231}}{\sigma } = - 2.878…\)     M1A1

\(\sigma = 1.04{\text{ grams}}\)     A1     N3

[6 marks]

(c)     \(X \sim {\text{B(100, 0.002)}}\)     (M1)

\({\text{P}}(X \leqslant 1) = 0.982…\)     (A1)

\({\text{P}}(X \geqslant 2) = 1 - {\text{P}}(X \leqslant 1) = 0.0174\)     A1

[3 marks]

Total [11 marks]

(a)     Boys can be chosen in \(\frac{{6 \times 5}}{2} = 15\) ways     (A1)

Girls can be chosen in \(\frac{{5 \times 4}}{2} = 10\) ways     (A1)

Total \( = 15 \times 10 = 150\) ways     A1

[3 marks]

(b)     Number of ways \( = 5 \times 4 = 20\)     (M1)A1

[2 marks]

(c)     \(\frac{{20}}{{150}}{\text{ }}\left( { = \frac{2}{{15}}} \right)\)     A1

[1 mark]

(d)     METHOD 1

\({\text{P}}(T) = \frac{1}{5};{\text{ P}}(A) = \frac{2}{5}\)     A1

P(T or A but not both) \( = {\text{P}}(T) \times {\text{P}}(A') + {\text{P}}(T') \times {\text{P}}(A)\)     M1A1

\( = \frac{1}{5} \times \frac{3}{5} + \frac{4}{5} \times \frac{2}{5} = \frac{{11}}{{25}}\)     A1

METHOD 2

Number of selections including Fred \( = 5 \times \left( {\begin{array}{*{20}{c}}
  5 \\
  2
\end{array}} \right) = 50\)     A1

Number of selections including Tim but not Anna \( = \left( {\begin{array}{*{20}{c}}
  4 \\
  2
\end{array}} \right) = 6\)     A1

Number of selections including Anna but not Tim \( = 4 \times 4 = 16\)

Note: Both statements are needed to award A1.

P(T or A but not both) \( = \frac{{6 + 16}}{{50}} = \frac{{11}}{{25}}\)     M1A1

[4 marks]

Total [10 marks]

Part A was well done by many candidates although an arithmetic penalty was often awarded in (b)(i) for giving the new value of the mean to too many significant figures.

Candidates are known, however, to be generally uncomfortable with combinatorial mathematics and Part B caused problems for many candidates. Even some of those candidates who solved (a) and (b) correctly were then unable to deduce the answer to (c), sometimes going off on some long-winded solution which invariably gave the wrong answer. Very few correct solutions were seen to (d).