Alan and Brian are athletes specializing in the long jump. When Alan jumps, the length of his jump is a normally distributed random variable with mean 5.2 metres and standard deviation 0.1 metres. When Brian jumps, the length of his jump is a normally distributed random variable with mean 5.1 metres and standard deviation 0.12 metres. For both athletes, the length of a jump is independent of the lengths of all other jumps. During a training session, Alan makes four jumps and Brian makes three jumps. Calculate the probability that the mean length of Alan’s four jumps is less than the mean length of Brian’s three jumps.

Colin joins the squad and the coach wants to know the mean length, \(\mu \) metres, of his jumps. Colin makes six jumps resulting in the following lengths in metres.

5.21, 5.30, 5.22, 5.19, 5.28, 5.18

(i)     Calculate an unbiased estimate of both the mean \(\mu \) and the variance of the lengths of his jumps.

(ii)     Assuming that the lengths of these jumps are independent and normally distributed, calculate a 90 % confidence interval for \(\mu \) .

let \(\bar A,{\text{ }}\bar B\) denote the means of Alan’s and Brian’s jumps

attempting to find the distributions of \(\bar A,{\text{ }}\bar B\)     (M1)

\(\bar A{\text{ is N}}\left( {5.2,\frac{{{{0.1}^2}}}{4}} \right)\)     A1

\(\bar B{\text{ is N}}\left( {5.1,\frac{{{{0.12}^2}}}{3}} \right)\)     A1

attempting to find the distribution of \(\bar A - \bar B\)     (M1)

\(\bar A - \bar B{\text{ is N}}\left( {5.2 - 5.1,\frac{{{{0.1}^2}}}{4} + \frac{{{{0.12}^2}}}{3}} \right)\)     (A1)(A1)

i.e. \({\text{N}}(0.1,{\text{ }}0.0073)\)     A1

\({\text{P}}(\bar A < \bar B) = {\text{P}}(\bar A - \bar B < 0)\)     M1

\( = 0.121\)     A1

[9 marks]

(i)     \(\sum {x = 31.38,{\text{ }}\sum {{x^2} = 164.1294} } \)

\(\bar x = \frac{{31.38}}{6} = 5.23\)     (M1)A1

EITHER

\(s_{n - 1}^2 = \frac{{164.1294}}{5} - \frac{{{{31.38}^2}}}{{5 \times 6}} = 0.00240\)     (M1)(A1)A1

OR

\({s_{n - 1}} = 0.04899 \Rightarrow s_{n - 1}^2 = 0.00240\)     (M1)(A1)A1

Note: Accept the exact answer 0.0024 without an arithmetic penalty.

(ii)     using the t-distribution with DF = 5     (A1)

critical value of t = 2.015     A1

90 % confidence limits are \(5.23 \pm 2.015\sqrt {\frac{{0.0024}}{6}} \)     M1A1

giving [5.19, 5.27]     A1     N5

[10 marks]

In (a), it was disappointing to note that many candidates failed to realise that the question was concerned with the mean lengths of the jumps and worked instead with the sums of the lengths.

Most candidates obtained correct estimates in (b)(i), usually directly from the GDC. In (b)(ii), however, some candidates found a z-interval instead of a t-interval.