| Date | November 2014 | Marks available | 3 | Reference code | 14N.3sp.hl.TZ0.5 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Find | Question number | 5 | Adapted from | N/A |
Question
Two species of plant, \(A\) and \(B\), are identical in appearance though it is known that the mean length of leaves from a plant of species \(A\) is \(5.2\) cm, whereas the mean length of leaves from a plant of species \(B\) is \(4.6\) cm. Both lengths can be modelled by normal distributions with standard deviation \(1.2\) cm.
In order to test whether a particular plant is from species \(A\) or species \(B\), \(16\) leaves are collected at random from the plant. The length, \(x\), of each leaf is measured and the mean length evaluated. A one-tailed test of the sample mean, \(\bar X\), is then performed at the \(5\% \) level, with the hypotheses: \({H_0}:\mu = 5.2\) and \({H_1}:\mu < 5.2\).
Find the critical region for this test.
It is now known that in the area in which the plant was found \(90\% \) of all the plants are of species \(A\) and \(10\% \) are of species \(B\).
Find the probability that \(\bar X\) will fall within the critical region of the test.
If, having done the test, the sample mean is found to lie within the critical region, find the probability that the leaves came from a plant of species \(A\).
Markscheme
\(\bar X \sim N\left( {5.2,{\text{ }}\frac{{{{1.2}^2}}}{{16}}} \right)\) (M1)
critical value is \(5.2 - 1.64485 \ldots \times \frac{{1.2}}{4} = 4.70654 \ldots \) (A1)
critical region is \(] - \infty ,{\text{ }}4.71]\) A1
Note: Allow follow through for the final A1 from their critical value.
Note: Follow through previous values in (b), (c) and (d).
[3 marks]
\(0.9 \times 0.05 + 0.1 \times (1 - 0.361 \ldots ) = 0.108875997 \ldots = 0.109\) M1A1
Note: Award M1 for a weighted average of probabilities with weights \(0.1,0.9\).
[2 marks]
attempt to use conditional probability formula M1
\(\frac{{0.9 \times 0.05}}{{0.108875997 \ldots }}\) (A1)
\( = 0.41334 \ldots = 0.413\) A1
[3 marks]
Total [10 marks]
Examiners report
Solutions to this question were generally disappointing.
In (a), the standard error of the mean was often taken to be \(\sigma (1.2)\) instead of \(\frac{\sigma }{{\sqrt n }}(0.3)\) and the solution sometimes ended with the critical value without the critical region being given.
In (c), the question was often misunderstood with candidates finding the weighted mean of the two means, ie \(0.9 \times 5.2 + 0.1 \times 4.6 = 5.14\) instead of the weighted mean of two probabilities.
Without having the solution to (c), part (d) was inaccessible to most of the candidates so that very few correct solutions were seen.
