June 2025 Paper 3 Q18
18 The cholesterol level, \(X\), of an adult can be modelled by a normal distribution with mean 5.7 mmol/l and standard deviation 1.2 mmol/l.
It is given that the \(\mathrm{P}(X \leqslant a) = 0.25\) and \(\mathrm{P}(X \geqslant b) = 0.25\)
Find the interquartile range of \(X\)
[4 marks]A random sample of 160 adults were given the new dietary supplement for a trial period of one year.
After the trial, their cholesterol level was found to have a mean of 5.6 mmol/l.
Carry out a hypothesis test at the 5% significance level to investigate whether the mean cholesterol level has reduced after taking the dietary supplement.
It can be assumed that the cholesterol level after taking the dietary supplement follows a normal distribution with an unchanged variance.
[6 marks]| Scheme | Marks | AO |
|---|---|---|
| States the \(z\) value \((\pm)\) 0.67(45) for the inverse normal distribution PI by [4.89, 4.9] or [6.5, 6.51] seen | M1 | 3.4 |
| Obtains correct lower quartile AWFW [4.89, 4.9] or obtains correct upper quartile AWFW [6.5, 6.51] CAO | A1 | 1.1b |
| Uses a complete method to find the IQR e.g. \(b - a\) or \(2 \times (5.7 - a)\) or \(2 \times (b - 5.7)\) or states or calculates their \(b\) − their \(a\) | M1 | 3.4 |
| Obtains AWFW [1.6, 1.62] Condone missing units | A1 | 1.1b |
| (4) |
Typical solution
\[a = 4.89\]\[b = 6.51\]\[\text{IQR} = 6.51 - 4.89\]\[= 1.62 \text{ mmol/l}\]| Scheme | Marks | AO |
|---|---|---|
| States \(\mathrm{H}_0: \mu = 5.7\) \(\mathrm{H}_1: \mu \lt 5.7\) | B1 | 2.5 |
| States or uses correct model PI by normal with mean 5.7 and variance \(\dfrac{1.2^2}{160}\) or 0.009 or standard deviation \(\dfrac{1.2}{\sqrt{160}}\) or 0.095 or better OE or by correct probability AWFW [0.145, 0.147] or test statistic \((\pm)\dfrac{5.6 - 5.7}{1.2 \div \sqrt{160}}\) or test statistic value AWFW (±) [1.05, 1.1] or critical value AWFW [5.54, 5.55] | M1 | 1.1a |
| Obtains AWFW [0.145, 0.147] or the correct test statistic value AWFW − [1.05, 1.1] or the correct critical value AWFW [5.54, 5.55] May be seen in expression for critical/acceptance regions | A1 | 1.1b |
| Correctly compares their value of \(P(\lt \text{ or } \leqslant 5.6)\) with 0.05 or correctly compares their negative test statistic with AWFW − [1.64, 1.65] or correctly compares 5.6 with their acceptance region or critical region or critical value May be seen on a diagram | M1 | 3.5a |
| Infers \(\mathrm{H}_0\) not rejected or \(\mathrm{H}_0\) accepted FT their comparison Condone reject \(\mathrm{H}_1\) | A1F | 2.2b |
| Concludes, from a fully correct comparison, in context by stating that there is insufficient evidence to suggest the mean cholesterol level has reduced after taking the dietary supplement. Conclusion must not be definite, eg use of ‘suggest’, ‘support’ etc To be awarded R1, marks M1A1M1A1 must be scored as the minimum | R1 | 3.2a |
| (6) | ||
| (10 marks) |
Typical solution
\[\mathrm{H}_0: \mu = 5.7\]\[\mathrm{H}_1: \mu \lt 5.7\]\[\bar{X} \sim \mathrm{N}\left(5.7,\ \frac{1.2^2}{160}\right)\]\[\mathrm{P}(\bar{X} \lt 5.6) = 0.146\]\[0.146 \gt 0.05\]Do not reject \(\mathrm{H}_0\)
There is insufficient evidence to suggest that the mean cholesterol level has reduced after taking the dietary supplement.