S4 June 2007 Q7
7. A doctor wishes to study the level of blood glucose in males. The level of blood glucose is normally distributed. The doctor measured the blood glucose of 10 randomly selected male students from a school. The results, in mmol/litre, are given below.
4.7 3.6 3.8 4.7 4.1 2.2 3.6 4.0 4.4 5.0
(a) Calculate a 95% confidence interval for the mean. (7)
(b) Calculate a 95% confidence interval for the variance. (4)
A blood glucose reading of more than 7 mmol/litre is counted as high.
(c) Use appropriate confidence limits from parts (a) and (b) to find the highest estimate of the proportion of male students in the school with a high blood glucose level. (4)
| Scheme | Marks |
|---|---|
| \(\bar{x} = 4.01\) | B1 |
| \(s = 0.7992\ldots\) | M1 A1 |
| \(4.01 \pm t_9(2.5\%)\dfrac{0.7992..}{\sqrt{10}} = 4.01 \pm 2.262\dfrac{0.7992..}{\sqrt{10}}\) 2.262 | B1 |
| their \(\bar{x}\) and \(s\) and \(\sqrt{10}\) | M1 A1ft |
| \(= 4.5816\ldots\) and \(3.4383\ldots\) awrt 4.58 and 3.44 | A1 |
| (7) |
Notes
\(s^2 = 0.63877\ldots\)
| Scheme | Marks |
|---|---|
| \(2.700 \lt \dfrac{9 \times 0.7992..^2}{\sigma^2} \lt 19.023\) 2.7, 19.023 | B1 B1 |
| \(9 \times s^2/\sigma^2\) | M1 |
| \(\sigma^2 \lt 2.13, \quad \sigma^2 \gt 0.302\) both awrt 2.13, 0.302 | A1 |
| (4) |
Notes
(corrected from the printed mark scheme: the scheme prints the denominator of \(\dfrac{9 \times 0.7992..^2}{s^2}\) as \(s^2\); it is \(\sigma^2\))
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(X \gt 7) = \mathrm{P}\left(Z \gt \dfrac{7 - \mu}{\sigma}\right)\) needs to be as high as possible | M1 |
| Therefore \(\mu\) and \(\sigma\) must be as big as possible | M1 |
| \(= \mathrm{P}\left(Z \gt \dfrac{7 - 4.581}{\sqrt{2.13}}\right)\) | A1ft |
| \(= 1 - 0.9515\) \(= 0.0485\) \(= 4.85\%\) 4.8 to 4.9 | A1 |
| (4) |
Notes
M1 may be implied by them using their highest \(\mu\) and \(\sigma\).