S3 June 2009 Q4
4. A sample of size 8 is to be taken from a population that is normally distributed with mean 55 and standard deviation 3. Find the probability that the sample mean will be greater than 57. (5)
| Scheme | Marks |
|---|---|
| \(X \sim \mathrm{N}(55, 3^2)\) therefore \(\overline{X} \sim \mathrm{N}\left(55, \dfrac{9}{8}\right)\) | B1 B1 |
| \(\mathrm{P}(\overline{X} \gt 57) = \mathrm{P}\left(Z \gt \dfrac{57 - 55}{\sqrt{\frac{9}{8}}}\right) \quad = \mathrm{P}(Z \gt 1.8856\ldots)\) | M1 |
| \(= 1 - 0.9706\) | M1 |
| \(= 0.0294\) 0.0294~0.0297 | A1 |
| (5 marks) |
Notes
1st B1 for \(\overline{X}\) ~ normal and \(\mu = 55\), may be implied but must be \(\overline{X}\)
2nd B1 for Var(\(\overline{X}\)) or st. dev of \(\overline{X}\) e.g. \(\overline{X} \sim \mathrm{N}(55, \tfrac{9}{8})\) or \(\overline{X} \sim \mathrm{N}\left(55, \left(\tfrac{3}{\sqrt{8}}\right)^2\right)\) for B1B1
Condone use of \(X\) if they clearly mean \(\overline{X}\) so \(X \sim \mathrm{N}\left(55, \tfrac{9}{8}\right)\) is OK for B1B1
1st M1 for an attempt to standardize with 57 and mean of 55 and their st. dev. \(\ne 3\)
2nd M1 for 1 - tables value. Must be trying to find a probability < 0.5
A1 for answers in the range 0.0294~0.0297
ALT \(\displaystyle\sum_1^8 X_i \sim \mathrm{N}\left(8 \times 55, 8 \times 3^2\right)\)
1st B1 for \(\sum X\) ~ normal and mean \(= 8 \times 55\)
2nd B1 for variance \(= 8 \times 3^2\)
1st M1 for attempt to standardise with \(57 \times 8\), mean of \(55 \times 8\) and their st dev \(\ne 3\)