S3 June 2015 Q5
5.
Rebecca buys 5 bottles of Burxton’s water and one bottle of Hargate’s water.
Find the probability that the total volume in the 5 bottles of Burxton’s water is more than 5 times the volume in the bottle of Hargate’s water. (5)
Ankit believes that \(\mathrm{P}(U_1 \gt \bar{U} + \sigma) = 0.181\) correct to 3 decimal places, for any random sample \(U_1, U_2, U_3, U_4, U_5\) taken from a normal population with mean \(\mu\) and standard deviation \(\sigma\).
| Scheme | Marks |
|---|---|
| Let \(R = B_1 + B_2 + B_3 + B_4 + B_5 - 5H\) so \(\mathrm{E}(R) = -25\) (o.e.) | B1 |
| \(\mathrm{Var}(R) = 5 \times 6^2 + 5^2 \times 4^2\) \(R \sim \mathrm{N}\left(-25, \sqrt{580}^{\,2}\right)\) | M1A1 |
| \(\mathrm{P}(R \gt 0) = \mathrm{P}\left(Z \gt \frac{0 - -25}{\sqrt{580}}\right) = \mathrm{P}(Z \gt 1.04)\), \(= 0.149619\ldots\) (calc) or 0.1492 (tables) | dM1 A1 |
| (5) |
Notes
1st B1 for \(\mathrm{E}(R) = -25\) (or 25 if their \(R\) is defined the other way around)
1st M1 for an attempt at \(\mathrm{Var}(R) = 5\mathrm{Var}(B) + 25\mathrm{Var}(H)\). Condone swapping of \(6^2\) and \(4^2\)
1st A1 for normal and correct variance (ft their mean)
2nd dM1 for attempting the correct probability and standardising with their mean and sd.
This mark is dependent on 1st M1 so if \(R\) is not being used or M0 for variance score M0
If their method is not crystal clear then they must be attempting \(\mathrm{P}(Z \gt +\text{ve value})\) o.e
2nd A1 for answer in the range [0.149, 0.150]
| Scheme | Marks |
|---|---|
| \(\bar{X} \sim \mathrm{N}\left(\mu, \frac{\sigma^2}{5}\right)\) | B1 |
| \(\mathrm{Var}(D) = \sigma^2 + \text{"}\tfrac{\sigma^2}{5}\text{"} \left[= \tfrac{6\sigma^2}{5}\right]\), so \(D \sim \mathrm{N}\left(0, \dfrac{6\sigma^2}{5}\right)\) | M1, A1 |
| (3) |
Notes
B1 for correct distribution of \(\bar{X}\) (may be implied for a correct answer for \(D\))
M1 for correct attempt at \(\mathrm{Var}(D)\) (ft their \(\mathrm{Var}(\bar{X})\)) [A1 needs must be fully correct]
| Scheme | Marks |
|---|---|
| \(\mathrm{P}(Y_1 \gt \bar{X} + \sigma) = \mathrm{P}(D \gt \sigma) = \mathrm{P}\left(Z \gt \dfrac{\sigma}{\sqrt{\frac{6}{5}}\sigma}\right)\) | M1 |
| \(= \mathrm{P}(Z \gt 0.912\ldots) = 0.181\) (3 dp) (*) | A1cso |
| (2) |
Notes
M1 for expressing probability in terms of \(D\) and standardising
A1cso for seeing \(\mathrm{P}(Z \gt 0.912..)\) or prob of 1 – 0.8186 (tables) or 0.180655…(calc)
| Scheme | Marks |
|---|---|
| Since \(U_1\) and \(\bar{U}\) are not independent (so variance formula cannot be used) Can be implied e.g. \(U_1\) used to calculate \(\bar{U}\), \(U_1\) and \(\bar{U}\) from same sample o.e. | B1 |
| (1) |
Notes
B1 correct statement that should mention \(U_1\) and \(\bar{U}\)
| Scheme | Marks |
|---|---|
| Let \(F = U_1 - \bar{U} = U_1 - \dfrac{(U_1 + U_2 + U_3 + U_4 + U_5)}{5}\), \(= \dfrac{4U_1 - (U_2 + U_3 + U_4 + U_5)}{5}\) | M1, A1 |
| \(\mathrm{Var}(F) = \dfrac{4^2\sigma^2 + 4\sigma^2}{5^2} = 0.8\sigma^2\) ,so \(F \sim \mathrm{N}(0, 0.8\sigma^2)\) | dM1, A1 |
| \(\mathrm{P}(F \gt \sigma) = \mathrm{P}\left(Z \gt \dfrac{\sigma}{\sigma\sqrt{0.8}}\right) = \mathrm{P}(Z \gt 1.118..)\) | M1 |
| \(= 0.1314\) (tables) or \(0.131776\ldots\) (calc) awrt 0.131~0.132 | A1cso |
| (6) | |
| (17 marks) |
Notes
1st M1 for forming an expression in terms of \(U_1 \ldots U_5\) only
1st A1 for collecting \(U_1\) terms and getting in a form where \(\mathrm{Var}(aX \pm bY)\) can be used.
2nd dM1 for a correct expression for Var(their \(F\)). Dependent on 1st M1.
2nd A1 for a correct distribution for \(F\)
3rd M1 attempting a correct prob and standardising using their \(\mathrm{Var}(F)\), \(\sigma\) must cancel
3rd A1cso for awrt 0.131 or 0.132