S3 June 2011 Q6
6. The lifetimes of batteries from manufacturer \(A\) are normally distributed with mean 20 hours and standard deviation 5 hours when used in a camera.
Judy uses a camera that takes one battery at a time. She takes a pack of 6 batteries from manufacturer \(A\) to use in her camera on holiday.
The lifetimes of batteries from manufacturer \(B\) are normally distributed with mean 35 hours and standard deviation 8 hours when used in a camera.
| Scheme | Marks |
|---|---|
| \(L = A_1 + A_2 + \ldots + A_6\) Mean is \(\mathrm{E}(L) = 6 \times 20 = 120\) | B1 |
| Standard deviation is \(\sqrt{\mathrm{Var}(W)} = \sqrt{6 \times 5^2} = 5\sqrt{6} = 12.247\ldots\) awrt 12.2 | B1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(P(L \gt 110) \quad = P\left(Z \gt \left(\dfrac{110 - 120}{12.247\ldots}\right)\right)\) | M1 |
| \(= P(Z \lt 0.8164\ldots)\) \(= 0.7939\) (or 0.7929 using interpolation or 0.79289 by calc) | A1 |
| (2) |
Notes
M1 for identifying a correct probability (they must have the 110) and attempting to standardise with their mean and sd. This can be implied by the correct answer.
A1 for awrt 0.794 or 0.793
| Scheme | Marks |
|---|---|
| Let \(X = 4B - \sum_1^6 A_i\) \(\mathrm{E}(X) = 140 - 120 = 20\) | B1 |
| \(\mathrm{Var}(X) = 16 \times 8^2 + 6 \times 5^2 = 1174\) | M1M1A1 |
| \(\mathrm{P}(X \lt 0) = \mathrm{P}\left(Z \lt \dfrac{-20}{\sqrt{1174}}\right) = \mathrm{P}(Z \lt -0.583\ldots)\) | M1 |
| \(= 0.2797\) (or 0.2810 if no interpolation) or 0.27971 by calc. | A1 |
| (6) | |
| (10 marks) |
Notes
B1 Accept ±20 for B mark. Only award for probability statement if 2 terms in var
1st M1 for 1024, 2nd M1 for 150
3rd M for standardising with their mean and 2 term sd and finding probability <0.5
2nd A1 for awrt 0.280 or 0.281