| Date | May 2008 | Marks available | 14 | Reference code | 08M.3sp.hl.TZ1.3 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ1 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
A shop sells apples and pears. The weights, in grams, of the apples may be assumed to have a \({\text{N}}(200,{\text{ 1}}{{\text{5}}^2})\) distribution and the weights of the pears, in grams, may be assumed to have a \({\text{N}}(120,{\text{ 1}}{{\text{0}}^2})\) distribution.
(a) Find the probability that the weight of a randomly chosen apple is more than double the weight of a randomly chosen pear.
(b) A shopper buys 3 apples and 4 pears. Find the probability that the total weight is greater than 1000 grams.
Markscheme
(a) Let X, Y (grams) denote respectively the weights of a randomly chosen apple, pear.
Then
\(X - 2Y{\text{ is N}}(200 - 2 \times 120,{\text{ }}{15^2} + 4 \times {10^2}),\) (M1)(A1)(A1)
i.e. \({\text{N}}( - 40,{\text{ }}{25^2})\) A1
We require
\({\text{P}}(X > 2Y) = {\text{P}}(X - 2Y > 0)\) (M1)(A1)
\( = 0.0548\) A2
[8 marks]
(b) Let \(T = {X_1} + {X_2} + {X_3} + {Y_1} + {Y_2} + {Y_3} + {Y_4}\) (grams) denote the total weight.
Then
\(T{\text{ is N}}(3 \times 200 + 4 \times 120,{\text{ }}3 \times {15^2} + 4 \times {10^2}),\) (M1)(A1)(A1)
i.e. \({\text{N(1080, 1075)}}\) A1
\({\text{P}}(T > 1000) = 0.993\) A2
[6 marks]
Total [14 marks]
Examiners report
The response to this question was disappointing. Many candidates are unable to differentiate between quantities such as \(3X{\text{ and }}{X_1} + {X_2} + {X_3}\) . While this has no effect on the mean, there is a significant difference between the variances of these two random variables.
