| Date | May 2007 | Marks available | 5 | Reference code | 07M.1.hl.TZ0.6 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
The weights, \(X\) kg , of male birds of a certain species are normally distributed with mean \(4.5\) kg and standard deviation \(0.2\) kg . The weights, \(Y\) kg , of female birds of this species are normally distributed with mean \(2.5\) kg and standard deviation \(0.15\) kg .
(i) Find the mean and variance of \(2Y - X\) .
(ii) Find the probability that the weight of a randomly chosen male bird is more than twice the weight of a randomly chosen female bird.
Two randomly chosen male birds and three randomly chosen female birds are placed together on a weighing machine for which the recommended maximum weight is \(16\) kg . Find the probability that this maximum weight is exceeded.
Markscheme
(i) \({\rm{E}}(2Y - X) = 2 \times 2.5 - 4.5 = 0.5\) A1
\(Var(2Y - X) = 4 \times 0.1{5^2} + {0.2^2} = 0.13\) M1A1
(ii) We require \({\rm{P}}(X > 2Y) = {\rm{P}}(2Y - X < 0)\) M1
\(0.0828\) A2
Note: Using tables, answer is \(0.0823\).
[6 marks]
Let \(S\) denote the total weight of the \(5\) birds.
Then,
\({\rm{E}}(S) = 2 \times 4.5 + 3 \times 2.5 = 16.5\) A1
\(Var(S) = 2 \times 0.{2^2} + 3 \times 0.1{5^2} = 0.1475\) M1A1
\({\rm{P}}(S > 16) = 0.904\) A2
Note: Using tables, answer is \(0.903\).
[5 marks]
