| Date | May 2018 | Marks available | 6 | Reference code | 18M.3sp.hl.TZ0.1 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Find | Question number | 1 | Adapted from | N/A |
Question
The weights, X kg, of the males of a species of bird may be assumed to be normally distributed with mean 4.8 kg and standard deviation 0.2 kg.
The weights, Y kg, of female birds of the same species may be assumed to be normally distributed with mean 2.7 kg and standard deviation 0.15 kg.
Find the probability that a randomly chosen male bird weighs between 4.75 kg and 4.85 kg.
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 on a weighing machine that has a weight limit of 18 kg. Find the probability that the total weight of these five birds is greater than the weight limit.
Markscheme
Note: In question 1, accept answers that round correctly to 2 significant figures.
P(4.75 < X < 4.85) = 0.197 A1
[1 mark]
Note: In question 1, accept answers that round correctly to 2 significant figures.
consider the random variable X − 2Y (M1)
E(X − 2Y) = − 0.6 (A1)
Var(X − 2Y) = Var(X) + 4Var(Y) (M1)
= 0.13 (A1)
X − 2Y ∼ N(−0.6, 0.13)
P(X − 2Y > 0) (M1)
= 0.0480 A1
[6 marks]
Note: In question 1, accept answers that round correctly to 2 significant figures.
let W = X1 + X2 + Y1 + Y2 + Y3 be the total weight
E(W) = 17.7 (A1)
Var(W) = 2Var(X) + 3Var(Y) = 0.1475 (M1)(A1)
W ∼ N(17.7, 0.1475)
P(W > 18) = 0.217 A1
[4 marks]
