| Date | May 2017 | Marks available | 6 | Reference code | 17M.1.hl.TZ0.4 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The weights of male students in a college are modelled by a normal distribution with mean 80 kg and standard deviation 7 kg.
The weights of female students in the college are modelled by a normal distribution with mean 54 kg and standard deviation 5 kg.
The college has a lift installed with a recommended maximum load of 550 kg. One morning, the lift contains 3 male students and 6 female students. You may assume that the 9 students are randomly chosen.
Find the probability that the weight of a randomly chosen male student is more than twice the weight of a randomly chosen female student.
Determine the probability that their combined weight exceeds the recommended maximum.
Markscheme
let \(M\), \(F\) denote the weights of the male, female
consider \(D = M - 2F\) (M1)
\({\text{E}}(D) = 80 - 2 \times 54 = - 28\) A1
\({\text{Var}}(D) = {7^2} + 4 \times {5^2}\) (M1)
\( = 149\) A1
\({\text{P}}(M > 2F) = {\text{P}}(D > 0)\) (M1)
\( = 0.0109\) A1
Note: Accept any answer that rounds correctly to 0.011.
[6 marks]
consider \({\text{S}} = \sum\limits_{i = 1}^3 {{M_i} + \sum\limits_{i = 1}^6 {{F_i}} } \) (M1)
Note: Condone the use of the incorrect notation \(3M + 6F\).
\({\text{E}}(S) = 3 \times 80 + 6 \times 54 = 564\) A1
\({\text{Var}}(S) = 3 \times {7^2} + 6 \times {5^2}\) (M1)
\( = 297\) A1
\({\text{P}}(S > 550) = 0.792\) A1
Note: Accept any answer that rounds correctly to 0.792.
[5 marks]
