| Date | May 2017 | Marks available | 3 | Reference code | 17M.2.hl.TZ1.5 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Find | Question number | 5 | Adapted from | N/A |
Question
When carpet is manufactured, small faults occur at random. The number of faults in Premium carpets can be modelled by a Poisson distribution with mean 0.5 faults per 20\(\,\)m2. Mr Jones chooses Premium carpets to replace the carpets in his office building. The office building has 10 rooms, each with the area of 80\(\,\)m2.
Find the probability that the carpet laid in the first room has fewer than three faults.
Find the probability that exactly seven rooms will have fewer than three faults in the carpet.
Markscheme
\(\lambda = 4 \times 0.5\) (M1)
\(\lambda = 2\) (A1)
\({\text{P}}(X \leqslant 2) = 0.677\) A1
[3 marks]
\(Y \sim B(10,{\text{ }}0,677)\) (M1)(A1)
\({\text{P}}(Y = 7) = 0.263\) A1
Note: Award M1 for clear recognition of binomial distribution.
[3 marks]
