Date | May 2014 | Marks available | 2 | Reference code | 14M.2.hl.TZ1.9 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
The number of birds seen on a power line on any day can be modelled by a Poisson distribution with mean 5.84.
Find the probability that during a certain seven-day week, more than 40 birds have been seen on the power line.
On Monday there were more than 10 birds seen on the power line. Show that the probability of there being more than 40 birds seen on the power line from that Monday to the following Sunday, inclusive, can be expressed as:
\(\frac{{{\text{P}}(X > 40) + \sum\limits_{r = 11}^{40} {{\text{P}}(X = r){\text{P}}(Y > 40 - r)} }}{{{\text{P}}(X > 10)}}\) where \(X \sim {\text{Po}}(5.84)\) and \(Y \sim {\text{Po}}(35.04)\).
Markscheme
mean for week is 40.88 (A1)
\({\text{P}}(S > 40) = 1 - {\text{P}}(S \leqslant 40) = 0.513\) A1
[2 marks]
\(\frac{{{\text{probability there were more than 10 on Monday AND more than 40 over the week}}}}{{{\text{probability there were more than 10 on Monday}}}}\) M1
possibilities for the numerator are:
there were more than 40 birds on the power line on Monday R1
11 on Monday and more than 29 over the course of the next 6 days R1
12 on Monday and more than 28 over the course of the next 6 days … until
40 on Monday and more than 0 over the course of the next 6 days R1
hence if X is the number on the power line on Monday and Y, the number on the power line Tuesday – Sunday then the numerator is M1
\({\text{P}}(X > 40) + {\text{P}}(X = 11) \times {\text{P}}(Y > 29) + {\text{P}}(X = 12) \times {\text{P}}(Y > 28) + \ldots \)
\( + {\text{P}}(X = 40) \times {\text{P}}(Y > 0)\)
\( = {\text{P}}(X > 40) + \sum\limits_{r = 11}^{40} {{\text{P}}(X = r){\text{P}}(Y > 40 - r)} \)
hence solution is \(\frac{{{\text{P}}(X > 40) + \sum\limits_{r = 11}^{40} {{\text{P}}(X = r){\text{P}}(Y > 40 - r)} }}{{{\text{P}}(X > 10)}}\) AG
[5 marks]