Date | May 2010 | Marks available | 15 | Reference code | 10M.2.hl.TZ1.12 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Calculate and Find | Question number | 12 | Adapted from | N/A |
Question
Casualties arrive at an accident unit with a mean rate of one every 10 minutes. Assume that the number of arrivals can be modelled by a Poisson distribution.
(a) Find the probability that there are no arrivals in a given half hour period.
(b) A nurse works for a two hour period. Find the probability that there are fewer than ten casualties during this period.
(c) Six nurses work consecutive two hour periods between 8am and 8pm. Find the probability that no more than three nurses have to attend to less than ten casualties during their working period.
(d) Calculate the time interval during which there is a 95 % chance of there being at least two casualties.
Markscheme
Note: Accept exact answers in parts (a) to (c).
(a) number of patients in 30 minute period = X (A1)
\(X \sim {\text{Po(3)}}\) (M1)A1
[3 marks]
(b) number of patients in working period = Y (A1)
\(Y \sim {\text{Po(12)}}\) (M1)A1
[3 marks]
(c) number of working period with less than 10 patients = W (M1)(A1)
\(W \sim {\text{B}}(6,{\text{ }}0.2424 \ldots )\) (M1)A1
[4 marks]
(d) number of patients in t minute interval = X
\(X \sim {\text{Po}}(T)\)
\({\text{P}}(X \geqslant 2) = 0.95\)
\({\text{P}}(X = 0) + {\text{P}}(X = 1) = 0.05\) (M1)(A1)
\({{\text{e}}^{ - T}}(1 + T) = 0.05\) (M1)
\(T = 4.74\) (A1)
t = 47.4 minutes A1
[5 marks]
Total [15 marks]
Examiners report
Parts (a) and (b) were well answered, but many students were unable to recognise the Binomial distribution in part (c) and were unable to form the correct equation in part (d). There were many accuracy errors in this question.