| Date | November 2011 | Marks available | 2 | Reference code | 11N.2.hl.TZ0.3 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
The number of vehicles passing a particular junction can be modelled using the Poisson distribution. Vehicles pass the junction at an average rate of 300 per hour.
Find the probability that no vehicles pass in a given minute.
Find the expected number of vehicles which pass in a given two minute period.
Find the probability that more than this expected number actually pass in a given two minute period.
Markscheme
\(m = \frac{{300}}{{60}} = 5\) (A1)
\({\text{P}}(X = 0) = 0.00674\) A1
or \({{\text{e}}^{ - 5}}\)
[2 marks]
\({\text{E}}(X) = 5 \times 2 = 10\) A1
[1 mark]
\({\text{P}}(X > 10) = 1 - {\text{P}}(X \leqslant 10)\) (M1)
= 0.417 A1
[2 marks]
Examiners report
Parts (a) and (b) were answered successfully by many candidates. Some candidates had difficulty obtaining the correct inequality in (c).
Parts (a) and (b) were answered successfully by many candidates. Some candidates had difficulty obtaining the correct inequality in (c).
Parts (a) and (b) were answered successfully by many candidates. Some candidates had difficulty obtaining the correct inequality in (c).
