| Date | November 2009 | Marks available | 6 | Reference code | 09N.2.hl.TZ0.6 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
Tim goes to a popular restaurant that does not take any reservations for tables. It has been determined that the waiting times for a table are normally distributed with a mean of \(18\) minutes and standard deviation of \(4\) minutes.
(a) Tim says he will leave if he is not seated at a table within \(25\) minutes of arriving at the restaurant. Find the probability that Tim will leave without being seated.
(b) Tim has been waiting for \(15\) minutes. Find the probability that he will be seated within the next five minutes.
Markscheme
the waiting time, \(X{\text{ ~ }}N\)(\(18\), \({4^2}\))
(a) \({\text{P}}(X > 25) = 0.0401\) (M1)A1
(b) \({\text{P}}(X < 20|X > 15)\) (A1)
\( = \frac{{{\text{P}}(15 < X < 20)}}{{{\text{P}}(X > 15)}}\) (A1)
Note: Only one of the above A1 marks can be implied.
\( = \frac{{0.4648...}}{{0.7733...}} = 0.601\) (M1)A1
[6 marks]
Examiners report
Part (a) was well answered, whilst few candidates managed to correctly use conditional probability for part (b).
