| Date | May 2009 | Marks available | 3 | Reference code | 09M.2.sl.TZ2.9 |
| Level | SL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 9 | Adapted from | N/A |
Question
A van can take either Route A or Route B for a particular journey.
If Route A is taken, the journey time may be assumed to be normally distributed with mean 46 minutes and a standard deviation 10 minutes.
If Route B is taken, the journey time may be assumed to be normally distributed with mean \(\mu \) minutes and standard deviation 12 minutes.
For Route A, find the probability that the journey takes more than \(60\) minutes.
For Route B, the probability that the journey takes less than \(60\) minutes is \(0.85\).
Find the value of \(\mu \) .
The van sets out at 06:00 and needs to arrive before 07:00.
(i) Which route should it take?
(ii) Justify your answer.
On five consecutive days the van sets out at 06:00 and takes Route B. Find the probability that
(i) it arrives before 07:00 on all five days;
(ii) it arrives before 07:00 on at least three days.
Markscheme
\(A \sim N(46{\text{, }}{10^2})\) \(B \sim N(\mu {\text{, }}{12^2})\)
\({\rm{P}}(A > 60) = 0.0808\) A2 N2
[2 marks]
correct approach (A1)
e.g. \({\rm{P}}\left( {Z < \frac{{60 - \mu }}{{12}}} \right) = 0.85\) , sketch
\(\frac{{60 - \mu }}{{12}} = 1.036 \ldots \) (A1)
\(\mu = 47.6\) A1 N2
[3 marks]
(i) route A A1 N1
(ii) METHOD 1
\({\rm{P}}(A < 60) = 1 - 0.0808 = 0.9192\) A1
valid reason R1
e.g. probability of A getting there on time is greater than probability of B
\(0.9192 > 0.85\) N2
METHOD 2
\({\rm{P}}(B > 60) = 1 - 0.85 = 0.15\) A1
valid reason R1
e.g. probability of A getting there late is less than probability of B
\(0.0808 < 0.15\) N2
[3 marks]
(i) let X be the number of days when the van arrives before 07:00
\({\rm{P}}(X = 5) = {(0.85)^5}\) (A1)
\( = 0.444\) A1 N2
(ii) METHOD 1
evidence of adding correct probabilities (M1)
e.g. \({\rm{P}}(X \ge 3) = {\rm{P}}(X = 3) + {\rm{P}}(X = 4) + {\rm{P}}(X = 5)\)
correct values \(0.1382 + 0.3915 + 0.4437\) (A1)
\({\rm{P}}(X \ge 3) = 0.973\) A1 N3
METHOD 2
evidence of using the complement (M1)
e.g. \({\rm{P}}(X \ge 3) = 1 - {\rm{P}}(X \le 2)\) , \(1 - p\)
correct values \(1 - 0.02661\) (A1)
\({\rm{P}}(X \ge 3) = 0.973\) A1 N3
[5 marks]
Examiners report
A significant number of students clearly understood what was asked in part (a) and used the GDC to find the result.
In part (b), many candidates set the standardized formula equal to the probability (\(0.85\)), instead of using the corresponding z-score. Other candidates used the solver on their GDC with the inverse norm function.
A common incorrect approach in part (c) was to attempt to use the means and standard deviations for justification, although many candidates successfully considered probabilities.
A pleasing number of candidates recognized the binomial probability and made progress on part (d).
