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Date May 2013 Marks available 4 Reference code 13M.2.hl.TZ1.10
Level HL only Paper 2 Time zone TZ1
Command term Find Question number 10 Adapted from N/A

Question

A ferry carries cars across a river. There is a fixed time of T minutes between crossings. The arrival of cars at the crossing can be assumed to follow a Poisson distribution with a mean of one car every four minutes. Let X denote the number of cars that arrive in T minutes.

Find T, to the nearest minute, if \({\text{P}}(X \leqslant 3) = 0.6\).

[3]
a.

It is now decided that the time between crossings, T, will be 10 minutes. The ferry can carry a maximum of three cars on each trip.

One day all the cars waiting at 13:00 get on the ferry. Find the probability that all the cars that arrive in the next 20 minutes will get on either the 13:10 or the 13:20 ferry.

[4]
b.

Markscheme

\(X \sim {\text{Po(0.25T)}}\)     (A1)

Attempt to solve \({\text{P}}(X \leqslant 3) = 0.6\)     (M1)

\(T = 12.8453 \ldots  = 13{\text{ (minutes)}}\)     A1

Note: Award A1M1A0 if T found correctly but not stated to the nearest minute.

 

[3 marks]

a.

let \({X_1}\) be the number of cars that arrive during the first interval and \({X_2}\) be the number arriving during the second.

\({X_1}\) and \({X_2}\) are Po(2.5)     (A1)

P (all get on) \( = {\text{P}}({X_1} \leqslant 3) \times {\text{P}}({X_2} \leqslant 3) + {\text{P}}({X_1} = 4) \times {\text{P}}({X_2} \leqslant 2)\)

\( + {\text{P}}({X_1} = 5) \times {\text{P}}({X_2} \leqslant 1) + {\text{P}}({X_1} = 6) \times {\text{P}}({X_2} = 0)\)     (M1)

\( = 0.573922 \ldots + 0.072654 \ldots + 0.019192 \ldots + 0.002285 \ldots \)     (M1)

\( = 0.668{\text{ }}(053 \ldots )\)     A1 

[4 marks]

b.

Examiners report

There were some good answers to part (a), although poor calculator use frequently let down the candidates.

a.

Very few candidates were able to access part (b).

b.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.6 » Poisson distribution, its mean and variance.
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