S2 January 2011 Q6

EdexcelOld spec16 marksPoisson Distribution

6. Cars arrive at a motorway toll booth at an average rate of 150 per hour.

(a) Suggest a suitable distribution to model the number of cars arriving at the toll booth, \(X\), per minute. (2)

(b) State clearly any assumptions you have made by suggesting this model. (2)

Using your model,

(i) no cars arrive,

(ii) more than 3 cars arrive. (3)

(d) In any given 4 minute period, find \(m\) such that \(\mathrm{P}(X \gt m) = 0.0487\) (3)

(e) Using a suitable approximation find the probability that fewer than 15 cars arrive in any given 10 minute period. (6)

Scheme	Marks
\(X \sim \mathrm{Po}(2.5)\)	M1A1
	(2)

Mark scheme (b)

Scheme	Marks
Cars arrive at the toll booth independently/randomly Cars arrive one at a time The rate of arrival at a toll booth remains constant at 2.5 per minute	B1 B1
	(2)

Notes

Any two of the statements or equivalent. At least one must be in context. Need words that imply “cars arrive” or “rate of arrival.” SC no context but 2 correct reasons B1B0
No context but 1 correct reason B0B0

Mark scheme (c)

Scheme	Marks
(i) \(\mathrm{P}\left(X = 0\right) = \mathrm{e}^{-2.5} = 0.0821\)	B1
	(1)
(ii) \(\mathrm{P}(X \gt 3) = 1 - \mathrm{P}(X \leqslant 3)\)	M1
\(= 0.2424\)	A1
	(2)

Notes

(i) B1 awrt 0.0821
(ii) M1 for writing or finding \(1 - \mathrm{P}(X \leqslant 3)\)
A1 awrt 0.242

Scheme	Marks
Use of Po(10)	M1
\(1 - 0.0487 = 0.9513\)	M1
\(m = 15\)	A1 cao
	(3)

Mark scheme (e)

Scheme	Marks
\(Y \sim \mathrm{N}\left(25, 25\right)\)	B1B1
\(\mathrm{P}\left(X \lt 15\right) = \mathrm{P}(Y \leqslant 14.5)\)	M1
\(= \mathrm{P}\left(Z \leqslant \dfrac{14.5 - 25}{5}\right)\)	M1
\(= \mathrm{P}\left(Z \leqslant -2.1\right)\)	A1
\(= 0.01786\)	A1
	(6)
	(16 marks)

Notes

B1 use of normal
B1 using or seeing mean and variance of 25
These first two marks may be given if the following are seen in the correct places in the standardisation formula : 25 and \(\sqrt{25}\) or 5

M1 for attempting a continuity correction (14 \(\pm\) 0.5) or (15 \(\pm\) 0.5)
M1 for standardising using their mean and their standard deviation and using [14.5, 14, 13.5, 15 or 15.5] accept \(\pm\) z.
A1 correct z value \(\pm\)2.1 or \(\pm\dfrac{14.5 - 25}{5}\),
A1 awrt 0.0179
NB use of calculator gets full marks if the answer is awrt 0.0179.