S2 June 2013 (R) Q5

EdexcelOld spec13 marksPoisson Distribution

5. In a village shop the customers must join a queue to pay. The number of customers joining the queue in a 10 minute interval is modelled by a Poisson distribution with mean 3

Find the probability that

(a) exactly 4 customers join the queue in the next 10 minutes, (2)

(b) more than 10 customers join the queue in the next 20 minutes. (3)

When a customer reaches the front of the queue the customer pays the assistant. The time each customer takes paying the assistant, \(T\) minutes, has a continuous uniform distribution over the interval [0, 5]. The random variable \(T\) is independent of the number of people joining the queue.

In a random sample of 5 customers, the random variable \(C\) represents the number of customers who took more than 3.5 minutes paying the assistant.

(d) Find \(\mathrm{P}(C \geqslant 3)\) (3)

Bethan has just reached the front of the queue and starts paying the assistant.

(e) Find the probability that in the next 4 minutes Bethan finishes paying the assistant and no other customers join the queue. (4)

Mark scheme (a)

Scheme	Marks
[\(X\) = number of customers joining the queue in the next 10 mins \(\sim \mathrm{Po}(3)\)]
\(\mathrm{P}(X = 4) = \mathrm{P}(X \leqslant 4) - \mathrm{P}(X \leqslant 3)\) or \(\dfrac{\mathrm{e}^{-3}3^4}{4!}\)	M1
\(0.8153 - 0.6472 = 0.1681\) or \(0.1680313\ldots\) (awrt 0.168)	A1
	(2)

Notes

M1 for a correct method. May use incorrect \(\lambda\)

A1 for awrt 0.168

Mark scheme (b)

Scheme	Marks
\(Y\) [= number of customers joining the queue in the next 20 mins] \(\sim \mathrm{Po}(6)\)	B1
\(\mathrm{P}(Y \gt 10) = 1 - \mathrm{P}(Y \leqslant 10)\)	M1
\(= 1 - 0.9574 = 0.0426(209\ldots)\) (awrt 0.0426)	A1
	(3)

Notes

B1 for writing or using Po(6)

M1 for writing or using \(1 - \mathrm{P}(Y \leqslant 10)\)

A1 for awrt 0.0426

Scheme	Marks
\(\mathrm{P}(T \gt 3.5) = \underline{\mathbf{0.3}}\)	B1
	(1)

Mark scheme (d)

Scheme	Marks
\(C \sim \mathrm{B}(5, 0.3)\)	M1
\(\mathrm{P}(C \geqslant 3) = 1 - \mathrm{P}(C \leqslant 2)\)	M1
\(= 1 - 0.8369 = 0.1631\) (or \(0.16308..\)) (awrt 0.163)	A1
	(3)

Notes

1st M1 for identifying that \(C \sim \mathrm{B}(5, 0.3)\). Follow through their 0.3. May be implied

2nd M1 for writing or using \(1 - \mathrm{P}(C \leqslant 2)\)

A1 for awrt 0.163

SC if they use normal distribution they may get M0 M1 A0 if they find \(\mathrm{P}(C \geqslant 2.5)\)

Mark scheme (e)

Scheme	Marks
P(Bethan is served in < 4 minutes) = 0.8 (o.e.)	B1
\(J\) = number joining the queue in 4 mins has \(J \sim \mathrm{Po}(1.2)\)	M1
\(\mathrm{P}(J = 0) = \mathrm{e}^{-1.2} = 0.30119\ldots\)	A1
P(Bethan is served and \(J = 0\)) \(= 0.8 \times \mathrm{e}^{-1.2} = 0.240955\ldots\) (awrt 0.241)	A1
	(4)
	(13 marks)

Notes

B1 for 0.8 for P(Bethan is served in the next 4 minutes)

M1 for identifying Po(1.2)

A1 for \(\mathrm{e}^{-1.2}\) or awrt 0.301...

A1 for awrt 0.241