S2 June 2017 Q2

EdexcelOld spec13 marksIncludes hypothesis testingPoisson Distribution

2. A company receives telephone calls at random at a mean rate of 2.5 per hour.

(a) Find the probability that the company receives

(i) at least 4 telephone calls in the next hour,

(ii) exactly 3 telephone calls in the next 15 minutes. (5)

(b) Find, to the nearest minute, the maximum length of time the telephone can be left unattended so that the probability of missing a telephone call is less than 0.2 (3)

The company puts an advert in the local newspaper. The number of telephone calls received in a randomly selected 2 hour period after the paper is published is 10

(c) Test at the 5% level of significance whether or not the mean rate of telephone calls has increased. State your hypotheses clearly. (5)

Mark scheme (a)

Scheme	Marks
(i) \(X \sim \mathrm{Po}(2.5)\) \(\mathrm{P}(X \geqslant 4) = 1 - \mathrm{P}(X \leqslant 3)\) \(= 1 - 0.7576\)	M1
\(= 0.2424\)	A1
(ii) \(X \sim \mathrm{Po}(0.625)\)	B1
\(\mathrm{P}(X = 3) = \dfrac{\mathrm{e}^{-0.625}0.625^3}{3!}\)	M1
\(= 0.02177\ldots\)	A1
	(5)

Notes

(i) M1 writing or using \(1 - \mathrm{P}(X \leqslant 3)\) implied by awrt 0.242
A1 awrt 0.242

(ii) B1 Using Po(0.625)
M1 finding \(\mathrm{P}(X = 3)\) with any \(\lambda\) e.g \(\dfrac{\mathrm{e}^{-\lambda}\lambda^3}{3!}\) or \(\mathrm{P}(X \leqslant 3) - \mathrm{P}(X \leqslant 2)\) – may be implied by awrt 0.0218
A1 awrt 0.0218

Mark scheme (b)

Scheme	Marks
\(1 - \mathrm{P}(X = 0) \lt 0.2\) \(\mathrm{P}(X = 0) \gt 0.8\)	M1
\(\mathrm{e}^{-2.5t} \gt 0.8\) \(t \lt 0.089\ldots\) hours = 5.36 mins	M1
\([t \lt]\) 5 mins	A1cso
	(3)

Notes

1st M1 for writing or using \(1 - \mathrm{P}(X = 0) \lt 0.2\) or \(\mathrm{P}(X = 0) \gt 0.8\) oe allow use of = instead of > or <. May be implied by \(\mathrm{e}^{-\lambda} = 0.8\) or \(\mathrm{e}^{-\lambda} \gt 0.8\) or by awrt 5.36 or 0.089

2nd M1 writing an inequality of the form \(\mathrm{e}^{-\lambda} \gt 0.8\) using any \(\lambda\). May be implied by or by awrt 5.36 or 0.089 Do not allow \(\mathrm{e}^{-\lambda} = 0.8\)

A1cso both the method marks must be awarded. Accept 5 or \(t = 5\) or \(t \lt 5\)

Mark scheme (c)

Scheme	Marks
\(\mathrm{H}_0 : \lambda = 2.5\ (\lambda = 5)\) \(\mathrm{H}_1 : \lambda \gt 2.5\ (\lambda \gt 5)\)	B1
\(\mathrm{P}(X \geqslant 10) = 1 - \mathrm{P}(X \leqslant 9)\) \(= 1 - 0.9682\)	M1
\(= 0.0318\)	A1
Sufficient evidence to reject \(\mathrm{H}_0\), Accept \(\mathrm{H}_1\), significant. 10 does lie in the Critical region.	M1d
There is sufficient evidence that the mean rate of telephone calls has increased (oe)	A1cso
	(5)
	(13 marks)

Notes

B1 both hypotheses using \(\lambda\) or \(\mu\) - allow 5 or 2.5 and it must be clear which is \(\mathrm{H}_0\) and which is \(\mathrm{H}_1\)

1st M1 writing or using Po(5) and \(1 - \mathrm{P}(X \leqslant 9)\) May be implied by a correct CR. Do not allow for writing \(\mathrm{P}(X \geqslant 10)\)

1st A1 awrt 0.0318. Allow CR \(X \geqslant 10\) or \(X \gt 9\)

NB allow M1A1 if not using CR route for \(\mathrm{P}(X \leqslant 9)\) = awrt 0.968

2nd M1 dependent on previous M being awarded. A correct statement (do not allow if there are contradicting non-contextual statements). ft their Prob/CR compared with 0.05/10 (0.95 if using 0.968)

2nd A1 A correct contextual statement must include the word calls and the idea the rate has increased. (do not allow “it has changed” on its own oe). All previous marks must be awarded for this mark to be awarded.
M1A1 is awarded for a correct contextual statement on its own provided previous marks have been awarded