A2 June 2023 Q2

EdexcelCurrent spec11 marksIncludes hypothesis testingPoisson Distribution

2. Telephone calls arrive at a call centre randomly, at an average rate of 1.7 per minute.
After the call centre was closed for a week, in a random sample of 10 minutes there were 25 calls to the call centre.

(a) Carry out a suitable test to determine whether or not there is evidence that the rate of calls arriving at the call centre has changed.
Use a 5% level of significance and state your hypotheses clearly. (4)

Only 1.2% of the calls to the call centre last longer than 8 minutes.

One day Tiang has 70 calls.

(b) Find the probability that out of these 70 calls Tiang has more than 2 calls lasting longer than 8 minutes. (3)

The call centre records show that 95% of days have at least one call lasting longer than 30 minutes.
On Wednesday 900 calls arrived at the call centre and none of them lasted longer than 30 minutes.

(c) Use a Poisson approximation to estimate the proportion of calls arriving at the call centre that last longer than 30 minutes. (4)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{H}_0: \lambda = 1.7 \qquad \mathrm{H}_1: \lambda \neq 1.7\)	B1	2.5
[\(X\) = no. of calls in 10 mins] \(\quad X \sim \mathrm{Po}(17)\)	M1	3.3
[\(\mathrm{P}(X \geqslant 25) = 1 - \mathrm{P}(X \leqslant 24)\)] \(= 0.0406463\ldots\) or CR: \(X \geqslant 27\)	A1	3.4
[\(0.04.. \gt 0.025\) / 25 is not in CR so not significant] insufficient evidence of a change in rate of calls	A1	2.2b
	(4)

Notes

B1 for both hypotheses correct which must be attached to \(\mathrm{H}_0\) and \(\mathrm{H}_1\)
must be in terms of \(\lambda\) or \(\mu\) allow either 1.7 or 17

M1 for stating or using the correct Poisson model.
may be implied by sight of awrt 0.0406/7 or awrt 0.959 or 0.9747… or better

1^st A1 for correct prob of awrt 0.04
or for correct CR found \(X \geqslant 27\) (\(X \gt 26\)) (ignore lower tail CR if found) allow CV \(X = 27\)

2^nd A1 (dep on M1A1) for a correct conclusion in context mentioning “rate of calls” o.e.
Allow e.g. ‘The rate of calls is 1.7 per minute/17 per 10 minutes’
Must be rate o.e. not “number”
A0 if inconsistent comments are seen e.g. “reject \(\mathrm{H}_0\), no change in rate of calls”

Mark scheme (b)

Scheme	Marks	AO
[\(T\) = no. of calls longer than 8 minutes] \(\quad T \sim \mathrm{B}(70, 0.012)\)	M1	3.3
[\(\mathrm{P}(T \gt 2) =\)] \(\mathrm{P}(T \geqslant 3) = 1 - \mathrm{P}(T \leqslant 2) = 1 - 0.947725\ldots\)	M1	3.4
= awrt 0.0523	A1	1.1b
	(3)

Notes

1^st M1 for sight or use of the correct binomial model.
may be implied by sight of awrt: 0.0523 or 0.948 or 0.795 or 0.205

2^nd M1 for correct interpretation of more than 2 (allow 1 – 0.95 or better)

A1 for awrt 0.0523 (correct answer only scores 3 out of 3)

SC: Use of \(\mathrm{Po}(70 \times 0.012)\) leading to an answer of 0.0533(45…) and scores M1M1A0

Mark scheme (c)

Scheme	Marks	AO
[\(C\) = no. of calls out of 900 longer than 30 mins] [\(C \sim \mathrm{B}(900, p)\)] \(\quad C \approx\sim \mathrm{Po}(900p)\)	M1	3.3
\(\mathrm{P}(C = 0) \approx \mathrm{e}^{-900p} = 0.05\)	M1	3.4
\(900p = -\ln(0.05)\ [= 2.9957\ldots]\)	M1	1.1b
\(p = 0.003328\ldots\) awrt 0.00333	A1	1.1b
	(4)
	(11 marks)

Notes

1^st M1 for sight or use of \(\mathrm{Po}(900p)\) (as a suitable approx. to \(\mathrm{B}(900, p)\))
(may be implied by correct answer awrt 0.00333)

2^nd M1 for a correct equation in \(p\) or correct use of \(\mathrm{P}(C = 0)\) from Po e.g. \(\mathrm{e}^{-\lambda} = 0.05\)

3^rd M1 for a correct method to solve for \(p\) (allow \(p = \pm\ln(0.05)/900\))
or to solve for \(\lambda\), i.e. \(\lambda =\) awrt 3(.00)

A1 for \(p\) = awrt 0.00333 Must see Po used condone \(\dfrac{1}{300}\) o.e.
Allow standard form (awrt \(3.33 \times 10^{-3}\)) or percentage (awrt 0.333%)

SC: Use of Binomial gives 0.003323… awrt 0.00332 scores M0M0M0A1