S2 June 2018 Q1

EdexcelOld spec12 marksPoisson Distribution

1. In a call centre, the number of telephone calls, \(X\), received during any 10-minute period follows a Poisson distribution with mean 9

(a) Find

(i) \(\mathrm{P}(X \gt 5)\)

(ii) \(\mathrm{P}(4 \leqslant X \lt 10)\) (4)

The length of a working day is 7 hours.

(b) Using a suitable approximation, find the probability that there are fewer than 370 telephone calls in a randomly selected working day. (5)

A week, consisting of 5 working days, is selected at random.

Mark scheme (a)

Throughout the paper the candidates may use different letters to the ones given in the mark scheme

Scheme	Marks
\(X\) represents number of telephone calls per min \(\Rightarrow X \sim \mathrm{Po}(9)\)
(i) \(\mathrm{P}(X \gt 5) = 1 - \mathrm{P}(X \leqslant 5)\)	M1
\(= 0.8843\) awrt 0.884	A1
(ii) \(\mathrm{P}(4 \leqslant X \lt 10) = \mathrm{P}(X \leqslant 9) - \mathrm{P}(X \leqslant 3)\)	M1
\(= 0.5874 - 0.0212\) \(= 0.5662\) awrt 0.566	A1
	(4)

Notes

(i) M1 for using or writing \(1 - \mathrm{P}(X \leqslant 5)\) or \(1 - \mathrm{P}(X \lt 6)\) may be implied by awrt 0.884

(ii) M1 for using or writing \(\mathrm{P}(X \leqslant 9) - \mathrm{P}(X \leqslant 3)\) oe may be implied by awrt 0.566

Mark scheme (b)

Scheme	Marks
\(D\) represents number of telephone calls per day
Normal approximation \(\mu = \dfrac{7\times60\times9}{10} = 378\) and \(\sigma^2 = 378\)	M1
\(\mathrm{P}(D \lt 370) \approx \mathrm{P}\left(Z \lt \dfrac{369.5 - 378}{\sqrt{378}}\right)\) standardise, \(\pm 0.5\)	M1, M1d A1ft
\(\approx \mathrm{P}(Z \lt -0.44)\) \(= 1 - 0.670\) \(= 0.33\) or 0.330 or awrt 0.331	A1
	(5)

Notes

M1 Using normal approximation with mean = variance = 378 or sd = \(\sqrt{378}\) (awrt 19.4) or writing N(378,378) May be seen in standardisation.

M1 \(\pm\left(\dfrac{(369 \text{ or } 370 \text{ or } 369.5 \text{ or } 370.5) - \text{their mean}}{\text{their sd}}\right)\) If they have not given a mean and variance they must be correct in here. (allow 1 – standardisation)

M1d dep on previous method mark being awarded. Using a continuity correction \(370 \pm 0.5\)

A1ft standardisation with correct CC ie \(\pm\dfrac{369.5 - \text{"their 378"}}{\sqrt{\text{"their 378"}}}\) or awrt \(\pm\)0.44 or implied by 0.330 or 0.331 (allow 1 – standardisation) (0.33 must be from correct standardisation) NB 0.33 with no working gains NO marks. 0.330 or 0.331 with no working gains full marks.

Mark scheme (c)

Scheme	Marks
\(W\) represents number of days which have fewer than 370 telephone calls
\(W \sim \mathrm{B}(5, \text{"}0.33\text{"})\) \| \(W \sim \mathrm{B}(5, \text{"}0.67\text{"})\)	M1
\(\mathrm{P}(W = 4) + \mathrm{P}(W = 5)\) \| \(\mathrm{P}(W = 0) + \mathrm{P}(W = 1)\) \(= 5(\text{"}0.33\text{"})^4(1 - \text{"}0.33\text{"}) + (\text{"}0.33\text{"})^5\) \| \(= (1 - \text{"}0.67\text{"})^5 + 5(\text{"}0.67\text{"})(\text{"}1 - 0.67\text{"})^4\)	M1
\(= 0.0436\) awrt 0.044	A1
	(3)
	(12 marks)

Notes

M1 writing \(\mathrm{B}(5, \text{"}0.33\text{"})\) or \(\mathrm{B}(5, 1 - \text{"}0.33\text{"})\) or seeing \({}^5C_n(\text{"}0.33\text{"})^n(1 - \text{"}0.33\text{"})^{5-n}\) where \(1 \leqslant n \leqslant 4\) Allow if \({}^nC_r\) calculated or in factorial form

M1 \(1 - (1 - \text{"}0.33\text{"})^5 - 5(\text{"}0.33\text{"})^1(1 - \text{"}0.33\text{"})^4 - 10(\text{"}0.33\text{"})^2(1 - \text{"}0.33\text{"})^3 - 10(\text{"}0.33\text{"})^3(1 - \text{"}0.33\text{"})^2\) oe Allow if using \({}^nC_r\) form or factorial form

NB awrt 0.044 with no incorrect working gains M1M1A1