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Date	November 2018	Marks available	1	Reference code	18N.2.AHL.TZ0.H_10
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Give	Question number	H_10	Adapted from	N/A

Question

Willow finds that she receives approximately 70 emails per working day.

She decides to model the number of emails received per working day using the random variable $X$ , where $X$ follows a Poisson distribution with mean 70.

In order to test her model, Willow records the number of emails she receives per working day over a period of 6 months. The results are shown in the following table.

From the table, calculate

Archie works for a different company and knows that he receives emails according to a Poisson distribution, with a mean of $\lambda$ emails per day.

Using this distribution model, find ${\text{P}}\left( {X < 60} \right)$ .

[2]

a.i.

Using this distribution model, find the standard deviation of $X$ .

[2]

a.ii.

an estimate for the mean number of emails received per working day.

[3]

b.i.

an estimate for the standard deviation of the number of emails received per working day.

[2]

b.ii.

Give one piece of evidence that suggests Willow’s Poisson distribution model is not a good fit.

[1]

Suppose that the probability of Archie receiving more than 10 emails in total on any one day is 0.99. Find the value of λ.

[3]

Now suppose that Archie received exactly 20 emails in total in a consecutive two day period. Show that the probability that he received exactly 10 of them on the first day is independent of λ.

[5]

Markscheme

${\text{P}}\left( {X < 60} \right)$

$= {\text{P}}\left( {X \leqslant 59} \right)$ (M1)

= 0.102 A1

[2 marks]

a.i.

standard deviation = $\sqrt {70}$ (= 8.37) (M1)A1

[2 marks]

a.ii.

use of midpoints (accept consistent use of 45, 55 etc.) (M1)

$\frac{{44.5 \times 2 + 54.5 \times 15 + 64.5 \times 40 + 74.5 \times 53 + 94.5 + 104.5 \times 3 + 114.5 \times 6}}{{2 + 15 + 40 + 53 + 0 + 1 + 3 + 6}}$ (M1)

$= \frac{{8530}}{{120}}\left( { = 71.1} \right)$ A1

Note: If 45, 55, etc. are used consistently instead of midpoints (implied by the answer 71.58…) award M1M1A0.

[3 marks]

b.i.

13.9 (M1)A1

[2 marks]

b.ii.

valid reason given to include the examples below R1

variance is 192 which is not close to the mean (accept not equal to) standard deviation too high (using parts (a)(ii) and (b)(ii))

relative frequency of $X$ ≤ 59 is 0.142 which is too high (using part (a)(i))

Poisson would give a frequency of roughly 14 for 80 ≤ $X$ ≤ 89

Note: Reasons which do not use values found in previous parts must be backed up with numerical evidence.

[1 mark]

${\text{P}}\left( {Y > 10} \right) = 0.99$

$1 - {\text{P}}\left( {Y \leqslant 10} \right) = 0.99 \Rightarrow {\text{P}}\left( {Y \leqslant 10} \right) = 0.01$ (M1)

attempt to solve a correct equation (M1)

λ = 20.1 A1

[3 marks]

in 1 day, no of emails is X ~ Po(λ)

in 2 days, no of emails is Y ~ Po(2λ) (A1)

P(10 on first day | 20 in 2 days) (M1)

$\frac{{{\text{P}}\left( {X = 10} \right) \times {\text{P}}\left( {X = 10} \right)}}{{{\text{P}}\left( {Y = 20} \right)}}$ (M1)

$= \frac{{{{\left( {\frac{{{\lambda ^{10}}{e^{ - \lambda }}}}{{10{\text{!}}}}} \right)}^2}}}{{\frac{{{{\left( {2\lambda } \right)}^{20}}{e^{ - 2\lambda }}}}{{20{\text{!}}}}}}$ A1

$= \frac{{{\lambda ^{20}}{e^{ - 2\lambda }}}}{{{2^{20}}{\lambda ^{20}}{e^{ - 2\lambda }}}} \times \frac{{20{\text{!}}}}{{{{\left( {10{\text{!}}} \right)}^2}}}$ A1