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Date	May 2019	Marks available	4	Reference code	19M.2.AHL.TZ2.H_10
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Show that	Question number	H_10	Adapted from	N/A

Question

Steffi the stray cat often visits Will’s house in search of food. Let $X$ be the discrete random variable “the number of times per day that Steffi visits Will’s house”.

The random variable $X$ can be modelled by a Poisson distribution with mean 2.1.

Let Y be the discrete random variable “the number of times per day that Steffi is fed at Will’s house”. Steffi is only fed on the first four occasions that she visits each day.

Find the probability that on a randomly selected day, Steffi does not visit Will’s house.

[2]

a.

Copy and complete the probability distribution table for Y.

[4]

b.

Hence find the expected number of times per day that Steffi is fed at Will’s house.

[3]

c.

In any given year of 365 days, the probability that Steffi does not visit Will for at most $n$ days in total is 0.5 (to one decimal place). Find the value of $n$ .

[3]

d.

Show that the expected number of occasions per year on which Steffi visits Will’s house and is not fed is at least 30.

[4]

e.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$X \sim {\text{Po}}\left( {2.1} \right)$

${\text{P}}\left( {X = 0} \right) = 0.122\left( { = {{\text{e}}^{ - 2.1}}} \right)$ (M1)A1

[2 marks]

a.

A1A1A1A1

Note: Award A1 for each correct probability for Y = 1, 2, 3, 4. Accept 0.162 for P(Y = 4).

[4 marks]

b.

${\text{E}}\left( Y \right) = \sum {y{\text{P}}\left( {Y = y} \right)}$ (M1)

$= 1 \times 0.257 \ldots + 2 \times 0.270 \ldots + 3 \times 0.189 + 4 \times 0.161 \ldots$ (A1)

$= 2.01$ A1

[3 marks]

c.

let $T$ be the no of days per year that Steffi does not visit

$T \sim B\left( {365{\text{,}}\,\,0.122 \ldots } \right)$ (M1)

require $0.45 \leqslant {\text{P}}\left( {T \leqslant n} \right) < 0.55$ (M1)

${\text{P}}\left( {T \leqslant 44} \right) = 0.51$

$n = 44$ A1

[3 marks]

d.

METHOD 1

let $V$ be the discrete random variable “number of times Steffi is not fed per day”

${\text{E}}\left( V \right) = 1 \times {\text{P}}\left( {X = 5} \right) + 2 \times {\text{P}}\left( {X = 6} \right) + 3 \times {\text{P}}\left( {X = 7} \right) + \ldots$ M1

$= 1 \times 0.0416 \ldots + 2 \times 0.0145 \ldots + 3 \times 0.00437 \ldots + \ldots$ A1

= 0.083979... A1

expected no of occasions per year > 0.083979... × 365 = 30.7 A1

hence Steffi can expect not to be fed on at least 30 occasions AG

Note: Candidates may consider summing more than three terms in their calculation for ${\text{E}}\left( V \right)$ .

METHOD 2

${\text{E}}\left( X \right) - {\text{E}}\left( Y \right) = 0.0903 \ldots$ M1A1

0.0903… × 365 M1

= 33.0 > 30 A1AG

[4 marks]

e.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

Syllabus sections

Topic 4—Statistics and probability » SL 4.7—Discrete random variables

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