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Date	November 2016	Marks available	3	Reference code	16N.2.AHL.TZ0.H_3
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Show that	Question number	H_3	Adapted from	N/A

Question

A discrete random variable $X$ follows a Poisson distribution ${\text{Po}}(\mu )$ .

Show that ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x),{\text{ }}x \in \mathbb{N}$ .

[3]

a.

Given that ${\text{P}}(X = 2) = 0.241667$ and ${\text{P}}(X = 3) = 0.112777$ , use part (a) to find the value of $\mu$ .

[3]

b.

Markscheme

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

METHOD 1

${\text{P}}(X = x + 1) = \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ - \mu }}$ A1

$= \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ - \mu }}$ M1A1

$= \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$ AG

METHOD 2

$\frac{\mu }{{x + 1}} \times {\text{P}}(X = x) = \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ - \mu }}$ A1

$= \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ - \mu }}$ M1A1

$= {\text{P}}(X = x + 1)$ AG

METHOD 3

$\frac{{{\text{P}}(X = x + 1)}}{{{\text{P}}(X = x)}} = \frac{{\frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ - \mu }}}}{{\frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ - \mu }}}}$ (M1)

$= \frac{{{\mu ^{x + 1}}}}{{{\mu ^x}}} \times \frac{{x!}}{{(x + 1)!}}$ A1

$= \frac{\mu }{{x + 1}}$ A1

and so ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$ AG

[3 marks]

a.

${\text{P}}(X = 3) = \frac{\mu }{3} \bullet {\text{P}}(X = 2){\text{ }}\left( {0.112777 = \frac{\mu }{3} \bullet 0.241667} \right)$ A1

attempting to solve for $\mu$ (M1)

$\mu = 1.40$ A1

[3 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Syllabus sections

Topic 4—Statistics and probability » SL 4.8—Binomial distribution

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Topic 4—Statistics and probability