16N.2.AHL.TZ0.H_3

pestleMathematicsAIHLPaper 216N· sl-4-8-binomial-distributionsource ↗

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]

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]

Markscheme / solution

* 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]

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

attempting to solve for $\mu$ (M1)

$\mu = 1.40$ A1

[3 marks]

Examiners’ report

[N/A]