| Date | November 2016 | Marks available | 3 | Reference code | 16N.2.hl.TZ0.3 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
A discrete random variable \(X\) follows a Poisson distribution \({\text{Po}}(\mu )\).
Show that \({\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x),{\text{ }}x \in \mathbb{N}\).
Given that \({\text{P}}(X = 2) = 0.241667\) and \({\text{P}}(X = 3) = 0.112777\), use part (a) to find the value of \(\mu \).
Markscheme
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]
