| Date | May 2017 | Marks available | 2 | Reference code | 17M.3sp.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Show that | Question number | 3 | Adapted from | N/A |
Question
The discrete random variable \(X\) has the following probability distribution.
\({\text{P}}(X = x) = \left\{ {\begin{array}{*{20}{l}}
{p{q^{\frac{x}{2}}}}&{{\text{for }}x = 0,{\text{ }}2,{\text{ }}4,{\text{ }}6 \ldots {\text{ where }}p + q = 1,{\text{ }}0 < p < 1.} \\
0&{{\text{otherwise}}}
\end{array}} \right.\)
Show that the probability generating function for \(X\) is given by \(G(t) = \frac{P}{{1 - q{t^2}}}\).
Hence determine \({\text{E}}(X)\) in terms of \(p\) and \(q\).
The random variable \(Y\) is given by \(Y = 2X + 1\). Find the probability generating function for \(Y\).
Markscheme
\(G(t) = \sum {P(X = x){t^x}} \) (M1)
\( = p + pq{t^2} + p{q^2}{t^4} + \ldots \)
(summing \(GP\)) \({u_1} = p,{\text{ }}r = q{t^2}\) A1
\( = \frac{p}{{1 - q{t^2}}}\) AG
[2 marks]
\(G’(t) = - \frac{p}{{{{(1 - q{t^2})}^2}}} \times - 2qt\) M1A1
\({\text{E}}(X) = G’(1)\) (M1)
\( = \frac{{2pq}}{{{{(1 - q)}^2}}}\,\,\,\left( { = \frac{{2q}}{p}} \right)\) A1
[4 marks]
METHOD 1
\({\text{PGF of }}Y = \sum {P(Y = y){t^y}} \) (M1)
\( = pt + pq{t^5} + p{q^2}{t^9} + \ldots \) A1
\( = \frac{{pt}}{{1 - q{t^4}}}\) A1
METHOD 2
\({\text{PGF of }}Y = {\text{E}}({t^Y})\) (M1)
\( = {\text{E}}({t^{2X + 1}})\)
\( = {\text{E}}\left( {{{({t^2})}^X}} \right) \times {\text{E}}(t)\) A1
\( = \frac{{pt}}{{1 - q{t^4}}}\) A1
[3 marks]
