| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Topic | Probability Generating Functions |
| Type | Use PGF to find mean and variance |
| Difficulty | Challenging +1.2 This is a standard PGF question requiring routine application of differentiation formulas for mean and variance, followed by binomial expansion to identify the distribution. Part (i) uses memorized formulas G'(1) and G''(1)+G'(1)-[G'(1)]², while part (ii) requires expanding (t+2/t)⁴ and recognizing the binomial pattern after transformation. The algebraic manipulation is moderately involved but follows established techniques with no novel insight required, placing it slightly above average difficulty. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) | \(-4\) | \(-2\) |
| \(y = \frac{1}{2}(x+4)\) | \(0\) | \(1\) |
| \(\mathrm{P}(X=x) = \mathrm{P}(Y=y)\) | \(\dfrac{16}{81}\) | \(\dfrac{32}{81}\) |
**Question 3(i) and 3(ii)**
**(i)**
$G'(t) = \dfrac{1}{81} \times 4\left(t + \dfrac{2}{t}\right)^3\left(1 - \dfrac{2}{t^2}\right)$ — M1
$\mathrm{E}(X) = G'(1) = \dfrac{4}{81} \times 27 \times (-1) = -\dfrac{4}{3}$ — A1
$G''(t) = \dfrac{4}{27}\left(t + \dfrac{2}{t}\right)^2\left(1 - \dfrac{2}{t^2}\right)^2 + \dfrac{4}{81}\left(t + \dfrac{2}{t}\right)^3 \times \dfrac{4}{t^3}$ — M1
$G''(1) = \dfrac{4}{27} \times 9 \times 1 + \dfrac{4}{81} \times 27 \times 4 = \dfrac{20}{3}$ — M1A1
$\mathrm{Var}(X) = \dfrac{20}{3} + \left(-\dfrac{4}{3}\right) - \dfrac{16}{9} = \dfrac{32}{9}$ — [5]
For information:
$G(t) = \dfrac{1}{81}\left(t^4 + 8t^2 + 24 + \dfrac{32}{t^2} + \dfrac{16}{t^4}\right)$
$G'(t) = \dfrac{1}{81}\left(4t^3 + 16t - \dfrac{64}{t^3} - \dfrac{64}{t^5}\right)$
$G''(t) = \dfrac{1}{81}\left(12t^2 + 16 + \dfrac{192}{t^4} + \dfrac{320}{t^6}\right)$
**[5]**
**(ii)**
| $x$ | $-4$ | $-2$ | $0$ | $2$ | $4$ |
|---|---|---|---|---|---|
| $y = \frac{1}{2}(x+4)$ | $0$ | $1$ | $2$ | $3$ | $4$ |
| $\mathrm{P}(X=x) = \mathrm{P}(Y=y)$ | $\dfrac{16}{81}$ | $\dfrac{32}{81}$ | $\dfrac{24}{81}$ | $\dfrac{8}{81}$ | $\dfrac{1}{81}$ |
— B1 ($x$ and probabilities), B1 ($y$)
Or: $G_{X+4}(t) = t^4 G_X(t)$; $G_{\frac{1}{2}(X+4)}(t) = G_{X+4}(\sqrt{t})$
Recognising as terms of the expansion of $\left(\dfrac{2}{3} + \dfrac{1}{3}\right)^4$
State $n = 4$ and $p = \dfrac{1}{3}$ — B1B1 (Independent of method BUT max 3 if binomial not shown)
**[4]**3 The probability generating function of the random variable $X$ is $\frac { 1 } { 81 } \left( t + \frac { 2 } { t } \right) ^ { 4 }$.\\
(i) Use the probability generating function to find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.\\
(ii) The random variable $Y$ is defined by $Y = \frac { 1 } { 2 } ( X + 4 )$. By finding the probability distribution of $X$, or otherwise, show that $Y \sim \mathrm {~B} ( n , p )$, stating the values of $n$ and $p$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2015 Q3 [14]}}