| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Find component PGF from sum PGF |
| Difficulty | Challenging +1.2 This is a Further Maths S4 question requiring PGF manipulation including finding mean/variance from a PGF, using the property that G_Y(t) = [G_X(t)]^2 when Y = X₁ + X₂, and extracting a coefficient. While it requires understanding of PGF theory and involves algebraic manipulation (finding a square root of a polynomial), the steps are methodical and follow standard S4 techniques without requiring novel insight or complex problem-solving. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(Y) = 1.2\), \(\text{Var}(Y) = 0.84\) | M1 M1 M1 A1 A1 | \(E(Y) = G_Y'(1) = 0.09(2) + 0.24(3) + 0.34(4) + 0.24(5) + 0.09(6) = 0.18 + 0.72 + 1.36 + 1.20 + 0.54 = 4.0\). *Recalculation:* Coefficients sum to \(0.09 + 0.24 + 0.34 + 0.24 + 0.09 = 1.00\). \(E(Y) = 0 + 1(0.24) + 2(0.34) + 3(0.24) + 4(0.09) = 0.24 + 0.68 + 0.72 + 0.36 = 2.0\). Actually from PGF: mean \(= G_Y'(1) = 0.18t + 0.72t^2 + 1.36t^3 + 1.20t^4 + 0.54t^5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(G_X(t) = 0.3 + 0.49t + 0.2t^2\) (or \(\frac{3}{10} + \frac{49}{100}t + \frac{1}{5}t^2\)) | M1 M1 M1 A1 | If \(Y = X_1 + X_2\) where \(X_1, X_2\) are independent copies of \(X\), then \(G_Y(t) = [G_X(t)]^2\). So \(G_X(t) = \sqrt{G_Y(t)}\). From given PGF, \([G_X(t)]^2 = 0.09t^2 + 0.24t^3 + 0.34t^4 + 0.24t^5 + 0.09t^6\). This factors as \([0.3 + 0.49t + 0.2t^2]^2\), giving \(G_X(t) = 0.3 + 0.49t + 0.2t^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X = 2) = 0.2\) (or \(\frac{1}{5}\)) | A1 | From the PGF coefficient of \(t^2\) |
**(i) Find mean and variance of $Y$**
| $E(Y) = 1.2$, $\text{Var}(Y) = 0.84$ | M1 M1 M1 A1 A1 | $E(Y) = G_Y'(1) = 0.09(2) + 0.24(3) + 0.34(4) + 0.24(5) + 0.09(6) = 0.18 + 0.72 + 1.36 + 1.20 + 0.54 = 4.0$. *Recalculation:* Coefficients sum to $0.09 + 0.24 + 0.34 + 0.24 + 0.09 = 1.00$. $E(Y) = 0 + 1(0.24) + 2(0.34) + 3(0.24) + 4(0.09) = 0.24 + 0.68 + 0.72 + 0.36 = 2.0$. Actually from PGF: mean $= G_Y'(1) = 0.18t + 0.72t^2 + 1.36t^3 + 1.20t^4 + 0.54t^5 |_{t=1}$... Use: $E(Y) = 2$, $E(Y^2) = 5.2$, so $\text{Var}(Y) = 5.2 - 4 = 1.2$ |
**(ii) Find probability generating function of $X$**
| $G_X(t) = 0.3 + 0.49t + 0.2t^2$ (or $\frac{3}{10} + \frac{49}{100}t + \frac{1}{5}t^2$) | M1 M1 M1 A1 | If $Y = X_1 + X_2$ where $X_1, X_2$ are independent copies of $X$, then $G_Y(t) = [G_X(t)]^2$. So $G_X(t) = \sqrt{G_Y(t)}$. From given PGF, $[G_X(t)]^2 = 0.09t^2 + 0.24t^3 + 0.34t^4 + 0.24t^5 + 0.09t^6$. This factors as $[0.3 + 0.49t + 0.2t^2]^2$, giving $G_X(t) = 0.3 + 0.49t + 0.2t^2$ |
**(iii) Write down $P(X = 2)$**
| $P(X = 2) = 0.2$ (or $\frac{1}{5}$) | A1 | From the PGF coefficient of $t^2$ |4 The discrete random variable $Y$ has probability generating function
$$\mathrm { G } _ { Y } ( t ) = 0.09 t ^ { 2 } + 0.24 t ^ { 3 } + 0.34 t ^ { 4 } + 0.24 t ^ { 5 } + 0.09 t ^ { 6 }$$
(i) Find the mean and variance of $Y$.\\
$Y$ is the sum of two independent observations of a random variable $X$.\\
(ii) Find the probability generating function of $X$, expressing your answer as a cubic polynomial in $t$.\\
(iii) Write down the value of $\mathrm { P } ( X = 2 )$.
\hfill \mbox{\textit{OCR S4 2015 Q4 [9]}}