| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moment generating functions |
| Type | Derive MGF from PDF |
| Difficulty | Standard +0.8 This is a Further Maths S4 question requiring integration to derive an MGF, then using Taylor series expansion to extract moments—both non-trivial techniques. While the uniform distribution is conceptually simple, the algebraic manipulation of exponentials and series expansion to obtain variance requires careful multi-step work beyond standard A-level, placing it moderately above average difficulty. |
| Spec | 4.08a Maclaurin series: find series for function5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_a^b \frac{1}{b-a} e^{tx}\, dx\) | M1 | |
| \(\dfrac{e^{bt} - e^{at}}{t(b-a)}\) AG | A1 | Need \(\left[\dfrac{e^{tx}}{t(b-a)}\right]_a^b\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{(1+bt+\frac{b^2t^2}{2}+\frac{b^3t^3}{6}+(\ldots))-(1+at+\frac{a^2t^2}{2}+\frac{a^3t^3}{6}+(\ldots))}{t(b-a)}\) | M1 | As far as terms in \(t^2\). Allow num only or sign error. Use of \(M''(0)-(M'(0))^2\); M1A1 as main scheme. \([\frac{1}{2}(b^2-a^2)+\frac{1}{3}t(b^3-a^3)]/(b-a)\) A1 |
| \(1 + \dfrac{(b^2-a^2)}{(b-a)}\dfrac{t}{2} + \dfrac{(b^3-a^3)}{(b-a)}\dfrac{t^2}{6}\) allow \((b-a)/(b-a)\) for 1 | A1 | \(\dfrac{b^3-a^3}{3(b-a)}\) A1 |
| Simplify 3rd term to \(\frac{1}{6}(b^2+ab+a^2)t^2\) | A1 | \(\dfrac{b^3-a^3}{3(b-a)} - \dfrac{(b+a)^2}{4}\) M1 |
| \(E(Y) = \dfrac{b+a}{2}\) AG | A1 | \(\dfrac{(b-a)^2}{12}\) A1 CWO |
| \([E(Y^2)] = \dfrac{b^2+ab+a^2}{3}\) oe | A1 | |
| Use \(\text{Var}(Y) = E(Y^2) - (E(Y))^2\) | M1 | |
| \(\dfrac{(b-a)^2}{12}\) AG CWO | A1 | |
| [7] |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_a^b \frac{1}{b-a} e^{tx}\, dx$ | M1 | |
| $\dfrac{e^{bt} - e^{at}}{t(b-a)}$ AG | A1 | Need $\left[\dfrac{e^{tx}}{t(b-a)}\right]_a^b$ |
| **[2]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{(1+bt+\frac{b^2t^2}{2}+\frac{b^3t^3}{6}+(\ldots))-(1+at+\frac{a^2t^2}{2}+\frac{a^3t^3}{6}+(\ldots))}{t(b-a)}$ | M1 | As far as terms in $t^2$. Allow num only or sign error. Use of $M''(0)-(M'(0))^2$; M1A1 as main scheme. $[\frac{1}{2}(b^2-a^2)+\frac{1}{3}t(b^3-a^3)]/(b-a)$ A1 |
| $1 + \dfrac{(b^2-a^2)}{(b-a)}\dfrac{t}{2} + \dfrac{(b^3-a^3)}{(b-a)}\dfrac{t^2}{6}$ allow $(b-a)/(b-a)$ for 1 | A1 | $\dfrac{b^3-a^3}{3(b-a)}$ A1 |
| Simplify 3rd term to $\frac{1}{6}(b^2+ab+a^2)t^2$ | A1 | $\dfrac{b^3-a^3}{3(b-a)} - \dfrac{(b+a)^2}{4}$ M1 |
| $E(Y) = \dfrac{b+a}{2}$ AG | A1 | $\dfrac{(b-a)^2}{12}$ A1 CWO |
| $[E(Y^2)] = \dfrac{b^2+ab+a^2}{3}$ oe | A1 | |
| Use $\text{Var}(Y) = E(Y^2) - (E(Y))^2$ | M1 | |
| $\dfrac{(b-a)^2}{12}$ AG CWO | A1 | |
| **[7]** | | |
---4 The continuous random variable $Y$ has a uniform (rectangular) distribution on $[ a , b ]$, where $a$ and $b$ are constants.\\
(i) Show that the moment generating function $\mathrm { M } _ { Y } ( \mathrm { t } )$ of $Y$ is $\frac { \left( \mathrm { e } ^ { b t } - \mathrm { e } ^ { a t } \right) } { t ( b - a ) }$.\\
(ii) Use the series expansion of $\mathrm { e } ^ { x }$ to show that the mean and variance of $Y$ are $\frac { 1 } { 2 } ( a + b )$ and $\frac { 1 } { 12 } ( b - a ) ^ { 2 }$, respectively.
\hfill \mbox{\textit{OCR S4 2016 Q4 [9]}}