| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Calculate probabilities and expectations |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics question on continuous uniform distributions. Part (a) is trivial sketching, part (b) requires solving a linear inequality with the uniform distribution (routine manipulation), and part (c) is direct integration of x³ over [-3,5] with the constant pdf—all standard textbook exercises requiring no novel insight, though the algebraic manipulation in parts (b) and (c) elevates it slightly above average difficulty. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Rectangle shape with height \(\frac{1}{8}\) | B1 | For correct shape |
| Labels \(-3\) and \(5\) on axis | B1 | For correct labels |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X < 2(k-X)) = P\left(X < \frac{2}{3}k\right)\) | M1 | For simplifying to \(P\left(X < \frac{2}{3}k\right)\); alternatively for understanding \(2[k-x] = -1\) and \(x = -1\) |
| \(\frac{\frac{2}{3}k - (-3)}{5-(-3)} = 0.25\) | M1 | For equating probability expression to 0.25; alternatively for substitution and attempt to solve |
| \(k = -\frac{3}{2}\) | A1 | For \(-\frac{3}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X^3) = \int_{-3}^{5} \frac{1}{5-(-3)} x^3 dx\) | M1 | For integrating \(x^3 f(x)\); dependent on previous M1 for correct limits |
| \(= \left[\frac{1}{32}x^4\right]_{-3}^{5} = \frac{1}{32}(5^4 - (-3)^4)\) | dM1 | |
| \(= 17\) * | A1*cso | For fully correct solution leading to given answer with no errors seen |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Rectangle shape with height $\frac{1}{8}$ | B1 | For correct shape |
| Labels $-3$ and $5$ on axis | B1 | For correct labels |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X < 2(k-X)) = P\left(X < \frac{2}{3}k\right)$ | M1 | For simplifying to $P\left(X < \frac{2}{3}k\right)$; alternatively for understanding $2[k-x] = -1$ and $x = -1$ |
| $\frac{\frac{2}{3}k - (-3)}{5-(-3)} = 0.25$ | M1 | For equating probability expression to 0.25; alternatively for substitution and attempt to solve |
| $k = -\frac{3}{2}$ | A1 | For $-\frac{3}{2}$ |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X^3) = \int_{-3}^{5} \frac{1}{5-(-3)} x^3 dx$ | M1 | For integrating $x^3 f(x)$; dependent on previous M1 for correct limits |
| $= \left[\frac{1}{32}x^4\right]_{-3}^{5} = \frac{1}{32}(5^4 - (-3)^4)$ | dM1 | |
| $= 17$ * | A1*cso | For fully correct solution leading to given answer with no errors seen |\begin{enumerate}
\item The continuous random variable $X$ is uniformly distributed over the interval $[ - 3,5 ]$.\\
(a) Sketch the probability density function $\mathrm { f } ( \mathrm { x } )$ of X .\\
(b) Find the value of k such that $\mathrm { P } ( \mathrm { X } < 2 [ \mathrm { k } - \mathrm { X } ] ) = 0.25$\\
(c) Use algebraic integration to show that $\mathrm { E } \left( \mathrm { X } ^ { 3 } \right) = 17$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FS2 AS Q4 [8]}}