| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | PDF from CDF |
| Difficulty | Challenging +1.2 This is a Further Maths statistics question requiring differentiation of CDF to get PDF, then standard integration for expectation and variance of a transformation, plus change of variables for finding PDF of Y=X³. While it involves multiple parts and transformations of random variables (a Further Maths topic), each step follows routine procedures without requiring novel insight—the techniques are direct applications of standard formulas once the PDF is obtained. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(6) - F(3) = \frac{9}{20}\) or \(0.45\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = \begin{cases} \frac{1}{30}x, & 2 \leq x \leq 8 \\ 0, & \text{otherwise} \end{cases}\) | B1 | May be implied. Only need to see \(f(x) = \frac{1}{30}x\) |
| \(E(\sqrt{X}) = \int_2^8 \frac{1}{30}x^{\frac{3}{2}}dx = \left[\frac{1}{75}x^{\frac{5}{2}}\right]\) | M1 | Integrated |
| \(= 2.34\) | A1 | \(2.338\ldots\); answer of \(2.34\) without \(\frac{1}{75}x^{\frac{5}{2}}\) scores B1B1 (2/3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Var}(\sqrt{X}) = \left[\int_2^8 \frac{1}{30}x^2 dx = \left[\frac{1}{90}x^3\right]\right] - E\!\left(\sqrt{X}\right)^2\) | M1 | |
| \(= 5.6 - 2.338^2 = 0.133\) or \(0.134\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G(y) = \frac{1}{60}y^{\frac{2}{3}} - \frac{1}{15}\) | M1 | CDF for \(Y\) |
| \(g(y) = \frac{1}{90}y^{-\frac{1}{3}}\) for \(8 \leq y \leq 512\) (0 otherwise) | M1 A1 | Differentiate to find PDF for \(Y\); correct \(g(y)\) and correct range seen anywhere |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(6) - F(3) = \frac{9}{20}$ or $0.45$ | B1 | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = \begin{cases} \frac{1}{30}x, & 2 \leq x \leq 8 \\ 0, & \text{otherwise} \end{cases}$ | B1 | May be implied. Only need to see $f(x) = \frac{1}{30}x$ |
| $E(\sqrt{X}) = \int_2^8 \frac{1}{30}x^{\frac{3}{2}}dx = \left[\frac{1}{75}x^{\frac{5}{2}}\right]$ | M1 | Integrated |
| $= 2.34$ | A1 | $2.338\ldots$; answer of $2.34$ without $\frac{1}{75}x^{\frac{5}{2}}$ scores B1B1 (2/3) |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Var}(\sqrt{X}) = \left[\int_2^8 \frac{1}{30}x^2 dx = \left[\frac{1}{90}x^3\right]\right] - E\!\left(\sqrt{X}\right)^2$ | M1 | |
| $= 5.6 - 2.338^2 = 0.133$ or $0.134$ | A1 | |
---
## Question 4(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G(y) = \frac{1}{60}y^{\frac{2}{3}} - \frac{1}{15}$ | M1 | CDF for $Y$ |
| $g(y) = \frac{1}{90}y^{-\frac{1}{3}}$ for $8 \leq y \leq 512$ (0 otherwise) | M1 A1 | Differentiate to find PDF for $Y$; correct $g(y)$ and correct range seen anywhere |
---4 The continuous random variable $X$ has cumulative distribution function F given by
$$F ( x ) = \begin{cases} 0 & x < 2 \\ \frac { 1 } { 60 } x ^ { 2 } - \frac { 1 } { 15 } & 2 \leqslant x \leqslant 8 \\ 1 & x > 8 \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( 3 \leqslant X \leqslant 6 )$.
\item Find $\mathrm { E } ( \sqrt { X } )$.
\item Find $\operatorname { Var } ( \sqrt { X } )$.
\item The random variable $Y$ is defined by $Y = X ^ { 3 }$. Find the probability density function of $Y$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q4 [9]}}