| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Expectation of function of X |
| Difficulty | Standard +0.8 This Further Maths question requires finding E(√X) via integration, deriving a new pdf using the transformation Y=X², and finding a percentile. While the techniques are standard for Further Stats, the pdf transformation and the non-standard expectation E(√X) require careful algebraic manipulation and integration skills beyond typical A-level, placing it moderately above average difficulty. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles5.07a Non-parametric tests: when to use |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(\sqrt{X}) = \int \sqrt{x} \cdot \frac{3}{8}\left(1+\frac{1}{x^2}\right)dx = \int_1^3 \frac{3}{8}(x^{0.5}+x^{-1.5})dx\) | M1 | |
| \(\frac{3}{8}\left[\frac{2}{3}x^{1.5} - 2x^{-0.5}\right]\) | A1 | |
| \(\frac{3}{8}\left(2\sqrt{3} - \frac{2}{\sqrt{3}} - \frac{2}{3} + 2\right) = \frac{1}{2}(1+\sqrt{3})\ (=1.37)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(x) = \frac{3}{8}\left(x - \frac{1}{x}\right)\) | M1 | Integrate |
| \(G(y) = \frac{3}{8}\left(\sqrt{y} - \frac{1}{\sqrt{y}}\right)\) | M1 | Change variable throughout in their CDF |
| \(g(y) = \frac{3}{16}\left(y^{-0.5} + y^{-1.5}\right), \quad (1 \leq y \leq 9)\) | A1 | Differentiate to find \(g(y)\) |
| \(0\) otherwise | A1 | \(1 \leq y \leq 9\) and 0 otherwise, dependent on second M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(p) = 0.4\) so \(\frac{3}{8}\left(p - \frac{1}{p}\right) = 0.4\) | M1* | Correct method |
| \(15p^2 - 16p - 15 = 0\); \((5p+3)(3p-5)=0\), \(p = \frac{5}{3}\) | M1 dep | Obtain quadratic and solve |
| 40th percentile of \(Y\) is \(\frac{25}{9}\) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(\sqrt{X}) = \int \sqrt{x} \cdot \frac{3}{8}\left(1+\frac{1}{x^2}\right)dx = \int_1^3 \frac{3}{8}(x^{0.5}+x^{-1.5})dx$ | M1 | |
| $\frac{3}{8}\left[\frac{2}{3}x^{1.5} - 2x^{-0.5}\right]$ | A1 | |
| $\frac{3}{8}\left(2\sqrt{3} - \frac{2}{\sqrt{3}} - \frac{2}{3} + 2\right) = \frac{1}{2}(1+\sqrt{3})\ (=1.37)$ | A1 | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(x) = \frac{3}{8}\left(x - \frac{1}{x}\right)$ | M1 | Integrate |
| $G(y) = \frac{3}{8}\left(\sqrt{y} - \frac{1}{\sqrt{y}}\right)$ | M1 | Change variable throughout in their CDF |
| $g(y) = \frac{3}{16}\left(y^{-0.5} + y^{-1.5}\right), \quad (1 \leq y \leq 9)$ | A1 | Differentiate to find $g(y)$ |
| $0$ otherwise | A1 | $1 \leq y \leq 9$ and 0 otherwise, dependent on second M1 |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(p) = 0.4$ so $\frac{3}{8}\left(p - \frac{1}{p}\right) = 0.4$ | M1* | Correct method |
| $15p^2 - 16p - 15 = 0$; $(5p+3)(3p-5)=0$, $p = \frac{5}{3}$ | M1 dep | Obtain quadratic and solve |
| 40th percentile of $Y$ is $\frac{25}{9}$ | A1 | |
---4 The continuous random variable $X$ has probability density function f given by
$$f ( x ) = \begin{cases} \frac { 3 } { 8 } \left( 1 + \frac { 1 } { x ^ { 2 } } \right) & 1 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { E } ( \sqrt { X } )$.\\
The random variable $Y$ is given by $Y = X ^ { 2 }$.
\item Find the probability density function of $Y$.
\item Find the 40th percentile of $Y$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q4 [10]}}