| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Interquartile range calculation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths statistics question requiring standard techniques: solving F(x) = 0.25 and 0.75 for the IQR, finding E(X³) by integrating x³f(x) where f(x) = F'(x), and using the transformation method for the pdf of Y = √X. All are routine applications of well-defined procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| UQ: \(F(U) = 0.75\): \(u^2 - 16u + 45 = 0\) | M1 | Obtain quadratic equation for \(u\) or \(l\) |
| \(u = 8 - \sqrt{19}\ (= 3.64)\) | A1 | Value of UQ |
| LQ: \(l^2 - 16l + 15 = 0\), \(l = 1\) | A1 | Value of LQ |
| \(IQR = UQ - LQ = 7 - \sqrt{19} = 2.64\) | A1 FT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = \begin{cases} \frac{1}{30}(8-x), & 0 \leq x \leq 6 \\ 0, & \text{otherwise} \end{cases}\) | B1 | May be implied, only need to see \(f(x) = \frac{1}{30}(8-x)\) |
| \(\frac{1}{30}\int_0^6 (8x^3 - x^4)\,dx = \left[2x^4 - \frac{1}{5}x^5\right]_0^6\) | M1 A1 | Attempt integration (with PDF not CDF), limits not required; Correct integrated expression with limits |
| \(= 34.56\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G(y) = \frac{1}{60}(16y^2 - y^4)\) | M1 | CDF for \(Y\) |
| \(g(y) = \frac{1}{60}(32y - 4y^3)\) for \(0 \leq y \leq \sqrt{6}\) | M1A1 | Differentiate to find PDF for \(Y\); Correct \(g(y)\) and correct range seen anywhere |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| UQ: $F(U) = 0.75$: $u^2 - 16u + 45 = 0$ | M1 | Obtain quadratic equation for $u$ or $l$ |
| $u = 8 - \sqrt{19}\ (= 3.64)$ | A1 | Value of UQ |
| LQ: $l^2 - 16l + 15 = 0$, $l = 1$ | A1 | Value of LQ |
| $IQR = UQ - LQ = 7 - \sqrt{19} = 2.64$ | A1 FT | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = \begin{cases} \frac{1}{30}(8-x), & 0 \leq x \leq 6 \\ 0, & \text{otherwise} \end{cases}$ | B1 | May be implied, only need to see $f(x) = \frac{1}{30}(8-x)$ |
| $\frac{1}{30}\int_0^6 (8x^3 - x^4)\,dx = \left[2x^4 - \frac{1}{5}x^5\right]_0^6$ | M1 A1 | Attempt integration (with PDF not CDF), limits not required; Correct integrated expression with limits |
| $= 34.56$ | A1 | |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G(y) = \frac{1}{60}(16y^2 - y^4)$ | M1 | CDF for $Y$ |
| $g(y) = \frac{1}{60}(32y - 4y^3)$ for $0 \leq y \leq \sqrt{6}$ | M1A1 | Differentiate to find PDF for $Y$; Correct $g(y)$ and correct range seen anywhere |6 The continuous random variable $X$ has cumulative distribution function F given by
$$F ( x ) = \begin{cases} 0 & x < 0 \\ \frac { 1 } { 60 } \left( 16 x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 6 \\ 1 & x > 6 \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find the interquartile range of $X$.
\item Find $\mathrm { E } \left( X ^ { 3 } \right)$.\\
The random variable $Y$ is such that $Y = \sqrt { X }$.
\item Find the probability density function of $Y$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q6 [11]}}