| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF of transformed variable |
| Difficulty | Standard +0.8 This is a Further Maths question requiring integration of a non-trivial pdf involving fractional powers, then applying the transformation technique for continuous random variables. While the integration is straightforward (x^{1/2} and x^{-1/2}), the transformation Y=√X requires careful application of the Jacobian method and domain considerations. The multi-step nature and Further Maths context place it moderately above average difficulty. |
| Spec | 5.03e Find cdf: by integration5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| 6(i) | F(x) = ∫ f(x) dx = (1/80) (2x3/2 – 16x1/2) [+ c] | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| or x3/2/40 – x1/2/5 + 1/5 (AEF) | A1 | State F(x) for other values of x |
| F(x) = 0 (x < 4), F(x) = 1 (x > 16) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(ii) | EITHER: G(y) [= P(Y < y) = P(√X < y) = P(X < y2) ] | |
| = F(y 2) = (1/40) (y 3 – 8y + 8) (AEF) | M1A1 | Find or state G(y) for 2 ⩽ y ⩽ 4 from Y = √X |
| Answer | Marks | Guidance |
|---|---|---|
| f(x) = (1/80) (3y – 8/y) and dx/dy = 2y | (M1A1) | Find f(x) and dx/dy for use in g(y) = f(x) × dx/dy |
| Answer | Marks | Guidance |
|---|---|---|
| [ for 2 ⩽ y ⩽ 4, g(y) = 0 otherwise ] | A1 | Find g(y) in simplified form for 2 ⩽ y ⩽ 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
--- 6(i) ---
6(i) | F(x) = ∫ f(x) dx = (1/80) (2x3/2 – 16x1/2) [+ c] | M1 | Find or state distribution function F(x) for 4 ⩽ x ⩽ 16
using F(4) = 0 or F(16) = 1 to find c if necessary
= (1/80) (2x3/2 – 16x1/2 + 16)
or (1/40) (x3/2 – 8x1/2 + 8)
or x3/2/40 – x1/2/5 + 1/5 (AEF) | A1 | State F(x) for other values of x
F(x) = 0 (x < 4), F(x) = 1 (x > 16) | A1
3
--- 6(ii) ---
6(ii) | EITHER: G(y) [= P(Y < y) = P(√X < y) = P(X < y2) ]
= F(y 2) = (1/40) (y 3 – 8y + 8) (AEF) | M1A1 | Find or state G(y) for 2 ⩽ y ⩽ 4 from Y = √X
(allow < or ⩽ throughout; FT on constant term)
OR: Use x = y2 to find
f(x) = (1/80) (3y – 8/y) and dx/dy = 2y | (M1A1) | Find f(x) and dx/dy for use in g(y) = f(x) × dx/dy
g(y) [= G′(y)] = (1/40) (3y2 – 8) or (3/40) y2 – 1/5
[ for 2 ⩽ y ⩽ 4, g(y) = 0 otherwise ] | A1 | Find g(y) in simplified form for 2 ⩽ y ⩽ 4
3
Question | Answer | Marks | GuidanceThe continuous random variable $X$ has probability density function $f$ given by
$$f(x) = \begin{cases}
\frac{1}{80}\left(3\sqrt{x} - \frac{8}{\sqrt{x}}\right) & 4 \leqslant x \leqslant 16, \\
0 & \text{otherwise}.
\end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Find the distribution function of $X$. [3]
\end{enumerate}
The random variable $Y$ is defined by $Y = \sqrt{X}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the probability density function of $Y$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2018 Q6 [6]}}