| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| 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 statistics question requiring integration to find a CDF, then transformation of random variables using the Jacobian method, and finally solving for a median. While the integration itself is straightforward, the transformation technique and careful handling of the change of variables places this above average difficulty, though it follows a standard FM2 template without requiring novel insight. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| 8(i) | F(x) = ∫ f(x) dx = x2/8 – x/4 [+ c] | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| = x2/8 – x/4 or {(x – 1)2 – 1}/8 (AEF) | A1 | State F(x) for other values of x |
| F(x) = 0 (x < 2), F(x) = 1 (x > 4) | A1 | |
| Total: | 3 |
| Answer | Marks |
|---|---|
| 8(ii) | EITHER: |
| Answer | Marks | Guidance |
|---|---|---|
| = (1 + y1/3)2 /8 – (1 + y1/3)/4 or (y 2/3 – 1)/8 | (M1 A1) | Find or state G(y) for 2 ⩽x ⩽ 4 from Y = (X – 1)3 |
| Answer | Marks | Guidance |
|---|---|---|
| f(x) = ¼ y1/3 and dx/dy = ⅓ y –2/3 | (M1 A1) | Find f(x) and dx/dy for use in g(y) = f(x) × dx/dy |
| g(y) [= G′(y)] = (1/12) y –1/3 or 1 / (12 y1/3) | A1 | Find g(y) in simplified form |
| for 1 ⩽ y ⩽ 27 [g(y) = 0 otherwise] | A1 | State corresponding range of y for G(y) or g(y) |
| Total: | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| 8(iii) | (m2/3 – 1)/8 = ½ | M1 |
| m2/3 = 5, m = √125 or 5√5 or 11⋅2 | M1 A1 | |
| Total: | 3 | |
| Question | Answer | Marks |
Question 8:
--- 8(i) ---
8(i) | F(x) = ∫ f(x) dx = x2/8 – x/4 [+ c] | M1 | Find or state distribution function F(x) for 2 ⩽ x ⩽ 4
using F(2) = 0 or F(4) = 1 to find c if necessary
= x2/8 – x/4 or {(x – 1)2 – 1}/8 (AEF) | A1 | State F(x) for other values of x
F(x) = 0 (x < 2), F(x) = 1 (x > 4) | A1
Total: | 3
--- 8(ii) ---
8(ii) | EITHER:
G(y) = P(Y < y) = P((X – 1)3 < y)
= P(X < 1 + y1/3) = F(1 + y1/3)
= (1 + y1/3)2 /8 – (1 + y1/3)/4 or (y 2/3 – 1)/8 | (M1 A1) | Find or state G(y) for 2 ⩽x ⩽ 4 from Y = (X – 1)3
(allow < or ⩽ throughout)
OR:
Use x = 1 + y1/3 to find
f(x) = ¼ y1/3 and dx/dy = ⅓ y –2/3 | (M1 A1) | Find f(x) and dx/dy for use in g(y) = f(x) × dx/dy
g(y) [= G′(y)] = (1/12) y –1/3 or 1 / (12 y1/3) | A1 | Find g(y) in simplified form
for 1 ⩽ y ⩽ 27 [g(y) = 0 otherwise] | A1 | State corresponding range of y for G(y) or g(y)
Total: | 4
--- 8(iii) ---
8(iii) | (m2/3 – 1)/8 = ½ | M1 | Find median value m of Y from G(m) = ½
m2/3 = 5, m = √125 or 5√5 or 11⋅2 | M1 A1
Total: | 3
Question | Answer | Marks | GuidanceThe continuous random variable $X$ has probability density function f given by
$$\text{f}(x) = \begin{cases} \frac{1}{4}(x - 1) & 2 \leqslant x \leqslant 4, \\ 0 & \text{otherwise.} \end{cases}$$
\begin{enumerate}[label=(\roman*)]
\item Find the distribution function of $X$. [3]
\end{enumerate}
The random variable $Y$ is defined by $Y = (X - 1)^3$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the probability density function of $Y$. [4]
\item Find the median value of $Y$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2017 Q8 [10]}}