| 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 | Challenging +1.2 This is a standard transformation of random variables question requiring the chain rule formula for transforming PDFs (f_Y(y) = f_X(x)|dx/dy|) and then computing E(Y) by integration. While it involves Further Maths content and requires careful algebraic manipulation of the given CDF, the technique is routine and follows a well-practiced method. The main challenge is correctly differentiating the CDF and applying the transformation formula, making it moderately above average difficulty. |
| Spec | 5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | EITHER: G(y) [= P(Y < y) = P(X2 < y) | |
| = P(X < y1/2) = F(y1/2)] = (1/90) (y + y2) | M1A1 | Find or state G(y) for 0 ⩽ x ⩽ 3 from Y = X2 |
| Answer | Marks | Guidance |
|---|---|---|
| and dx/dy = 1 / 2y1/2 | (M1A1) | Find f(x) and dx/dy for use in g(y) = f(x) × dx/dy |
| g(y) [= G′(y)] = (1/90) (1 + 2y) | A1 | Find g(y) in simplified form |
| for 0 ⩽ y ⩽ 9 [g(y) = 0 otherwise] | A1 | State corresponding range of y at any stage |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | E(Y) = (1/90) ∫ (y + 2y 2) dy | M1 |
| Answer | Marks |
|---|---|
| 0 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
--- 7(i) ---
7(i) | EITHER: G(y) [= P(Y < y) = P(X2 < y)
= P(X < y1/2) = F(y1/2)] = (1/90) (y + y2) | M1A1 | Find or state G(y) for 0 ⩽ x ⩽ 3 from Y = X2
(allow < or ⩽ throughout)
OR: Use x = y1/2 to find
f(x) = (1/90) (2x + 4x3) = (1/90) (2y1/2 + 4y3/2)
and dx/dy = 1 / 2y1/2 | (M1A1) | Find f(x) and dx/dy for use in g(y) = f(x) × dx/dy
g(y) [= G′(y)] = (1/90) (1 + 2y) | A1 | Find g(y) in simplified form
for 0 ⩽ y ⩽ 9 [g(y) = 0 otherwise] | A1 | State corresponding range of y at any stage
4
--- 7(ii) ---
7(ii) | E(Y) = (1/90) ∫ (y + 2y 2) dy | M1 | Find mean of Y from ∫ y g(y) dy
= (1/90) [½ y 2 + ⅔ y 3] 9 = 117/20 or 5.85
0 | A1
2
Question | Answer | Marks | GuidanceThe continuous random variable $X$ has distribution function given by
$$\text{F}(x) = \begin{cases} 0 & x < 0, \\ \frac{1}{90}(x^2 + x^4) & 0 \leqslant x \leqslant 3, \\ 1 & x > 3. \end{cases}$$
The random variable $Y$ is defined by $Y = X^2$.
\begin{enumerate}[label=(\roman*)]
\item Find the probability density function of $Y$.
[4]
\item Find the mean value of $Y$.
[2]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2018 Q7 [6]}}