| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Standard +0.8 This is a Further Maths question requiring transformation of random variables using the Jacobian method, then finding median and expectation. While the transformation Y=X² is straightforward and the integration is manageable, students must know the formula g(y)=f(x)|dx/dy| and apply it correctly. The multi-step nature and Further Maths context place it moderately above average difficulty. |
| Spec | 5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks |
|---|---|
| (ii) | Find F(x) for 1 ⩽ x ⩽ 4: F(x) = (x 3 – 1)/63 B1 |
| Answer | Marks |
|---|---|
| = 341/35 or 9⋅74 M1 A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 9 | |
| Page 6 | Mark Scheme | Syllabus |
| Cambridge International A Level – October/November 2015 | 9231 | 21 |
Question 7:
--- 7
(i)
(ii) ---
7
(i)
(ii) | Find F(x) for 1 ⩽ x ⩽ 4: F(x) = (x 3 – 1)/63 B1
Find G(y) from Y = X 2 for 1 ⩽ x ⩽ 4: G(y) = P(Y < y) = P(X 2 < y)
1/2 1/2
= P(X < y ) = F(y )
(result may be stated) = (y 3/2 – 1)/63 M1 A1
Find g(y) for corresponding range of y: g(y) = y 1/2 /42 A.G. A1
Find or state corresponding range of y : 1 ⩽ y ⩽ 16 A.G. B1
3/2
Find median value m of Y: (m – 1)/63 = ½
m = 32⋅5 2/3 = 10⋅2 M1 A1
Find E(Y) [or equivalently E(X 2 )]: E(Y) = ∫ y g(y) dy = ∫ y 3/2 dy /42
5/2 16
= [y ]1 /105 = 1023/105
= 341/35 or 9⋅74 M1 A1 | 5
2
2 | 9
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2015 | 9231 | 21The continuous random variable $X$ has probability density function given by
$$f(x) = \begin{cases}
\frac{1}{2t}x^2 & 1 \leq x \leq 4, \\
0 & \text{otherwise}.
\end{cases}$$
The random variable $Y$ is defined by $Y = X^2$. Show that $Y$ has probability density function given by
$$g(y) = \begin{cases}
\left(\frac{1}{2t}\right)y^{\frac{1}{2}} & 1 \leq y \leq 16, \\
0 & \text{otherwise}.
\end{cases}$$ [5]
Find
\begin{enumerate}[label=(\roman*)]
\item the median value of $Y$, [2]
\item the expected value of $Y$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2015 Q7 [9]}}