| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Power transformation (Y = X^n, n≥2) |
| Difficulty | Standard +0.8 This is a Further Maths probability transformation question requiring the Jacobian method to derive a new pdf, then computing median and expectation. While the calculus is straightforward, the transformation technique and careful handling of limits elevates this above standard A-level, though it's a textbook FP2 exercise rather than requiring novel insight. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks |
|---|---|
| Find \(F(x)\) for \(1 \leq x \leq 4\): \(F(x) = \frac{x^3 - 1}{63}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(G(y) = P(Y \leq y) = P(X^2 \leq y) = P(X \leq y^{1/2}) = F(y^{1/2}) = \frac{y^{3/2} - 1}{63}\) | M1 A1 | (result may be stated) |
| Find \(g(y)\) for corresponding range of \(y\): \(g(y) = \frac{y^{1/2}}{42}\) | A1 | A.G. |
| Find or state corresponding range of \(y\): \(1 \leq y \leq 16\) | B1 | A.G. |
| Answer | Marks |
|---|---|
| \(m = 32.5^{2/3} = 10.2\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(= \left[\frac{y^{5/2}}{1}\right]_1^{16}/105 = \frac{1023}{105} = \frac{341}{35}\) *or* \(9.74\) | M1 A1 |
## Question 7(i):
Find $F(x)$ for $1 \leq x \leq 4$: $F(x) = \frac{x^3 - 1}{63}$ | B1 |
Find $G(y)$ from $Y = X^2$ for $1 \leq x \leq 4$:
$G(y) = P(Y \leq y) = P(X^2 \leq y) = P(X \leq y^{1/2}) = F(y^{1/2}) = \frac{y^{3/2} - 1}{63}$ | M1 A1 | (result may be stated)
Find $g(y)$ for corresponding range of $y$: $g(y) = \frac{y^{1/2}}{42}$ | A1 | A.G.
Find or state corresponding range of $y$: $1 \leq y \leq 16$ | B1 | A.G.
**Total: 5 marks**
---
## Question 7(ii):
Find median value $m$ of $Y$:
$\frac{m^{3/2} - 1}{63} = \frac{1}{2}$
$m = 32.5^{2/3} = 10.2$ | M1 A1 |
**Total: 2 marks**
---
## Question 7(iii):
Find $E(Y)$ [or equivalently $E(X^2)$]:
$E(Y) = \int y\,g(y)\,dy = \int \frac{y^{3/2}}{42}\,dy$
$= \left[\frac{y^{5/2}}{1}\right]_1^{16}/105 = \frac{1023}{105} = \frac{341}{35}$ *or* $9.74$ | M1 A1 |
**Total: 2 marks**
---7 The continuous random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} \frac { 1 } { 21 } x ^ { 2 } & 1 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
The random variable $Y$ is defined by $Y = X ^ { 2 }$.\\
(i) Show that $Y$ has probability density function given by
$$g ( y ) = \begin{cases} \frac { 1 } { 42 } y ^ { \frac { 1 } { 2 } } & 1 \leqslant y \leqslant 16 \\ 0 & \text { otherwise } \end{cases}$$
(ii) Find the median value of $Y$.\\
(iii) Find the expected value of $Y$.\\
\hfill \mbox{\textit{CAIE FP2 2017 Q7 [9]}}