| 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 (showing the derived pdf using the Jacobian method), then finding median via integration and E(Y) via integration. The transformation proof requires understanding of the change-of-variables technique, which is non-trivial. The subsequent calculations are straightforward but the overall question demands solid understanding of advanced probability theory beyond standard A-level. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(x) = (x^3-1)/63\) for \(1 \leq x \leq 4\) | B1 | |
| \(G(y) = P(Y < y) = P(X^2 < y) = P(X < y^{1/2}) = F(y^{1/2}) = (y^{3/2}-1)/63\) | M1 A1 | Find \(G(y)\) from \(Y = X^2\) for \(1 \leq x \leq 4\) |
| \(g(y) = y^{1/2}/42\) | A1 | A.G. |
| \(1 \leq y \leq 16\) | B1 | Find or state corresponding range of \(y\) |
| (i) \((m^{3/2}-1)/63 = \frac{1}{2}\); \(m = 32.5^{2/3} = 10.2\) | M1 A1 | Find median value \(m\) of \(Y\) |
| (ii) \(E(Y) = \int y\,g(y)\,dy = \int y^{3/2}\,dy/42 = [y^{5/2}]_1^{16}/105 = 1023/105 = 341/35\) or \(9.74\) | M1 A1 | Find \(E(Y)\) [or equivalently \(E(X^2)\)] |
## Question 7:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(x) = (x^3-1)/63$ for $1 \leq x \leq 4$ | B1 | |
| $G(y) = P(Y < y) = P(X^2 < y) = P(X < y^{1/2}) = F(y^{1/2}) = (y^{3/2}-1)/63$ | M1 A1 | Find $G(y)$ from $Y = X^2$ for $1 \leq x \leq 4$ |
| $g(y) = y^{1/2}/42$ | A1 | **A.G.** |
| $1 \leq y \leq 16$ | B1 | Find or state corresponding range of $y$ | **A.G.** |
| **(i)** $(m^{3/2}-1)/63 = \frac{1}{2}$; $m = 32.5^{2/3} = 10.2$ | M1 A1 | Find median value $m$ of $Y$ |
| **(ii)** $E(Y) = \int y\,g(y)\,dy = \int y^{3/2}\,dy/42 = [y^{5/2}]_1^{16}/105 = 1023/105 = 341/35$ or $9.74$ | M1 A1 | Find $E(Y)$ [or equivalently $E(X^2)$] | Total: 9 |
---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 $Y = X ^ { 2 }$. Show that $Y$ has probability density function given by
$$\operatorname { g } ( y ) = \begin{cases} \frac { 1 } { 42 } y ^ { \frac { 1 } { 2 } } & 1 \leqslant y \leqslant 16 \\ 0 & \text { otherwise } \end{cases}$$
Find\\
(i) the median value of $Y$,\\
(ii) the expected value of $Y$.
\hfill \mbox{\textit{CAIE FP2 2015 Q7 [9]}}