CAIE FP2 2013 June — Question 8 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypePower transformation (Y = X^n, n≥2)
DifficultyStandard +0.8 This is a Further Pure 2 transformation of variables question requiring the Jacobian method (dy/dx = 3x², so dx/dy = 1/(3x²) = 1/(3y^(2/3))), substitution into the original pdf, and then integration of y^(2/3) to find E(Y). While the technique is standard for FP2, it requires careful algebraic manipulation and is more conceptually demanding than typical A-level statistics questions.
Spec5.03c Calculate mean/variance: by integration5.03g Cdf of transformed variables

8 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 6 } x & 2 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = X ^ { 3 }\). Show that \(Y\) has probability density function g given by $$g ( y ) = \begin{cases} \frac { 1 } { 18 } y ^ { - \frac { 1 } { 3 } } & 8 \leqslant y \leqslant 64 \\ 0 & \text { otherwise } \end{cases}$$ Find \(\mathrm { E } ( Y )\).

Question 8:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(F(x) = \int f(x)\,dx\)M1 A1 Integrate \(f(x)\) to find \(F(x)\) for \(2 \leq x \leq 4\)
\(G(y) = P(Y < y) = P(X^3 < y) = P(X < y^{1/3}) = F(y^{1/3})\) Find \(G(y)\) for \(8 \leq y \leq 64\)
\(= (y^{2/3} - 4)/12\)M1 A1
OR: \(G(y) = P(Y < y) = P(X^3 < y) = P(X < y^{1/3}) = \int_2^{y^{1/3}} f(x)\,dx\)(M1 A1)
\(= \big[x^2/12\big]_2^{y^{1/3}} = (y^{2/3} - 4)/12\)(M1 A1)
\(G(y) = 0\ (y < 8)\), \(1\ (y > 64)\)B1 State \(G(y)\) for other values of \(y\)
\(g(y) = y^{-1/3}/18\) A.G.B1 Differentiate to find \(g(y)\) for \(8 \leq y \leq 64\)
\(\frac{1}{18}\int_8^{64} y^{2/3}\,dy = \frac{1}{30}\big[y^{5/3}\big]_8^{64}\)M1 A1 Find \(E(Y)\) from \(\int yg(y)\); OR find \(E(Y)\) from \(E(X^3)\): \(\frac{1}{6}\int_2^4 x^4\,dx = \frac{1}{30}\big[x^5\big]_2^4\)
\(= (4^5 - 2^5)/30 = 992/30 = 496/15\) or \(33.1\)A1
Total: 9 marks
## Question 8:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $F(x) = \int f(x)\,dx$ | M1 A1 | Integrate $f(x)$ to find $F(x)$ for $2 \leq x \leq 4$ |
| $G(y) = P(Y < y) = P(X^3 < y) = P(X < y^{1/3}) = F(y^{1/3})$ | | Find $G(y)$ for $8 \leq y \leq 64$ |
| $= (y^{2/3} - 4)/12$ | M1 A1 | |
| OR: $G(y) = P(Y < y) = P(X^3 < y) = P(X < y^{1/3}) = \int_2^{y^{1/3}} f(x)\,dx$ | (M1 A1) | |
| $= \big[x^2/12\big]_2^{y^{1/3}} = (y^{2/3} - 4)/12$ | (M1 A1) | |
| $G(y) = 0\ (y < 8)$, $1\ (y > 64)$ | B1 | State $G(y)$ for other values of $y$ |
| $g(y) = y^{-1/3}/18$ **A.G.** | B1 | Differentiate to find $g(y)$ for $8 \leq y \leq 64$ |
| $\frac{1}{18}\int_8^{64} y^{2/3}\,dy = \frac{1}{30}\big[y^{5/3}\big]_8^{64}$ | M1 A1 | Find $E(Y)$ from $\int yg(y)$; OR find $E(Y)$ from $E(X^3)$: $\frac{1}{6}\int_2^4 x^4\,dx = \frac{1}{30}\big[x^5\big]_2^4$ |
| $= (4^5 - 2^5)/30 = 992/30 = 496/15$ or $33.1$ | A1 | |

**Total: 9 marks**

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8 The continuous random variable $X$ has probability density function f given by

$$f ( x ) = \begin{cases} \frac { 1 } { 6 } x & 2 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$

The random variable $Y$ is defined by $Y = X ^ { 3 }$. Show that $Y$ has probability density function g given by

$$g ( y ) = \begin{cases} \frac { 1 } { 18 } y ^ { - \frac { 1 } { 3 } } & 8 \leqslant y \leqslant 64 \\ 0 & \text { otherwise } \end{cases}$$

Find $\mathrm { E } ( Y )$.

\hfill \mbox{\textit{CAIE FP2 2013 Q8 [9]}}