CAIE FP2 2012 June — Question 9 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCumulative distribution functions
TypeCDF of transformed variable
DifficultyStandard +0.3 This is a straightforward application of standard techniques: finding a CDF by integration of a uniform distribution, then transforming it using the relationship Y = e^X. The final part requires solving P(Y ≥ k) = 0.25 using the derived CDF. While it involves multiple steps and the transformation of random variables (a Further Maths topic), each step follows directly from learned procedures without requiring novel insight or complex problem-solving.
Spec5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03g Cdf of transformed variables

9 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 2 a } & - a \leqslant x \leqslant a \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) is a positive constant. Find the distribution function of \(X\). The random variable \(Y\) is defined by \(Y = \mathrm { e } ^ { X }\). Find the distribution function of \(Y\). Given that \(a = 4\), find the value of \(k\) for which \(\mathrm { P } ( Y \geqslant k ) = 0.25\).

Question 9:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(F(x) = x/2a + c = (x+a)/2a\)M1 A1 Integrate \(f(x)\) to find \(F(x)\) for \(-a \le x \le a\) (AEF; finding \(F(x) = x/2a\) earns M1 A0)
\(F(x) = 0\ (x < -a)\), \(F(x) = 1\ (x > a)\)B1 State \(F(x)\) for other intervals
\(G(y) = P(Y < y) = P(e^X < y) = P(X < \ln y) = F(\ln y)\) Relate dist. fn. \(G(y)\) of \(Y\) to \(X\) for central interval (working may be omitted; ignore other intervals)
\(= (\ln y + a)/2a\); \((e^{-a} \le y \le e^a)\)M1 A1; A1
\(1 - G(k) = \frac{1}{4}\) with \(a = 4\): \((\ln k)/2a + \frac{1}{2} = \frac{3}{4}\)M1 A1 Find \(k\) from \(P(Y \ge k) = \frac{1}{4}\)
\(\ln k = 2\), \(k = e^2\) or \(7.39\)A1 (Using \(G(k) = \frac{1}{4}\) can earn M1)
Total: 9 marks
## Question 9:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(x) = x/2a + c = (x+a)/2a$ | M1 A1 | Integrate $f(x)$ to find $F(x)$ for $-a \le x \le a$ (AEF; finding $F(x) = x/2a$ earns M1 A0) |
| $F(x) = 0\ (x < -a)$, $F(x) = 1\ (x > a)$ | B1 | State $F(x)$ for other intervals |
| $G(y) = P(Y < y) = P(e^X < y) = P(X < \ln y) = F(\ln y)$ | | Relate dist. fn. $G(y)$ of $Y$ to $X$ for central interval (working may be omitted; ignore other intervals) |
| $= (\ln y + a)/2a$; $(e^{-a} \le y \le e^a)$ | M1 A1; A1 | |
| $1 - G(k) = \frac{1}{4}$ with $a = 4$: $(\ln k)/2a + \frac{1}{2} = \frac{3}{4}$ | M1 A1 | Find $k$ from $P(Y \ge k) = \frac{1}{4}$ |
| $\ln k = 2$, $k = e^2$ or $7.39$ | A1 | (Using $G(k) = \frac{1}{4}$ can earn M1) |

**Total: 9 marks**

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

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

where $a$ is a positive constant. Find the distribution function of $X$.

The random variable $Y$ is defined by $Y = \mathrm { e } ^ { X }$. Find the distribution function of $Y$.

Given that $a = 4$, find the value of $k$ for which $\mathrm { P } ( Y \geqslant k ) = 0.25$.

\hfill \mbox{\textit{CAIE FP2 2012 Q9 [9]}}