Question 7 - A-Level Maths

CAIE FP2 2010 June — Question 7 7 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2010
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Cumulative distribution functions
Type	Calculate probabilities from CDF
Difficulty	Challenging +1.2 This question requires understanding of CDFs and calculating P(2 ∈ (X, 4X)) by considering cases X ≤ 2 ≤ 4X, which translates to 1/2 ≤ X ≤ 2. The second part requires finding E(3X) = 3E(X), involving differentiation to get the pdf and integration for expectation. While it requires careful case analysis and multiple techniques (CDF manipulation, pdf derivation, expectation), these are standard Further Maths topics with straightforward execution once the setup is understood.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf

7 The continuous random variable $X$ has distribution function given by $$\mathrm { F } ( x ) = \begin{cases} 0 & x < 0 \\ 1 - \mathrm { e } ^ { - \frac { 1 } { 2 } x } & x \geqslant 0 \end{cases}$$ For a random value of $X$, find the probability that 2 lies between $X$ and $4 X$. Find also the expected value of the width of the interval ( $X , 4 X$ ).

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Question 7:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$P(\frac{1}{2} \leq X \leq 2) = F(2) - F(\frac{1}{2}) = (1-e^{-1}) - (1-e^{-1/4}) = 0.632 - 0.221 = 0.411$	M1 A1, A1	Relate $P(\frac{1}{2} \leq X \leq 2 \leq 4X)$ to $F(x)$ and evaluate
EITHER: $f(x) = \frac{1}{2}e^{-x/5x},\; E(X) = 2$	M1 A1	State $E(X)$ or find $f(x)$ for $x > 0$
$E(3X) = 3 \times E(X) = 6$	M1 A1	Find width of interval $(X, 4X)$
OR: Find $f(y)$ for $Y = 4X - X$: $F(y) = P(X < y/3) = 1 - e^{-y/6}$; $f(y) = e^{-y/6}/6$	(M1 A1)
$E(Y) = \int y(e^{-y/6}/6)\,dy = 6$	(M1 A1)	Find width of interval $(X, 4X)$

Total: 7 marks

## Question 7:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\frac{1}{2} \leq X \leq 2) = F(2) - F(\frac{1}{2}) = (1-e^{-1}) - (1-e^{-1/4}) = 0.632 - 0.221 = 0.411$ | M1 A1, A1 | Relate $P(\frac{1}{2} \leq X \leq 2 \leq 4X)$ to $F(x)$ and evaluate |
| **EITHER:** $f(x) = \frac{1}{2}e^{-x/5x},\; E(X) = 2$ | M1 A1 | State $E(X)$ or find $f(x)$ for $x > 0$ |
| $E(3X) = 3 \times E(X) = 6$ | M1 A1 | Find width of interval $(X, 4X)$ |
| **OR:** Find $f(y)$ for $Y = 4X - X$: $F(y) = P(X < y/3) = 1 - e^{-y/6}$; $f(y) = e^{-y/6}/6$ | (M1 A1) | |
| $E(Y) = \int y(e^{-y/6}/6)\,dy = 6$ | (M1 A1) | Find width of interval $(X, 4X)$ |

**Total: 7 marks**

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7 The continuous random variable $X$ has distribution function given by

$$\mathrm { F } ( x ) = \begin{cases} 0 & x < 0 \\ 1 - \mathrm { e } ^ { - \frac { 1 } { 2 } x } & x \geqslant 0 \end{cases}$$

For a random value of $X$, find the probability that 2 lies between $X$ and $4 X$.

Find also the expected value of the width of the interval ( $X , 4 X$ ).

\hfill \mbox{\textit{CAIE FP2 2010 Q7 [7]}}

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Q1 5 Q2 7 Q3 9 Q4 9 Q5 13 Q6 5 Q7 7 Q8 9 Q9 9 Q10 13 Q11 EITHER Q11 OR