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	Conditional probability with CDF
Difficulty	Standard +0.8 This question requires understanding that '2 lies between X and 4X' means finding P(X < 2 < 4X), which simplifies to P(X < 2) ∩ P(X > 0.5) = P(0.5 < X < 2). The second part requires finding E(4X - X) = 3E(X), needing integration by parts or recognition of the exponential distribution. While the concepts are standard for Further Maths, the probability interpretation and multi-step reasoning elevate this above routine exercises.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03c Calculate mean/variance: by integration

The continuous random variable $X$ has distribution function given by $$\text{F}(x) = \begin{cases} 0 & x < 0, \\ 1 - e^{-\frac{x}{4}} & x \geqslant 0. \end{cases}$$ For a random value of $X$, find the probability that $2$ lies between $X$ and $4X$. [3] Find also the expected value of the width of the interval $(X, 4X)$. [4]

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Question 7:

Answer	Marks
7	Relate P(X ≤ 2 ≤ 4X) to F(x): = P(½ ≤ X ≤ 2) = F(2) – F(½) M1 A1

-1 -¼

Evaluate: = (1 – e ) – (1 – e )

= 0⋅632 – 0⋅221 = 0⋅411 A1

-½x

EITHER: State E(X) or find using f(x) for x > 0: f(x) = ½e , E(X) = 2 M1 A1

Find width of interval (X, 4X): E(3X) = 3 × E(X) = 6 M1 A1

-y/6

OR: Find f(y) for Y = 4X – X: F(y) = P(X < y/3) = 1 – e

-y/6

f(y) = e /6 (M1 A1)

Answer	Marks
Find width of interval (X, 4X): E(Y) = ∫ y(e -y/6 /6) dy = 6 (M1 A1)	3
4	[7]

Question 7:
7 | Relate P(X ≤ 2 ≤ 4X) to F(x): = P(½ ≤ X ≤ 2) = F(2) – F(½) M1 A1
-1 -¼
Evaluate: = (1 – e ) – (1 – e )
= 0⋅632 – 0⋅221 = 0⋅411 A1
-½x
EITHER: State E(X) or find using f(x) for x > 0: f(x) = ½e , E(X) = 2 M1 A1
Find width of interval (X, 4X): E(3X) = 3 × E(X) = 6 M1 A1
-y/6
OR: Find f(y) for Y = 4X – X: F(y) = P(X < y/3) = 1 – e
-y/6
f(y) = e /6 (M1 A1)
Find width of interval (X, 4X): E(Y) = ∫ y(e -y/6 /6) dy = 6 (M1 A1) | 3
4 | [7]

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The continuous random variable $X$ has distribution function given by
$$\text{F}(x) = \begin{cases} 0 & x < 0, \\ 1 - e^{-\frac{x}{4}} & x \geqslant 0. \end{cases}$$

For a random value of $X$, find the probability that $2$ lies between $X$ and $4X$. [3]

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

\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 28