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Date	November 2019	Marks available	4	Reference code	19N.2.AHL.TZ0.H_10
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	H_10	Adapted from	N/A

Question

A random variable $X$ has probability density function

$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {3a}&{\text{,}}&{0 \leqslant x < 2}&{} \\ {a\left( {x - 5} \right)\left( {1 - x} \right)}&{\text{,}}&{2 \leqslant x \leqslant b}&{a{\text{, }}b \in {\mathbb{R}^ + }{\text{, }}3 < b \leqslant 5.} \\ 0&{\text{,}}&{{\text{otherwise}}}&{} \end{array}} \right.$

Consider the case where $b = 5$ .

Find the value of

Find, in terms of $a$ , the probability that $X$ lies between 1 and 3.

[4]

Sketch the graph of $f$ . State the coordinates of the end points and any local maximum or minimum points, giving your answers in terms of $a$ .

[4]

$a$ .

[4]

c.i.

${\text{E}}\left( X \right)$ .

[3]

c.ii.

the median of $X$ .

[4]

c.iii.

Markscheme

$\left( {{\text{P}}\left( {1 < X < 3} \right) = } \right)\int_1^2 {3a\,{\text{d}}x + a} \int_2^3 { - {x^2} + 6x - 5} \,{\text{d}}x$ (M1)(A1)(A1)

$= 3a + \frac{{11}}{3}a$

$= \frac{{20}}{3}a\,\left( { = 6.67a} \right)$ A1

[4 marks]

award A1 for (0, 3 $a$ ), A1 for continuity at (2, 3 $a$ ), A1 for maximum at (3, 4 $a$ ), A1 for (5, 0)

Note: Award A3 if correct four points are not joined by a straight line and a quadratic curve.

[4 marks]

${\text{P}}\left( {0 \leqslant X \leqslant 5} \right) = 6a + a\int_2^5 { - {x^2} + 6x - 5} \,{\text{d}}x$ (M1)

$= 15a$ (A1)

$15a = 1$ (M1)

$\Rightarrow a = \frac{1}{{15}}\left( { = 0.0667} \right)$ A1

[4 marks]

c.i.

${\text{E}}\left( X \right) = \frac{1}{5}\int_0^2 {x\,{\text{d}}x} + \frac{1}{{15}}\int_2^5 { - {x^3} + 6{x^2} - 5} x\,{\text{d}}x$ (M1)(A1)

= 2.35 A1

[3 marks]

c.ii.

attempt to use $\int_0^m {f\left( x \right)} {\text{d}}x = 0.5$ (M1)

$0.4 + a\int_2^m { - {x^2} + 6x - 5} \,{\text{d}}x = 0.5$ (A1)

$a\int_2^m { - {x^2} + 6x - 5} \,{\text{d}}x = 0.1$

attempt to solve integral using GDC and/or analytically (M1)

$\frac{1}{{15}}\left[ { - \frac{1}{3}{x^3} + 3{x^2} - 5x} \right]_2^m = 0.1$

$m = 2.44$ A1

[4 marks]

c.iii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

Syllabus sections

Topic 4—Statistics and probability » SL 4.7—Discrete random variables

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