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Date May 2018 Marks available 3 Reference code 18M.2.AHL.TZ1.H_10
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number H_10 Adapted from N/A

Question

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

f ( x ) = { 3 a x , 0 x < 0.5 a ( 2 x ) , 0.5 x < 2 0 , otherwise

 

Show that  a = 2 3 .

[3]
a.

Find  P ( X < 1 ) .

[3]
b.

Given that P ( s < X < 0.8 ) = 2 × P ( 2 s < X < 0.8 ) , and that 0.25 < s < 0.4 , find the value of s.

[7]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

 

a [ 0 0.5 3 x d x + 0.5 2 ( 2 x ) d x ] = 1      M1

Note: Award the M1 for the total integral equalling 1, or equivalent.

a ( 3 2 ) = 1      (M1)A1

a = 2 3      AG

[3 marks]

a.

EITHER

0 0.5 2 x d x + 2 3 0.5 1 ( 2 x ) d x      (M1)(A1)

= 2 3      A1

OR

2 3 1 2 ( 2 x ) d x = 1 3      (M1)

so  P ( X < 1 ) = 2 3       (M1)A1

[3 marks]

b.

P ( s < X < 0.8 ) = s 0.5 2 x d x + 2 3 0.5 0.8 ( 2 x ) d x      M1A1

= [ x 2 ] s 0.5 + 0.27

0.25 s 2 + 0.27      (A1)

P ( 2 s < X < 0.8 ) = 2 3 2 s 0.8 ( 2 x ) d x      A1

= 2 3 [ 2 x x 2 2 ] 2 s 0.8

2 3 ( 1.28 ( 4 s 2 s 2 ) )

equating

0.25 s 2 + 0.27 = 4 3 ( 1.28 ( 4 s 2 s 2 ) )      (A1)

attempt to solve for s      (M1)

s = 0.274      A1

[7 marks]

c.

Examiners report

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c.

Syllabus sections

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