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Date May 2018 Marks available 7 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)={3ax,0x<0.5a(2x),0.5x<20,otherwise

 

Show that a=23.

[3]
a.

Find P(X<1).

[3]
b.

Given that P(s<X<0.8)=2×P(2s<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[0.503xdx+20.5(2x)dx]=1     M1

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

a(32)=1     (M1)A1

a=23     AG

[3 marks]

a.

EITHER

0.502xdx+2310.5(2x)dx     (M1)(A1)

=23     A1

OR

2321(2x)dx=13     (M1)

so P(X<1)=23      (M1)A1

[3 marks]

b.

P(s<X<0.8)=0.5s2xdx+230.80.5(2x)dx     M1A1

=[x2]0.5s+0.27

0.25s2+0.27     (A1)

P(2s<X<0.8)=230.82s(2x)dx     A1

=23[2xx22]0.82s

23(1.28(4s2s2))

equating

0.25s2+0.27=43(1.28(4s2s2))     (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 » AHL 4.14—Properties of discrete and continuous random variables
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Topic 4—Statistics and probability

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