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Date May 2017 Marks available 3 Reference code 17M.2.AHL.TZ2.H_10
Level Additional Higher Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number H_10 Adapted from N/A

Question

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

f ( x ) = { x 2 a + b , 0 x 4 0 otherwise where  a  and  b  are positive constants.

It is given that P ( X 2 ) = 0.75 .

Eight independent observations of X are now taken and the random variable Y is the number of observations such that X 2 .

Show that a = 32 and b = 1 12 .

[5]
a.

Find E ( X ) .

[2]
b.

Find Var ( X ) .

[2]
c.

Find the median of X .

[3]
d.

Find E ( Y ) .

[2]
e.

Find P ( Y 3 ) .

[1]
f.

Markscheme

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

0 4 ( x 2 a + b ) d x = 1 [ x 3 3 a + b x ] 0 4 = 1 64 3 a + 4 b = 1     M1A1

2 4 ( x 2 a + b ) d x = 0.75 56 3 a + 2 b = 0.75     M1A1

 

Note:    0 2 ( x 2 a + b ) d x = 0.25 8 3 a + 2 b = 0.25 could be seen/used in place of either of the above equations.

 

evidence of an attempt to solve simultaneously (or check given a,b values are consistent)     M1

a = 32 ,   b = 1 12      AG

[5 marks]

a.

E ( X ) = 0 4 x ( x 2 32 + 1 12 ) d x     (M1)

E ( X ) = 8 3 ( = 2.67 )      A1

[2 marks]

b.

E ( X 2 ) = 0 4 x 2 ( x 2 32 + 1 12 ) d x      (M1)

Var ( X ) = E ( X 2 ) [ E ( X ) ] 2 = 16 15 ( = 1.07 )      A1

[2 marks]

c.

0 m ( x 2 32 + 1 12 ) d x = 0.5     (M1)

m 3 96 + m 12 = 0.5 ( m 3 + 8 m 48 = 0 )      (A1)

m = 2.91      A1

[3 marks]

d.

Y B ( 8 ,   0.75 )      (M1)

E ( Y ) = 8 × 0.75 = 6      A1

[2 marks]

e.

P ( Y 3 ) = 0.996      A1

[1 mark]

f.

Examiners report

[N/A]
a.
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b.
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c.
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d.
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e.
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f.

Syllabus sections

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