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Date May 2012 Marks available 2 Reference code 12M.1.sl.TZ2.4
Level SL only Paper 1 Time zone TZ2
Command term Find Question number 4 Adapted from N/A

Question

The random variable X has the following probability distribution, with P(X>1)=0.5P(X>1)=0.5 .


 

Find the value of r .

[2]
a.

Given that E(X)=1.4E(X)=1.4 , find the value of p and of q .

[6]
b.

Markscheme

attempt to substitute P(X>1)=0.5P(X>1)=0.5     (M1)

e.g. r+0.2=0.5r+0.2=0.5

r=0.3r=0.3     A1     N2

[2 marks]

a.

correct substitution into E(X)E(X) (seen anywhere)     (A1)

e.g. 0×p+1×q+2×r+3×0.20×p+1×q+2×r+3×0.2

correct equation     A1

e.g. q+2×0.3+3×0.2=1.4q+2×0.3+3×0.2=1.4 , q+1.2=1.4q+1.2=1.4

q=0.2q=0.2     A1      N1

evidence of choosing pi=1pi=1     M1

e.g. p+0.2+0.3+0.2=1p+0.2+0.3+0.2=1 , p+q=0.5p+q=0.5

correct working     (A1)

p+0.7=1p+0.7=1 , 10.20.30.210.20.30.2 , p+0.2=0.5p+0.2=0.5

p=0.3p=0.3     A1 N2

Note: Exception to the FT rule. Award FT marks on an incorrect value of q, even if q is an inappropriate value. Do not award the final A mark for an inappropriate value of p.

[6 marks]

b.

Examiners report

The majority of candidates were successful in earning full marks on this question.

a.

In part (b), a small number of candidates did not use the correct formula for E(X)E(X) , even though this formula is given in the formula booklet. There were also a few candidates who incorrectly assumed that p=0p=0 , forgetting that the sum of the probabilities must equal 1. There were a few candidates who left this question blank, which raises concerns about whether they had been exposed to probability distributions during the course.

b.

Syllabus sections

Topic 5 - Statistics and probability » 5.7 » Concept of discrete random variables and their probability distributions.
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