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Date November 2017 Marks available 4 Reference code 17N.2.sl.TZ0.4
Level SL only Paper 2 Time zone TZ0
Command term Find Question number 4 Adapted from N/A

Question

A discrete random variable X has the following probability distribution.

N17/5/MATME/SP2/ENG/TZ0/04

Find the value of k.

[4]
a.

Write down P(X=2).

[1]
b.

Find P(X=2|X>0).

[3]
c.

Markscheme

valid approach     (M1)

egtotal probability = 1

correct equation     (A1)

eg0.475+2k2+k10+6k2=1, 8k2+0.1k0.525=0

k=0.25     A2     N3

[4 marks]

a.

P(X=2)=0.025     A1     N1

[1 mark]

b.

valid approach for finding P(X>0)     (M1)

eg10.475, 2(0.252)+0.025+6(0.252), 1P(X=0), 2k2+k10+6k2

correct substitution into formula for conditional probability     (A1)

eg0.02510.475, 0.0250.525

0.0476190

P(X=2|X>0)=121 (exact), 0.0476     A1     N2

[3 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

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