Loading [MathJax]/jax/output/CommonHTML/fonts/TeX/fontdata.js

User interface language: English | Español

Date May 2008 Marks available 6 Reference code 08M.1.hl.TZ2.1
Level HL only Paper 1 Time zone TZ2
Command term Find Question number 1 Adapted from N/A

Question

The probability distribution of a discrete random variable X is defined by

P(X=x)=cx(5x), x=1, 2, 3, 4 .

(a)     Find the value of c.

(b)     Find E(X) .

Markscheme

(a)     Using P(X=x)=1     (M1)

4c+6c+6c+4c=1(20c=1)     A1

c=120(=0.05)     A1     N1

 

(b)     Using E(X)=xP(X=x)     (M1)

=(1×0.2)+(2×0.3)+(3×0.3)+(4×0.2)     (A1)

=2.5     A1     N1

Notes: Only one of the first two marks can be implied.

Award M1A1A1 if the x values are averaged only if symmetry is explicitly mentioned.

 

[6 marks]

Examiners report

This question was generally well done, but a few candidates tried integration for part (b).

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.5 » Concept of discrete and continuous random variables and their probability distributions.

View options