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(5−x), 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).