Date | May 2014 | Marks available | 14 | Reference code | 14M.3sp.hl.TZ0.3 |
Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
Command term | Determine, Show that, Verify, and Write down | Question number | 3 | Adapted from | N/A |
Question
(a) Consider the random variable X for which E(X)=aλ+b, where a and bare constants and λ is a parameter.
Show that X−ba is an unbiased estimator for λ.
(b) The continuous random variable Y has probability density function
f(y)={29(3+y−λ),0,forλ−3≤y≤λotherwise
where λ is a parameter.
(i) Verify that f(y) is a probability density function for all values of λ.
(ii) Determine E(Y).
(iii) Write down an unbiased estimator for λ.
Markscheme
(a) E(X−ba)=aλ+b−ba M1A1
=λ A1
(Therefore X−ba is an unbiased estimator for λ) AG
[3 marks]
(b) (i) f(y)⩾0 R1
Note: Only award R1 if this statement is made explicitly.
recognition or showing that integral of f is 1 (seen anywhere) R1
EITHER
∫λλ−329(3+y−λ)dy M1
=29[(3−λ)y+12y2]λλ−3 A1
=29(λ(3−λ)+12λ2−(3−λ)(λ−3)−12(λ−3)2) or equivalent A1
=1
OR
the graph of the probability density is a triangle with base length 3 and height 23 M1A1
its area is therefore 12×3×23 A1
=1
(ii) E(Y)=∫λλ−329y(3+y−λ)dy M1
=29[(3−λ)12y2+13y3]λλ−3 A1
=29((3−λ)12(λ2−(λ−3)2)+13(λ3−(λ−3)3)) M1
=λ−1 A1A1
Note: Award 3 marks for noting that the mean is 23rds the way along the base and then A1A1 for λ−1.
Note: Award A1 for λ and A1 for –1.
(iii) unbiased estimator: Y+1 A1
Note: Accept ˉY+1.
Follow through their E(Y) if linear.
[11 marks]
Total [14 marks]