Name: SOLUTION Statistics 305 Exam 2 part I
June 22nd, 2009
[60 points total]
Problem 1 [12 points total]
Suppose for three independent random variables X, Y and Z, we have that
X ~ N(μ = 10, σμ = 10, σ2 = 4), Y ~ N(μ = 5, σ), Y ~ N(μ = 10, σμ = 5, σ2 = 15) and Z ~ Poisson(μ = 10, σλ = 3).
Now define a two random variables, U = 2X – 2Y + Z + 5 and V = XY2 + ½ Z.
a) Calculate E[U]. [3 pts]
EU = 2(μ = 10, σEX) – 2(μ = 10, σEY) + EZ + 5
= 2(μ = 10, σ10) – 2(μ = 10, σ5) + 3 + 5
= 18
b) Calculate Var(μ = 10, σU). [3 pts]
Var(μ = 10, σU) = 22(μ = 10, σEX) + (μ = 10, σ– 2) 2 (μ = 10, σEY) + EZ
= 4), Y ~ N(μ = 5, σ(μ = 10, σ4), Y ~ N(μ = 5, σ) + 4), Y ~ N(μ = 5, σ(μ = 10, σ15) + 3
= 16 + 60 + 3
= 79
c) Calculate E[V]. [3 pts] **Note this is non-linear so we have to use propagation of error formulas**
EV = (μ = 10, σEX)(μ = 10, σEY)2 + ½ EZ
= 10(μ = 10, σ5)2 + ½ (μ = 10, σ3)
= 250 + 1.5 = 251.5
d) Calculate Var(μ = 10, σV). [3 pts]
2/12
2
Z
V
XY
Y
V
Y
X
V
Var(μ = 10, σV) = (μ = 10, σ(μ = 10, σEY)2)2Var(μ = 10, σX) + [2(μ = 10, σEX)(μ = 10, σEY)]2Var(μ = 10, σY) + (μ = 10, σ½)2 Var(μ = 10, σZ)
= 252(μ = 10, σ4), Y ~ N(μ = 5, σ) + 4), Y ~ N(μ = 5, σ(μ = 10, σ100)(μ = 10, σ25)(μ = 10, σ15) + ¼(μ = 10, σ3)
= 152500.8
