EXAM P SAMPLE SOLUTIONS

122. Solution: D
x
6
e −
The marginal distribution of Y is given by f2(y)
= e e dx = 6 e dx
–2y –y –2y –2y –3y
= −6 e e + 6e = 6 e – 6 e , 0 < y < ∞
∞
−2y −3y
Therefore, E(Y) = ∫ y f2(y)
dy = ∫ (6ye − 6 ye ) dy = 6
∞
6 –2y 6 –3y
2
2 ∫ ye dy − 3
3 ∫ y e dy
0
0
∞
0
∞
0
y
∫
0
–x –2y –2y
∞
∫
0
ye
−2
y
y
∫
0
∞
∫
–3y
dy – 6 y e dy =
∞
∞
But
–2y
2 y e dy and
–3y
3y
e dy are equivalent to the means of exponential random
∫
0
∫
0
∞
–2y
variables with parameters 1/2 and 1/3, respectively. In other words, 2 y e dy = 1/2
∞
–3y
and 3y
e dy = 1/3 . We conclude that E(Y) = (6/2) (1/2) – (6/3) (1/3) = 3/2 – 2/3 =
∫
0
9/6 − 4/6 = 5/6 = 0.83 .
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123. Solution: C
Observe
Pr 4< S < 8 = Pr⎡ ⎣4< S < 8 N = 1⎤ ⎦Pr N = 1 + Pr⎡ ⎣4< S < 8 N > 1⎤ ⎦Pr
1 −4 −8
1 −1
5 5 2 −1
=
3( e − e ) +
6(
e −e
) *
= 0.122
N > 1
*Uses that if X has an exponential distribution with mean μ
a b
1 t 1 t
Pr ( a X b) Pr ( X a) Pr ( X b) e dt e dt e e μ
∞ ∞
− −
− μ − μ μ
≤ ≤ = ≥ − ≥ = ∫ − = −
μ ∫ μ
[ ] [ ] [ ]
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a b
∫
0
0