EXAM P SAMPLE SOLUTIONS

99. Solution: C
We use the relationships Var aX + b
2
= a Var X , Cov aX , bY = ab Cov X , Y , and
( ) ( ) ( ) ( )
( X + Y) = ( X) + ( Y) + ( X Y)
= ( X+ Y) = + + ( XY) ( XY)
=
⎡⎣( X + ) + Y⎤⎦
= ar ⎡⎣( X + 1.1Y) + 100⎤⎦
[ X Y] X ⎡( ) Y⎤ ( X Y)
Var Var Var 2 Cov , . First we observe
17,000 Var 5000 10,000 2 Cov , , and so Cov , 1000.
We want to find Var 100 1.1 V
= Var + 1.1 = Var + Var ⎣ 1.1 ⎦+
2 Cov ,1.1
= Var X + 1.1 VarY + 2 1.1 Cov X, Y = 5000 + 12,100 + 2200 = 19,300.
2
( ) ( ) ( )
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100. Solution: B
Note
P(X = 0) = 1/6
P(X = 1) = 1/12 + 1/6 = 3/12
P(X = 2) = 1/12 + 1/3 + 1/6 = 7/12 .
E[X] = (0)(1/6) + (1)(3/12) + (2)(7/12) = 17/12
2 2
E[X ] = (0) (1/6) + (1) 2 (3/12) + (2) 2 (7/12) = 31/12
2
Var[X] = 31/12 – (17/12) = 0.58 .
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101. Solution: D
Note that due to the independence of X and Y
2
Var(Z) = Var(3X − Y − 5) = Var(3X) + Var(Y) = 3 Var(X) + Var(Y) = 9(1) + 2 = 11 .
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102. Solution: E
Let X and Y denote the times that the two backup generators can operate. Now the
2
variance of an exponential random variable with mean β is β . Therefore,
2
[ X] [ Y]
Var = Var = 10 = 100
Then assuming that X and Y are independent, we see
Var X+Y = Var X + Var Y = 100 + 100 = 200
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