 
Summary: Test One Mathematics 135.01 Fall 2007
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The average was 82 and the standard deviation was 12.0949.
1. 5 pts. Suppose X is a random variable with variance 5. Compute Var(3X + 9).
Solution.
Var(3X + 9) = Var(3X) = 32
Var(X) = 32
(5) = 45.
2. 5 pts. Suppose X and Y are independent variables with expectations 3 and 4, respectively. Compute
E(XY ).
Solution. Since X and Y are independent we have
E(XY ) = E(X)E(Y ) = 3 · 4 = 12.
3. 10 pts. How many 12 letter strings can be made with 3 A's, 4 B's and 5 C's?
Solution. From the multinomial formula we obtain
3 + 4 + 5
3, 4, 5
=
12!
