Quiz1-key - variable alone 0.21 2 Suppose you have a normally distributed random variable X with mean μ = 5 and standard deviation σ = 3 You want

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University of California, Davis ARE 106 - Winter 2011 Quiz 1 Name: Student Number: 1. Suppose Y is a binary random variable that denotes the outcome of a Sacramento Kings basketball game. It takes the value 0 if the Kings lose and the value 1 if the Kings win. The probability of Y taking on the value y i , (i.e. the probability that the Kings win a given game) is given by the probability density function f ( y i ) and can be found in the table below: Event y i f ( y i ) Kings lose 0 0.7 Kings win 1 0.3 Using the information in the table, compute the following. Each answer is worth 2 points. Show your work. (a) E ( Y ) = y i f ( y i ) = 0(0 . 7) + 1(0 . 3) = 0 . 3 (b) E ( Y 2 ) = y 2 i f ( y i ) = 0(0 . 7) + 1 2 (0 . 3) = 0 . 3 (c) E ( Y ) 2 From (a), E(Y) = 0.3; 0 . 3 2 = 0 . 09 (d) V ar ( Y ) Recall that V ar ( Y ) = E ( Y 2 ) - E ( Y ) 2 . Using answers from (b) and (c), 0 . 3 - 0 . 09 = 0 . 21 (e) V ar ( Y + 1) Variance of random variable plus a constant equals the variance of the random
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Unformatted text preview: variable alone, 0.21. 2. Suppose you have a normally distributed random variable, X , with mean μ = 5 and standard deviation σ = 3. You want to know the probability that X < 2. (a) Compute the standardized value of X necessary to look up the probability on the standard normal table. (2 points) Standardize X by subtracting the mean and dividing by the standard deviation: X-μ σ = 2-5 3 =-1 (b) On a standard normal table, you find that P ( Z < 1) is approximately 0.84. What is the probability that X < 2? (3 points) Since P ( Z < 1) = 0 . 84, then P ( Z > 1) = 1-. 84 = 0 . 16. Since the normal distribution is symmetric, P ( Z > 1) = P ( Z <-1), so the probability that X < 2 = 0 . 16....
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This note was uploaded on 01/13/2012 for the course ARE 32482 taught by Professor Havenner during the Spring '10 term at UC Davis.

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