131A_1_131amtsol

# 131A_1_131amtsol - EE 131A Midterm Exam Solution Winter...

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Unformatted text preview: EE 131A Midterm Exam Solution Winter 2007 K. Yao 1. a. p = 1 / 26 . b. p = 5 / 26 . c. p = 21 / 26 = 1- (5 / 26) . d. p = 3 / 26 = P ( board card ) . e. p = 6 / 26 = P ( board ) + P ( card )- P ( board card ) = (5 + 4- 3) / 26 . 2. a. No. At x=0, F(0) = 0, but at F(0+)=0.5 . At x=1, F(1) = 3/4, but F(1+) =1. This show F(x) is not continuous from the right at x = 0 and x = 1. But all cdfs must be continuous from the right for all values. Thus, this F(x) function is not a valid cdf. b. No. For example, at x=-0.5, F(-0.5) = -0.5 . In fact, F(x) is negative-valued for- 1 x < . Since a cdf must be non-negative valued, this F(x) function is not a valid cdf. c. Yes. This F(x)function is a valid cdf since it takes values from 0 to 1. F(x) is a non-decreasing function. At x = 0 and x = 1, there are no jumps, and F(x) is continuous from the right at both x-values. 3. a. There are 6 outcomes whose sum of the dots on the two dice equals 7 out of 36 possible outcomes. Thus, p = 6 / 36 = 1...
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## This note was uploaded on 06/09/2010 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

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