W7-slides1

# W7-slides1 - -9-8 8 9(i.e P X = k = 1 19 for-9 ≤ k ≤ 9...

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Lecture 19- Outline and Examples Continuous random variables (Ross Ch 5) -Exponential random variables (Ross § 5.5) -Gamma distribution (Ross § 5.6.1) Distribution of a Function of a random variables (Ross § 5.7)

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Exponential random variables (Ross § 5.5) Example. Earthquakes. Suppose that every three months, on average, an earthquake occurs in California. What is the probability that the next earthquake occurs after three but before 7 months?
Exponential random variables (Ross § 5.5) Example. Earthquakes. Suppose that every three months, on average, an earthquake occurs in California. What is the probability that the next earthquake occurs after three but before 7 months? Solution: Let X = time (in months) until next earthquake. Assume X Exp ( λ ). E [ X ] =average amount of time till next earthquake = 1 λ = 3 months. Therefore λ = 1 / 3. P (3 < X < 7) = F X (7) - F X (3) = (1 - e - 7 / 3 ) - (1 - e - 1 ) 0 . 27 .

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Function of random variable- discrete case Example. Let X be uniform on the 19 integers {- 9 , - 8 ,..., 8 , 9 } (i.e. P ( X = k ) = 1 / 19 for - 9 k 9.) Problem 1 . Find P ( X 2 5).
Function of random variable- discrete case Example. Let X be uniform on the 19 integers

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Unformatted text preview: {-9 ,-8 ,..., 8 , 9 } (i.e. P ( X = k ) = 1 / 19 for-9 ≤ k ≤ 9.) Problem 1 . Find P ( X 2 ≤ 5). Solution . P ( X 2 ≤ 5) = P (-√ 5 ≤ X ≤ √ 5) = P ( X ∈ {-2 ,-1 , , 1 , 2 } ) = 5 / 19. Function of random variable- discrete case Example. Let X be uniform on the 19 integers {-9 ,-8 ,..., 8 , 9 } (i.e. P ( X = k ) = 1 / 19 for-9 ≤ k ≤ 9.) Problem 1 . Find P ( X 2 ≤ 5). Solution . P ( X 2 ≤ 5) = P (-√ 5 ≤ X ≤ √ 5) = P ( X ∈ {-2 ,-1 , , 1 , 2 } ) = 5 / 19. Problem 2 . Find P ( | X-2 | ≤ 5). Function of random variable- discrete case Example. Let X be uniform on the 19 integers {-9 ,-8 ,..., 8 , 9 } (i.e. P ( X = k ) = 1 / 19 for-9 ≤ k ≤ 9.) Problem 1 . Find P ( X 2 ≤ 5). Solution . P ( X 2 ≤ 5) = P (-√ 5 ≤ X ≤ √ 5) = P ( X ∈ {-2 ,-1 , , 1 , 2 } ) = 5 / 19. Problem 2 . Find P ( | X-2 | ≤ 5). Solution. P ( | X-2 | ≤ 5) = P (-5 ≤ X-2 ≤ 5) = P (-3 ≤ X ≤ 7) = 11 / 19....
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## This note was uploaded on 06/05/2010 for the course STAT PStat 120a taught by Professor Rohinikumar during the Spring '10 term at UCSB.

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W7-slides1 - -9-8 8 9(i.e P X = k = 1 19 for-9 ≤ k ≤ 9...

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