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Unformatted text preview: Geostatistics Petroleum and Geosystems Engineering 337 (19065) Spring 2008 Homework Assignment #4 Section 4.1 (Page 201) 3. When a certain glaze is applied to a ceramic surface, the probability is 5% that there will be discoloration, 20% that there will be a crack, and 23% that there will be either discoloration or a crack, or both. Let X = 1 if there is discoloration, and let X = 0 otherwise. Let Y = 1 if there is a crack, and let Y = 0 otherwise. Let Z = 1 if there is either discoloration or a crack, or both, and let Z = 0 otherwise. Sol n This is Bernoulli distribution X = Discoloration Y = Crack Z = either crack or discoloration or both Hence; otherwise x x for for x p 1 05 . 1 05 . ) ( = = ⎪ ⎩ ⎪ ⎨ ⎧ − = otherwise y y for for y p 1 20 . 1 20 . ) ( = = ⎪ ⎩ ⎪ ⎨ ⎧ − = otherwise y y for for z p 1 23 . 1 23 . ) ( = = ⎪ ⎩ ⎪ ⎨ ⎧ − = a. Let p x denote the success probability for X. Find p x Ans From the explanation above X = 1, p x = 0.05 b. Let p y denote the success probability for Y. Find p y Ans Y = 1, p y = 0.20 c. Let p z denote the success probability for Z. Find p z Ans Z = 1, p z = 0.23 d. Is it possible for both X and Y to equal 1? Ans Yes. If X = 1; Y = 1, then Z = 1. e. Does p z = p x + p y ? Ans No. If X = 1, p x = 0.05 and Y = 1, p y = 0.20; then Z = 1, p z = 0.23 which is not equal to 0.05 + 0.20 (0.25). f. Does Z = X + Y? Explain Ans No, because if X = 1, Y = 1, then Z = 1 but X + Y = 2. 7. Let X and Y be Bernoulli random variable. Let Z = XY. a. Show that Z is a Bernoulli random Variable. Ans If X = 1 and Y = 1, then Z = (1)*(1) = 1 If X = 0 and Y = 0, then Z = (0) * (0)= 0 If X = 1 and Y = 0, then Z = (1) * (0)= 0 If X = 0 and Y = 1, then Z = (0) * (1)= 0 Hence, the possible outcomes for Z are only 0 and 1 which is Bernoulli Distribution. b. Show that if X and Y are independent, then p z = p x p y Ans For example p z = P(Z = 1) ; where Z = XY = P ( X Y = 1 ) = P(X = 1) * P(Y = 1) = p x p y Section 4.2 – Page 211 1. Let X ~ Bin (8, 0.4) Find Sol n ⎪ ⎩ ⎪ ⎨ ⎧ = − − = = = − otherwise x for p p x n x n x X P x p x n x 8 ,..., 2 , 1 , ) 1 ( )! ( ! ! ) ( ) ( a. P(X = 2) Ans 209 . ) 2 ( ) 4 . 1 ( 4 . )! 2 8 ( ! 2 ! 8 ) 2 ( 2 8 2 = = − − = = − X P X P b. P(X = 4) Ans 232 . ) 4 ( ) 4 . 1 ( 4 . )! 4 8 ( ! 4 ! 8 ) 4 ( 4 8 4 = = − − = = − X P X P c. P(X < 2) Ans 1064 . 0896 . 01680 . ) 4 . 1 ( 4 . )! 1 8 ( ! 1 ! 8 ) 4 . 1 ( 4 . )! 8 ( ! ! 8 ) 1 ( ) ( ) 2 ( 1 8 1 8 = + = − − + − − = = + = = < − − X P X P X P d. P(X > 6) Ans 008519 . 0006554 . 007864 . ) 4 . 1 ( 4 . )! 8 8 ( ! 8 ! 8 ) 4 . 1 ( 4 . )! 7 8 ( ! 7 ! 8 ) 8 ( ) 7 ( ) 6 ( 8 8 8 7 8 7 = + = − − + − − = = + = = > − − X P X P X P e....
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 Spring '08
 Jablonowski
 Normal Distribution, Standard Deviation, eger

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