Assignment 8 solutions

# Assignment 8 solutions - CE93-Engineering Data Analysis...

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CE93-Engineering Data Analysis Mark Hansen, Fall 2009 Assignment 8 Solutions 1. Problem 5.37 in Ross. If g = G±²³´±µ ℎ¶·¸¹ºG¶ ¸³»¼±¸ ¼ ²¼½ℎ±µ³ , g~exp ´¾ = 1º a) ¿´g > 2 ℎ¶·¸¹º = 1 − ¿´g ≤ 2º = 1 − À1 − ³ ÁÂÃ Ä = ³ ÁÅ = 0.135 1-expcdf(2,1) b) ¿´g > 3ug > 2º = Æ´ÇÈÉº Æ´ÇÈÅº = ÊÁÆ´ÇËÉº ÊÁÆ´ÇËÅº = ÊÁ´ÊÁÌ ÍÎ º Ï.ÊÉÐ = 0.368 (1-expcdf(3,1))/(1-expcdf(2,1)) 2. Problem 5.39 in Ross. g = G¶G¼Ñ µ·²Ò³¸ ¶Ó 1000¹ ¶Ó ²±Ñ³¹ If g~ exp Ô¾ = Ê ÅÏ Õ , ¿´g > 10 + 20ug > 10º = ¿´g > 30 ∩ g > 10º ¿´g > 10º = ¿´g > 30º ¿´g > 10º = 1 − ¿´g ≤ 30º 1 − ¿´g ≤ 10º = 1 − Ö1 − ³ Á Ê ÅÏ ∗ÉÏ × 1 − Ö1 − ³ Á Ê ÅÏ ∗ÊÏ × = 0.2231 0.6065 = 0.368 And exponential distribution has memoryless property ¿´g > 10 + 20ug > 10º = ¿´g > 30º ¿´g > 10º = ¿´g > 20º = 0.368 (1-expcdf(30,20))/(1-expcdf(10,20)) Or 1-expcdf(20,20) Assuming g~Ø´0,40º, Ù Ç ´Úº = Ú 40 ¿´g > 10 + 20ug > 10º =

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• Fall '10
• hansen
• Exponential distribution, Poisson process, Geometric distribution, Memorylessness, Mark Hansen, CE93-Engineering Data Analysis

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Assignment 8 solutions - CE93-Engineering Data Analysis...

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