Week 4 Extra Examples

8 and beta 04 find a the probability that this random

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Unformatted text preview: value between ln 3 = 1.099 and ln 6 = 1.792. Standardizing each of these values to get z1 and z2 Z1 = 1.099 – 1.5 / 0.4 =  ­1.0 Z2 = 1.792 – 1.5 / 0.4 = 0.73 Using table 3 we find the required probability to be 0.7673 – (1 – 0.8413) = 0.6086 If a random variable has the lognormal distribution with paramter alpha = 1.8 and beta = 0.4, find (a) the probability that this random variable will assume a value between 1 and 5 and (b)...
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This note was uploaded on 03/29/2011 for the course ENGR 9397 taught by Professor Susanhunt during the Winter '11 term at Memorial University.

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