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SECTION 5.6 5.6.10) X has mean and standard deviation 0.4. Therefore, 2.5( X ) has a standard normal distribution. So, P(| X | 1) P(2.5| X | 2.5) 2(2.5) 0.9876 . 5.6.14) Let Z X Y . Then we have, E(Z)=625-600=25. and Var ( Z ) 1 1 (100 100) (150 150 150) 100. 4 9 So, P(Z 0) 1 (2.5) 0.9938. 5.6.16) Note that X and Y are independent standard normal variables and hence their sum is a normal variable with mean 0 and variance 2. So, P( 2 X Y 2 2) (2) (1 (1)) 0.8186. 5.6.20) Let Y=log(X). Then Y has a normal distribution with mean 3 and variance 1.44. So, P( X 6.05) P(Y log(6.05)) P( 5.6.22) Y 3 log(6.05) 3 ) P( Z 0.99995) 0.1587 1.2 1.2 Since log(X ) has a normal distribution with mean and variance normal distribution with mean and variance with parameters and 2 . 2 , log(1/X)=-log(X) has a 2 . Hence 1/X has a lognormal distribution