Exercise Solution 3.37

Exercise Solution 3.37

The mean and standard deviation of X are μ = 1.05 and σ = 0.10. By [3.105] and [3.106], we calculate m = .044 and s = .095. Accordingly, X = exp(.095Z + .044), where Z ~ N(0,1). Denote the CDFs of X and Z as ΦX and ΦY. From the standard normal table

[s1]

[s2]

[s3]

[s4]

[s5]

Accordingly, the .75-quantile of X is 1.11.

 
FacebooktwitterlinkedinFacebooktwitterlinkedin
 
Comments are closed.

Powered by WordPress. Designed by WooThemes