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densities with overlapping area of 0.35

Let X be normally distributed with mean 0 and let f be it's density. Now the density of X+a will be f shifted right by a. Since the density is symmetric around mean it follows that the area of overlap of the two densities is exactly P(X>a) + P(X<-a).
So if X~N(0,1), we want P(X>a) + P(X<-a) = 2P(X<-a) = 0.35, so P(X<-a) = 0.175 which yields -a =  qnorm(0.175) = -0.9345893, so a = 0.9345893.
--- On Tue, 9/9/08, Lavan <rsumithran at yahoo.com> wrote: