# [R] densities with overlapping area of 0.35

Moshe Olshansky m_olshansky at yahoo.com
Tue Sep 9 04:37:38 CEST 2008

```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:

> From: Lavan <rsumithran at yahoo.com>
> Subject: [R]  densities with overlapping area of 0.35
> To: r-help at r-project.org
> Received: Tuesday, 9 September, 2008, 12:11 PM
> Hi,
>
> I like to generate two normal densities such that the
> overlapping area
> between them is 0.35. Is there any code/package available
> in R to do that??
>
> Regards,
>
> Lavan
> --
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