Normal distribution `N(mu, 1)>1.96`

Question

I try to write an R code to find value of mu s.t. the Normal distribution satisfies the probability P(N(mu, 1)>1.96)=0.95 (i.e., P(Z>1.96)=0.95 where Z~N(mu, 1) and mu is what I want to get). Is there any code for solving the parameter of distribution? It seems that this will be a integral equation about mu such that

int_{1.96}^{\infty} 1/\sqrt{2\pi} \exp(-(x-mu)^2/2)dx=0.95

We can sample Normal distribution from dnorm(x, mean, sd) or rnorm(n, mean, sd). But we need to first take value for mean and sd.

Answer 1

Here are some other options

• Using qnorm + uniroot (similar to the answer by @Rui Barradas)

> uniroot(function(m) qnorm(0.05, m) - 1.96, c(-1e3, 1e3))$root
[1] 3.604854

• Using erfinv from pracma package

library(pracma)
> 1.96 + sqrt(2) * pracma::erfinv(2 * 0.95-1)
[1] 3.604854