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.

user_8245406 · Answer

Use uniroot to find the point where the probability equals 0.95. The function f is a way of writing this. The interval's end point are arbitrary, chosen because I wasn't not expecting the solution to be much faraway from the origin.
Note the argument lower.tail set to FALSE.

f <- function(x) {
  pnorm(1.96, mean = x, sd = 1, lower.tail = FALSE) - 0.95
}
uniroot(f, c(-5, 5))
#> $root
#> [1] 3.604879
#> 
#> $f.root
#> [1] 2.594975e-06
#> 
#> $iter
#> [1] 12
#> 
#> $init.it
#> [1] NA
#> 
#> $estim.prec
#> [1] 6.103516e-05

mu <- uniroot(f, c(-5, 5))$root
mu
#> [1] 3.604879

pnorm(1.96, mu, 1)
#> [1] 0.04999741
curve(pnorm(x, mu, 1),  from = -5, to = 8)
abline(h = 0.95, lty = "dashed")
abline(v = 1.96, lty = "dashed")

^{Created on 2022-02-28 by the reprex package (v2.0.1)}

user_12158757 · Answer

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

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

2 Answers