标准正态分布的pth分位数-R

Question

I'm learning statistics and R from a book called "Discovering Statistics using R"... Although it's very informative, it seems to skip over areas even though it suggests no prior knowledge of statistics or R is needed. So to the problem:

How can you calculate in R the pth quantile of the Standard Normal Distribution using the Dichotomy (or division in halves) method? (and assuming no use of qnorm() ). that is:

    pnorm(x)   = p
    pnorm(x)-p = 0
    f (x)      = 0

Update:

Dichotomy is a method where you take an interval [a,b] which takes values of different signs at the end points of the interval and has a single root x within [a,b] . You then half if to find F(x1) , and if f(x1) != 0 it gives you [a,x1] and [x1,b] ... where the sequence x1, x2,..., converges to 0 .

Answer 1

Clumsy, but this works:

tolerance <- 1e-6
interval <- c(-1000,1000)
quantile <- 0.2

while(interval[2]-interval[1] > tolerance) {
  cat('current interval: ',interval,'\n')
  interval.left <- c(interval[1],mean(interval))
  interval.right <- c(mean(interval),interval[2])
  if(sum(sign(pnorm(interval.left)-quantile))==0) {
    interval <- interval.left
  } else {
    interval <- interval.right
  }
}
mean(interval)
qnorm(quantile)