A normal probability plot is defined as a plot of $n$ pairs:
($[100(i-0.5)/n]$ th $z$ percentile, $i$th observation).
Theoretically the points should fall close to a straight line with slope $\sigma$ and intercept $\mu$, the population sd and mean of the observed random variable. But as shown in simulation results with R, the intercept seems always lower than $\mu$, why?
Here is the R code that I used:
simu = function(n) {
y = sort(rnorm(n, 40, .1))
yperc = ((1:n)-.5)/n
x = qnorm(yperc)
plot(x,y)
}
simu(10000)
And the plot I get:
Also, the points in the middle always look denser than those on the two sides, why?
The link @whuber gave in the comments is really helpful. I will just write down some notes here for future reference. Imposing a linear percentage function on the orders, and under the restriction of symmetry, we have $f(i) = ai + b$, $f(n+1-i) = a(n+1-i) + b$, also $f(n+1-i) + f(i) = 1$, combining these, we get $a(n+1) + 2b = 1$, giving $a$ an arbitrary value of $1/n$, then $b$ has to be equal to $-0.5/n$.
((1:n)-0.5)/n
is reversed under the transformation $x\to 1-x$ and so is the inverse CDF of any symmetric distribution such as the Normal. So, what evidence do you have about a biased "intercept"? Are you computing it correctly? In your example the intercept looks perfectly accurate. (2) If the points in the middle were not denser, then the Normal distribution would be a uniform distribution. $\endgroup$