Of course I saw your second post. The information on the order statistics was useful. Still I can't get from that formulation of the pdf for the minimum and the maximum to the expression in the original link. Here's the problem:

In the orginal link the integrand involved only constants and powers of the normal CDF. The integrand you write for the expected value has the independant variable x, the pdf f(x), and powers of the CDF F(x) and the complement (1-F(x)). That is the remaining step that I'm missing. There is also the matter of the constant n! /(r-1)!*(n-r)! in the initial expression for the expectation where r = 1 for the minimum and r = n for the maximum.

Let

F1(x) is the CDF of Min &

F2(x) is the CDF of Max

E(Min)=(Integral ( [1-F1(x) ] dx, -%INF, %INF)

E(Max)=(Integral ( [1-F2(x) ] dx, -%INF, %INF)

E(Range) =E(max)-E(min)

= (Integral ( [1-F2(x) -1 + F1(x) ] dx, -%INF, %INF)

= (Integral ( [F1(x) - F2(x) ] dx, -%INF, %INF)

I have already explaind the formulation of CDF of Min and Max

Code:

```
let Ma is the maximum
P[Ma <=x ]= P[X1<=x,.....Xn<=x] = product(i=1:n)(P[Xi<=x]) = Fx^n ( identical)
For deriving Minimum, [B]use the Reliability function[/B]
Let Mi is the minimum
P(Min>x) = P[X1>x,.....Xn>x]
etc.
```

ie F1(x) = 1-{1 - F(x)}^n

F2(x) = F(x)^n

Replace this, you will get the answer.

you have to offer a beer for more explanation.