Question 367420
For each four-digit number x, p(x) = {{{1/10^4 = 1/10000}}}.  The mean:
{{{mu = (1/10000)*(0000) + (1/10000)*(0001) + (1/10000)*(0002)}}}+...+
{{{(1/10000)*(9998) + (1/10000)*(9999)}}}
=(1/10000)*(sum of integers from 1 to 9999)
= {{{(1/10000)*(((9999)*(10000))/2) = 9999/2=4999.5}}}.
The variance is given by 
{{{Var(X) = E(X^2) - (E(X))^2}}}
={{{(1/10000)*(((9999)*(10000)*(19999))/6)-(9999/2)^2}}}
={{{((9999)*(19999))/6 - (9999/2)^2}}}
={{{(9999/2)*(19999/3 - 9999/2)}}}
={{{(9999/2)*(10001/6)}}}
={{{((3333)*(10001))/4}}}.
The standard deviation would be the square root of this.