Question 437225
a.  {{{sum((3x+1)/k, x) = (3(1+3+5+7+9+11) + 6)/k = (108 + 6)/k = 114/k = 1}}}
==> k = 114.
b.  {{{mu = (1/114)sum(x(3x+1), x)=(1/114)(3(1^2 + 3^2 + 5^2 + 7^2 + 9^2 + 11^2) + (1+3+5+7+9+11)) = 146/19 }}}

{{{sigma^2 = E(X^2) - mu^2 = (1/114)sum(x^2(3x+1), x) - (146/19)^2}}}
={{{(1/114)(3(1^3 + 3^3 + 5^3 + 7^3 + 9^3 + 11^3) + (1^2+3^2+5^2+7^2+9^2+11^2)) - (146/19)^2 = 10.724838 }}}, to 6 decimal places.