When you compute the variance using probabilities…e.g, 0.4(0.5-.20)^2 0.3(0-.20)^2 0.3(0-.20)2 WHy is it that the sum, which in this case is .0600 not being divided by n-1 (3-1 =3) In computing the s.d… the just take the .06 ^ 1/2. Is it because the probabilities already take into account the degrees of freedom? Thanks much!

probabilities are provided. so you already have the size of sample built in, when that number is provided.