walker wrote:
C
\(\sigma=\sqrt{\frac{\sum(x-x_a)^2}{n}}\)
where, \(x\) - the sum of 3 numbers
\(x_a\) - the average sum.
probabilities for \(x\):
\(x=0:\) \(p_0=\frac18;\) \(x=1:\) \(p_1=\frac38;\) \(x=2:\) \(p_2=\frac38;\) \(x=3:\) \(p_3=\frac18\)
\(x_a=\frac32\)
\(n \to \infty\) (n=1000 is very large number)
we can write:
\(\sigma=\sqrt{\frac18*(0-\frac32)^2+\frac38*(1-\frac32)^2+\frac38*(2-\frac32)^2+\frac18*(3-\frac32)^2}\)
\(\sigma=\sqrt{\frac18*\frac94+\frac38*\frac14+\frac38*\frac14+\frac18*\frac94}\)
\(\sigma=\sqrt{\frac{24}{32}}=\sqrt{\frac34}=\frac{\sqrt3}{2}\)
good Q +1
Walker, can you point out the flaw in my approach. I know from binomial dist parameters that my approach is wrong... but i have forgotten why this approach is worng.
For the trial, prob mass funtion for all outcomes = 1/2
so E[x] = 0*1/2 + 1*1/2 ... first moment
E[x^2] = 0^2 * 1/2 + 1^2 *1/2 ... second moment
E[x^2] - (E[x])^2 = 1/4 ... second central moment = var[x]
SD = sqrt(var[x]) = 1/sqrt(x)..