SOLUTION: Find the variance of the following probability distribution. x P(x) 0 0.1052 1 0.0455 2 0.3337 3 0.3754 4 0.1401 variance =

Question 663910: Find the variance of the following probability distribution.
x P(x)
0 0.1052
1 0.0455
2 0.3337
3 0.3754
4 0.1401

variance =
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Mean or Expected Value:

E[X] = Sum( P(X = xi) * xi )

E[X] = [P(X = x1)*(x1)] + [P(X = x2)*(x2)] + [P(X = x3)*(x3)] + [P(X = x4)*(x4)] + [P(X = x5)*(x5)]

E[X] = [0.1052*(0)] + [0.0455*(1)] + [0.3337*(2)] + [0.3754*(3)] + [0.1401*(4)]

E[X] = 0.1052*(0) + 0.0455*(1) + 0.3337*(2) + 0.3754*(3) + 0.1401*(4)

E[X] = 0 + 0.0455 + 0.6674 + 1.1262 + 0.5604

E[X] = 2.3995

So mu = 2.3995
--------------------------------------------------------------------------------------------------

Variance:

sigma^2 = Sum( P(X = xi) * (xi - E(X) )^2 )

sigma^2 = [P(X = x1)*(x1 - mu)^2] + [P(X = x2)*(x2 - mu)^2] + [P(X = x3)*(x3 - mu)^2] + [P(X = x4)*(x4 - mu)^2] + [P(X = x5)*(x5 - mu)^2]

sigma^2 = [0.1052*(0 - 2.3995)^2] + [0.0455*(1 - 2.3995)^2] + [0.3337*(2 - 2.3995)^2] + [0.3754*(3 - 2.3995)^2] + [0.1401*(4 - 2.3995)^2]

sigma^2 = [0.1052*(-2.3995)^2] + [0.0455*(-1.3995)^2] + [0.3337*(-0.3995)^2] + [0.3754*(0.6005)^2] + [0.1401*(1.6005)^2]

sigma^2 = [0.1052*(5.75760025)] + [0.0455*(1.95860025)] + [0.3337*(0.15960025)] + [0.3754*(0.36060025)] + [0.1401*(2.56160025)]

sigma^2 = 0.1052*(5.75760025) + 0.0455*(1.95860025) + 0.3337*(0.15960025) + 0.3754*(0.36060025) + 0.1401*(2.56160025)

sigma^2 = 0.6056995463 + 0.089116311375 + 0.053258603425 + 0.13536933385 + 0.358880195025

sigma^2 = 1.242323989975

So the variance is roughly 1.242323989975
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