![E(X) = \displaystyle\sum x_iP(x_i)\\\\E(X) = 0(0.027) + 1(0.189) + 2(0.441) + 3(0.343)\\E(X) = 2.1\\\\E(X^2) = \displaystyle\sum x_i^2P(x_i)\\\\E(X^2) = 0^2(0.027) + 1^2(0.189) + 2^2(0.441) + 3^2(0.343)\\E(X^2) = 5.04\\\\Var(X) = E(X^2) -[E(X)]^2\\Var(X) = 5.04 - (2.1)^2\\Var(X) = 0.63](/tpl/images/0466/7572/4fafb.png)
For the following distribution,
x: 0,1,2,3
p(x): 0.027, 0.189, 0.441, 0.34...
Mathematics, 23.01.2020 04:31, xcasta213
For the following distribution,
x: 0,1,2,3
p(x): 0.027, 0.189, 0.441, 0.343
what is the variance of the distribution?
2.1
0.63
3.9
2.754
Answers: 1
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