A probability distrubution is given by
f(x) =
x, 0 < x < 1
2-x, 1 <= x <2
0, elsewhere
Find the variance of X.
I first calculate the expected value:
\(E(X) = \int^{1}_{0}x^2 dx + \int^{2}_{1}2x - x^2 dx = 1\)
Then the variance:
\(\sigma^2 = \int^{1}_{0}(x-1)^2x dx + \int^{2}_{1}(x-1)^2(2-x) dx = 1/6\)
f(x) =
x, 0 < x < 1
2-x, 1 <= x <2
0, elsewhere
Find the variance of X.
I first calculate the expected value:
\(E(X) = \int^{1}_{0}x^2 dx + \int^{2}_{1}2x - x^2 dx = 1\)
Then the variance:
\(\sigma^2 = \int^{1}_{0}(x-1)^2x dx + \int^{2}_{1}(x-1)^2(2-x) dx = 1/6\)
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