SOLUTION: Let X be a continuous random variable with density function
f(x) = { theta*x + 3/2*theta^{3/2}*x^2, for 0 < x < 1/sqrt( theta ); 0, otherwise }
where theta > 0. What is the expec
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-> SOLUTION: Let X be a continuous random variable with density function
f(x) = { theta*x + 3/2*theta^{3/2}*x^2, for 0 < x < 1/sqrt( theta ); 0, otherwise }
where theta > 0. What is the expec
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Question 1203892: Let X be a continuous random variable with density function
f(x) = { theta*x + 3/2*theta^{3/2}*x^2, for 0 < x < 1/sqrt( theta ); 0, otherwise }
where theta > 0. What is the expected value of X? Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
To find the expected value, we apply the integral to x*f(x) over the limits from -infinity to infinity.
E(x) = expected value
Because much of the real number line makes f(x) = 0, we only need to worry about the interval which is the same as where theta > 0
The goal has been reduced to computing
Using fairly elementary calculus integration rules, you should find that:
For some constant C.
(