Respuesta :
Answer:
[tex] E(X) =\frac{x^2}{2} -\frac{0.5}{3}x^3 \Big|_0^2 = (\frac{2^2}{2}- \frac{0.5}{3}*(2^2)) -0 = 0.667 = \frac{2}{3}[/tex]
For the variance we need to calculate first the second moment given by:
[tex] E(X^2) = \int_{0}^2 x^2 (1-0.5x) dx=\int_{0}^2 x^2-0.5x^3 dx [/tex]
And after solve the integral we got:
[tex] E(X^2) = \frac{x^3}{3} -\frac{0.5}{4} x^4 \Big|_0^2 = \frac{2^3}{3} -\frac{0.5}{4} 2^4 = \frac{8}{3} -2 = \frac{2}{3}[/tex]
And for this case the variance would be:
[tex] Var(X) = E(X^2) -[E(X)]^2 = \frac{2}{3} -(2/3)^2 = \frac{2}{9}[/tex]
And the deviation would be:
[tex] Sd(X) = \sqrt{\frac{2}{9}}= \frac{\sqrt{2}}{3}[/tex]
Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
For this case w ehave the following density function:
[tex] f(x) = 1 -0.5x , 0 < x< 2[/tex]
We can determine the mean with the following integral:
[tex] E(X) = \int_{0}^2 x(1-0.5x) dx[/tex]
And if we solve the integral we got:
[tex] E(X) =\frac{x^2}{2} -\frac{0.5}{3}x^3 \Big|_0^2 = (\frac{2^2}{2}- \frac{0.5}{3}*(2^2)) -0 = 0.667 = \frac{2}{3}[/tex]
For the variance we need to calculate first the second moment given by:
[tex] E(X^2) = \int_{0}^2 x^2 (1-0.5x) dx=\int_{0}^2 x^2-0.5x^3 dx [/tex]
And after solve the integral we got:
[tex] E(X^2) = \frac{x^3}{3} -\frac{0.5}{4} x^4 \Big|_0^2 = \frac{2^3}{3} -\frac{0.5}{4} 2^4 = \frac{8}{3} -2 = \frac{2}{3}[/tex]
And for this case the variance would be:
[tex] Var(X) = E(X^2) -[E(X)]^2 = \frac{2}{3} -(2/3)^2 = \frac{2}{9}[/tex]
And the deviation would be:
[tex] Sd(X) = \sqrt{\frac{2}{9}}= \frac{\sqrt{2}}{3}[/tex]
