VARIANCE OF THE VARIABLE x DISTRIBUTED ACCORDING T EXPONENTIAL LAW f(x)= ) 2e^-2x,&xgeqslant 0 0,&xlt 0 IS Select one: 1/2 -1/2 (Ln2)/2 1/4 -1/4

To find the variance of the random variable X, we need to first find the mean (expected value) of X, denoted as μ, and then use the formula for variance:<br /><br />Var(X) = E(X^2) - [E(X)]^2<br /><br />where E(X^2) is the expected value of X^2 and [E(X)]^2 is the square of the mean.<br /><br />Given the probability density function (pdf) of X:<br /><br />f(x) = { 2e^(-2x), x ≥ 0<br />0, x < 0<br /><br />we can find the mean by integrating the product of x and f(x) over the entire range of X:<br /><br />μ = ∫(x * f(x)) dx<br /><br />Integrating from 0 to ∞:<br /><br />μ = ∫(x * 2e^(-2x)) dx<br /><br />Let u = -2x, then du = -2 dx, and when x = 0, u = 0, and when x = ∞, u = -∞.<br /><br />μ = ∫(u/2 * 2e^u) du<br /><br />Integrating by parts:<br /><br />μ = [u * e^u]_(-∞)^0 - ∫(e^u) du<br /><br />μ = 0 - [e^u]_(-∞)^0<br /><br />μ = -1<br /><br />Now, we can find E(X^2) by integrating the product of x^2 and f(x entire range of X:<br /><br />E(X^2) = ∫(x^2 * f(x)) dx<br /><br />Integrating from 0 to ∞:<br /><br />E(X^2) = ∫(x^2 * 2e^(-2x)) dx<br /><br />Let u = -2x, then du = -2 dx, and when x = 0, u = 0, and when x = ∞, u = -∞.<br /><br />E(X^2) = ∫(u^2/4 * 2e^u) du<br /><br />Integrating by parts:<br /><br />E(X^2) = [u^2/8 * e^u]_(-∞)^0 - ∫(u/4 * e^u) du<br /><br />E(X^2) = 0 - [u^2/8 * e^u]_(-∞)^0 + ∫(u/4 * e^u) du<br /><br />E(X^2) = 0 - [u^2/8 * e^u]_(-∞)^0 + [u/8 * e^u]_(-∞)^0<br /><br />E(X^2) = 0 - 0 + 1/8<br /><br />E(X^2) = 1/8<br /><br />Now, we can find the variance using the formula:<br /><br />Var(X) = E(X^2) - [E(X)]^2<br /><br />Var(X) = 1/8 - (-1)^2<br /><br />Var(X) = 1/8 - 1<br /><br />Var = -7/8<br /><br />Therefore, the variance of the random variable X is -7/8.