MEDIAN OF THE VARIABLE X. DISTRIBUTED ACCORDING ; TO EXPONENT IAL LAW f(x)= ) 2e^-2x,&xgeqslant 0 0,&xlt 0 IS Select one: 1/2 -1/4 D 1/4 -1/2 (Ln2)/2

To find the median of the variable X, we need to integrate the probability density function (PDF) from 0 to infinity and set it equal to 0.5.<br /><br />The PDF is given by:<br />$f(x)=\begin{cases} 2e^{-2x}, & x\geqslant 0\\ 0, & x<0\end{cases}$<br /><br />Integrating the PDF from 0 to infinity:<br />$\int_{0}^{\infty} f(x) dx = \int_{0}^{\infty} 2e^{-2x} dx = 1$<br /><br />To find the median, we need to find the value of x such that the cumulative distribution function (CDF) is equal to 0.5. The CDF is given by:<br />$F(x) = \int_{-\infty}^{x} f(t) dt$<br /><br />Since the PDF is 0 for $x<0$, the CDF is 0 for $x<0$. Therefore, we only need to consider the integral from 0 to infinity:<br />$F(x) = \int_{0}^{x} 2e^{-2t} dt = 1 - e^{-2x}$<br /><br />Setting the CDF equal to 0.5 and solving for x:<br />$1 - e^{-2x} = 0.5$<br />$e^{-2x} = 0.5$<br />$-2x = \ln(0.5)$<br />$x = -\frac{\ln(2)}{2}$<br /><br />Therefore, the median of the variable X is $-\frac{\ln(2)}{2}$.<br /><br />So, the correct answer is $(\ln2)/2$.