If Y is a continuous random variable and m is the median of the distribution, then m is such that P(Y ≤m) = P(Y ≥ m) = 1/2. If Y₁, Y₂,..., Y are independent, exponentially dis- tributed random variables with mean ß and median m, Example 6.17 implies that Y(n) max (Y₁, Y₂, ..., Yn) does not have an exponential distribution. Use the general form of Fy) (y) to show that P(Y(n) > m) = 1-(.5)".

It is given that $Y_1, Y_2, \dots ,Y_n$ are independent random variables that follow an exponential distribution with mean $\beta$ and median $m$.

Thus the probability density function (PDF) of Y is as follows:

$f_Y\left(y\right)=\frac{1}{\beta }{e}^{-\frac{1}{\beta }y}, y\ge 0$

As it is given that Y has median m, thus

$\begin{array}{rcl}{F}_Y\left(m\right)& =& P\left(Y\le m\right)\\ & =& \frac{1}{2}\\ & =& 0.5\end{array}$