Let $X$ be a random variable of the continuous type with pdf given by:

` $$ \begin{aligned} f(x)& = \frac{1}{7}e^{-x/7}\\ &\qquad \text{ for }x > 0 \end{aligned} $$`

Use the cdf of $X$ to determine the median of the distribution.

#### Solution

Let $X$ be a random variable of the continuous type with pdf given by:

`$$ \begin{aligned} f(x)& = \frac{1}{7}e^{-x/7}\\ &\qquad \text{ for }x > 0 \end{aligned} $$`

The cumulative density function of $X$ is

`$$ \begin{eqnarray*} F(x) &=& P(X\leq x) \\ &=& \int_0^x f(x)\;dx\\ &=& \frac{1}{7} \int_0^x e^{-x/7}\;dx\\ &=& \frac{1}{7} \bigg[-7e^{-\frac{x}{7}}\bigg]_0^x \\ &=& 1-e^{-\frac{x}{7}}. \end{eqnarray*} $$`

Let $M$ be the median of $X$. Then

`$$ \begin{eqnarray*} P(X \leq M)= 0.5 &\Rightarrow & F(M) = 0.5 \\ &\Rightarrow & 1-e^{-\frac{M}{7}}= 0.5 \\ &\Rightarrow & \bigg[e^{-M/7}\bigg] = 1-0.5 \\ &\Rightarrow & e^{-M/7}= 0.5 \\ &\Rightarrow & -M/7= \log_e 0.5 \\ &\Rightarrow & -M/7= -0.69315 \\ &\Rightarrow & M= (-7)\times (-0.69315) \\ &\Rightarrow & M= 4.85205 \end{eqnarray*} $$`

Thus the median of the distribution is $M=4.85205$.