How to find the probability, mean and cdf using a pdf 
Let $X$ be a random variable with pdf $f_{x}(x)= \large \frac{1}{5} e^{\frac{-x}{5}}$, $x>0$
a. Sketch the graph of $f_x$. Use the pdf to find $P(X>5)$. Find the mean of $X$.
b. Find the cdf of $X$. Use it to compute $P(X>7)$. Also use the cdf to compute the median of the distribution. Compare it to the mean of the distribution to the median in terms of the skewness of the distribution.
c. Finally, compute the interquartile range which is the difference between the $75^{th}$ percentile and the $25^{th}$ percentile.

 A: a) Some calculus needed. For the probability that $X\gt 5$, you need to find $\int_5^\infty \frac{1}{5}e^{-x/5}\,dx$.  
For the mean, probably you are expected to use the fact that if $f_X(x)$ is the density function, then the mean is $\int_{-\infty}^\infty xf_X(x)\,dx$. In our case that comes down to evaluating $\int_0^\infty \frac{x}{5}e^{-x/5}\,dx$. The antiderivative you need is usually calculated by integration by parts.
b) he cumulative distribution function $F_X(x)$ of $X$ is $0$ if $x\lt 0$.  For $x\ge 0$, it is given by 
$$F_X(x)=\int_0^x \frac{1}{5}e^{-t/5}\,dt.$$
Now that we have $F_X$, note that $\Pr(X\gt 7)=1-\Pr(X\le 7)=1-F_X(7)$. 
To go on, unfortunately I must reveal that for $x\ge 0$, it turns out that $F_X(x)=1-e^{-x/5}$.
To find the median $m$, we need to solve the equation
$$1-e^{-m/5}=\frac{1}{2}.$$
In solving this, you will need to use the (natural) logarithm function.
c) The $25$-th percentile is the point $a$ such that $\Pr(X\le a)=0.25$.  To find $a$ you proceed much like in the calculation of the median. 
The $75$-th percentile $b$ is computed in a similar way. Finally, calculate $b-a$.
