Uniformly Distributed Random Variables Minimum Let A,B,C,D,E be independent random variables, each of which is uniformly distributed in the interval [0,18]. 
Let X=min{A,B,C,D,E}. What is the expected value of X? Enter your answer as a decimal. 
I know that I need to be attempting some work,but any direction would be greatly appreciated!
 A: We find the density function of $X$. For $0\lt x\lt 18$, we have $X\gt x$ if all of $A,B,C,D,E$  are $\gt x$. 
The probability that $A$ is $\gt x$ is $\frac{18-x}{18}$.  So the probability that all of $A,B,C,D,E$ are $\gt x$ is $\left(\frac{18-x}{18}\right)^5$. 
It follows that
$$F_X(x)=1-\left(\frac{18-x}{18}\right)^5.$$
Now we can differentiate to find the density function $f_X(x)$, and then calculate $E(X)$ as usual. The substitution $u=18-x$ will be useful for the evaluation of the integral. Or else you can use integration by parts.
Remark: There is a "fancier" way that bypasses finding the density function, if you happen to know an expression for the expectation in terms of the cdf. 
A: Let's suppose a set of identically uniformly distributed random variables $A,B,C,D,E$ and assume that we want to find the distribution $X=\min{\{A,B,C,D,E\}}$.
Then the probability $P(X≤x)=P(\min{\{A,B,C,D,E\}}≤x)$ implies that one can find at least one of the elements of $\{A,B,C,D,E\}$ smaller than $x$. Or one can say that the probability that at least one of them is smaller than $x$ equals to: 
$$\begin{eqnarray}P(\min{\{A,B,C,D,E\}}≤x)&=&1-P(A>x,B>x,C>x,D>x,E>x)=\\&=&1-P(A>x)P(B>x)P(C>x)P(D>x)P(E>x)
=\\&=&1-\left(P(Y>x)\right)^5
\end{eqnarray}$$
where $Y$ is any of the variables $\{A,B,C,D,E\}$ and $F_Y(y)$ its cdf (since they are identically distributed.)
Then one can conclude that the cdf of $X$ would look like:$F_X(x)=1-(1-F_Y(x))^5.$ 
Since you have $Y\sim \text{Uniform} (0,18)$, then the cdf $F_Y=\frac{x}{18}$. Then $F_X=1-(1-\frac{x}{18})^5$.
A: If you have a finite number of i.i.d random variable, their maximum, their minimum, and any $j$-rank is a RV with a closed form pdf and cdf. As such, it should be easy for you to find the expectation. These are called the rank statistics of a sample. Casella and Berger's "Statistical Inference" has a nice treatment of the subject. 
Sorry I'm not giving an explicit answer, I just figured you would benefit more from learning it in a general context than from seeing an equation.
