If the area of a square is uniformly distributed, then find the expectation and variance of the side length. I'm told that the area of a square (denoted $A$) is uniformly distributed across the interval $[15,20]$, I'm then asked to find the expectation and variance of the side length but I can't work out how to do this. 
I know the formula 
$$\text{Var}(X)=E(X^2)-[E(X)]^2.$$
But I fail to see how this is applicable.
Thanks!
 A: They tell you that $A\sim\text{Unif}(15,20)$, but 
$$A = L^2$$
where $L$ is the length of a square. This implies that $L = \sqrt A$, and hence
$$E[L] = E[\sqrt A]$$
and 
$$\text{Var}[L] = \text{Var}(\sqrt A) = E[(\sqrt A)^2] -\{E[\sqrt A]\}^2=E[A] -\{E[\sqrt A]\}^2.$$
Use integration to find $E[\sqrt A]$.
A: If $X \sim Unif(15, 20)$, then you can get $E(Y),\; V(Y)$ and other information about $Y$ by finding the distribution
of $Y = \sqrt{X}.$ Please look in your text under some heading such as 'transformations of random variables'.
One method, sometimes called the 'CDF method' is as follows:
To start, the support of $Y$ is the interval $(\sqrt{15}, \sqrt{20}).$
So for $y \in (\sqrt{15}, \sqrt{20})$, the CDF of $Y$ is
$$F_Y(y) = P(Y \le y) = P(\sqrt{X} \le y) = P(X \le y^2) = \dots.$$
For $y < \sqrt{15},\;F_Y(y) = 0$ and for $y > \sqrt{20},\;F_Y(y) = 1.$
You can use the density function $f_Y(y)$ of $Y$ to get $E(Y)$ and $V(Y).$ To get the density
take the derivative of $F_Y(y).$
The following simulation of 100,000 realizations of $Y$ can be
used to judge whether your results are correct.
 m = 100000;  x = runif(m, 15, 20)
 y = sqrt(x);  cutp = seq(sqrt(15), sqrt(20), len=20)
 mean(y);  var(y)
 ## 4.18102           # Approx. E(Y)
 ## 0.02993067        # Approx. V(Y)
 mean(x);  mean(y^2)
 ## 17.51086
 ## 17.51086          # Approx. E(Y^2)
 hist(y, br=cutp, prob=T, col="skyblue2")


