How to find variance and probability from a uniformly distributed random variable?

Question

Let $x$ be a random variable that describes the weight in pounds of a potato. Assume $X\sim U(.5,1.5)$; that is, $X$ is uniformly distributed between $0.5$ and $1.5$ with average value $1$ pound.

a) Find the variance $V(X)$;

b) Find the probability that a sack of 100 potatoes will weigh less than $97$ pounds?

So to get the variance I need to find the expected value. And that would be...

\begin{equation} fx(t)dt = ft(x) = \begin{cases} 0.5, \quad x < 0.5, \\ 1.5, \quad 0.5 < x < 1.5\quad (?) \end{cases} \end{equation}

Sorry. I've only done the uniform distribution of $(0,1)$.

Answer 1

One may recall that the PDF of the continuous uniform distribution over $[a,b]$ is $$ f(x)=\begin{cases} \frac{1}{b - a} & \mathrm{for}\ a \le x \le b, \\[8pt] 0 & \mathrm{for}\ x<a\ \mathrm{or}\ x>b, \end{cases} $$ hence $$ E(X)=\int_a^bxf(x)dx=\int_a^b \frac{x}{b-a}dx=? $$ $$ V(X)=\int_a^bx^2f(x)dx-(E(X))^2=\int_a^b \frac{x^2}{b-a}dx-(E(X))^2=? $$

Here $a=\frac12$, $b=\frac32$.

Can you take it from here?

Answer 2

Hint:

If $a>0$ and $U$ has uniform distribution on $(0,1)$ then $X:=aU+b$ has uniform distribution on $(b,a+b)$.

Backwards: if $X$ has uniform distribution on $(b,a+b)$ then $U:=\frac{X-b}{a}$ has uniform distribution on $(0,1)$.