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 $(0,1)$.
Best Answer
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 $$ E(X)=\int_a^bxf(x)dx=\int_a^b \frac{x}{b-a}dx=? $$ $$ V(X)=\int_a^bx^2f(x)dx-(E(X))^2=\int_a^b \frac{x^2}{b-a}dx-(E(X))^2=? $$
Can you take it from here?