Assume that Y the amount of cleaner dispensed per minute is

Assume that Y, the amount of cleaner dispensed per minute,
is a uniform random variable with probability density function
[p.d.f.], f(y), given below.

         {   5   4.95 <= y <= 5.15
f(y) = {
         {   0   otherwise

[(a)]


Find the expected value of Y, E(Y).

Find the variance value of Y, VAR(Y).

Solution

E(Y)=integral(x*f(x))dx=(5/2)*(5.15^2-4.95^2)=5.05

VAR(Y)=integral((x-5.05)^2*f(x))dx =(5/3)*(x-1)^3 = (5/3)*(0.1^3-0.1^3)=0