The age [in years] $X$ of sewing machines to be reconditioned is a random variable with the following probability distribution: $f(x)=(1/972)x(18-x)$ for $0<x<18,$ and $f(x)=0,$ elsewhere. The time [in months] for reconditioning a sewing machine, Y, is related to X by $Y=(X/9+.4).$
$(i)$ Determine $E(X)$ and $\sigma_x$
What I did: First,
$E(X)=\int_{-{\infty}}^\infty$ $xf(x)dx$
$E(X)=$ $f(x)=(1/972)x(18-x)$
$=1/972\int_0^{18}x(18x-x^2)dx$
$=1/972[18x^3/3-x^4/4]_0^{18}$
$E[X]=9$
Next, $E(X^2)=\int_{-{\infty}}^\infty$ $x^2f(x)dx$
$=1/972\int_0^{18} x^2(18x-x^2)dx$
$=1/972[18x^4/4-x^5/5]_0^{18}$
$E[x^2]=97.2$
$Var(x)=E[X^2]-E[X]^2=16.2$
$\sigma_x = 4.02$
$(ii)$ Determine $E(Y)$ and $\sigma_y$
Using $Y=(X/9+.4),$ we get $E[Y]=(E[X]/9+.4)$
So, $E[Y]=[9]/9+.4=1.4$
I am not sure how to calculate $\sigma_y$. Can someone please show me how this would be done with the working?
You can use the variance formulae to calculate the variance, and then take the square root to find the standard deviation.
$V(Y)=V(\frac{X}{9}+.4)$
$V(Y)=V(\frac{X}{9})$ using the variance rules.
$V(Y)=\frac{1}{81}V(X)$ again using variance laws.
$V(Y)=\frac{1}{81}\times 16.2$
$V(Y) = 0.2$.
You can now calculate the standard deviation.
Hope this helps!