Answer:
[tex]E(T)=115510ft^{3}[/tex]
[tex]V(T)=15154105ft^{6}[/tex]
Step-by-step explanation:
Let's start writing the random variables :
X1 : ''The number of 27 [tex]ft^{3}[/tex] containers shipped during a given week''
X2 : ''The number of 125 [tex]ft^{3}[/tex] containers shipped during a given week''
X3 : ''The number of 512 [tex]ft^{3}[/tex] containers shipped during a given week''
We assume that X1, X2 and X3 are independent (data from the exercise).
The random variable T : ''Total volume shipped during a given week'' has the following equation :
[tex]T=(27ft^{3})X_{1}+(125ft^{3})X_{2}+(512ft^{3})X_{3}[/tex]
We want to calculate E(T) and (σ^2)(T).
In the following equations : (σ^2)(X) = V(X)
The expected value operator works as a linear operator.
Then, we calculate E(T) as the following :
[tex]E(T)=E(27X1+125X2+512X3)[/tex]
[tex]E(T)=E(27X1)+E(125X2)+E(512X3)[/tex]
[tex]E(T)=27E(X1)+125E(X2)+512E(X3)[/tex]
We use the information from the exercise ⇒
[tex]E(T)=27.(230)+125.(260)+512.(150)=6210+32500+76800=115510[/tex]
[tex]E(T)=115510ft^{3}[/tex]
For the variance :
If X1, X2, X3, ... , Xn are independent random variables, then ⇒
[tex]V(a_{1}X_{1}+a_{2}X_{2}+...+a_{n}X_{n})=a_{1} ^{2}V(X_{1})+a_{2} ^{2}V(X_{2})+...+a_{n} ^{2}V(X_{n})[/tex]
Applying this to the exercise :
[tex]V(T)=V(27X_{1}+125X_{2}+512X_{3})\\[/tex]
[tex]V(T)=27^{2}V(X_{1})+125^{2}V(X_{2})+512^{2}V(X_{3})[/tex]
We square the standard deviations to obtain the variance of X1,X2 and X3
[tex]V(T)=(27^{2})(9^{2})+(125^{2})(12^{2})+(512^{2})(7^{2})=59049+2250000+12845056[/tex]
[tex]V(T)=15154105(ft^{3})^{2}=15154105ft^{6}[/tex]
