Question

A shipping company handles containers in three different sizes: (1) 27 ft3 (3 × 3 × 3), (2) 125 ft3, and (3) 512 ft3. Let Xi (i = 1, 2, 3) denote the number of type i containers shipped during a given week. With μi = E(Xi) and σi2 = V(Xi), suppose that the mean values and standard deviations are as follows: μ1 = 230 μ2 = 260 μ3 = 150 σ1 = 9 σ2 = 12 σ3 = 7 (a) Assuming that X1, X2, X3 are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume = 27X1 + 125X2 + 512X3.]

Answer 1

Answer:

$E(T)=115510ft^{3}$

$V(T)=15154105ft^{6}$

Step-by-step explanation:

Let's start writing the random variables :

X1 : ''The number of 27 $ft^{3}$ containers shipped during a given week''

X2 : ''The number of 125 $ft^{3}$ containers shipped during a given week''

X3 : ''The number of 512 $ft^{3}$ 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 :

$T=(27ft^{3})X_{1}+(125ft^{3})X_{2}+(512ft^{3})X_{3}$

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 :

$E(T)=E(27X1+125X2+512X3)$

$E(T)=E(27X1)+E(125X2)+E(512X3)$

$E(T)=27E(X1)+125E(X2)+512E(X3)$

We use the information from the exercise ⇒

$E(T)=27.(230)+125.(260)+512.(150)=6210+32500+76800=115510$

$E(T)=115510ft^{3}$

For the variance :

If X1, X2, X3, ... , Xn are independent random variables, then ⇒

$V(a_{1}X_{1}+a_{2}X_{2}+...+a_{n}X_{n})=a_{1} ^{2}V(X_{1})+a_{2} ^{2}V(X_{2})+...+a_{n} ^{2}V(X_{n})$

Applying this to the exercise :

$V(T)=V(27X_{1}+125X_{2}+512X_{3})$

$V(T)=27^{2}V(X_{1})+125^{2}V(X_{2})+512^{2}V(X_{3})$

We square the standard deviations to obtain the variance of X1,X2 and X3

$V(T)=(27^{2})(9^{2})+(125^{2})(12^{2})+(512^{2})(7^{2})=59049+2250000+12845056$

$V(T)=15154105(ft^{3})^{2}=15154105ft^{6}$