Question

A shipping company handles containers in three different sizes: (1) 27 ft3 (3 Ý 3 Ý 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

?1 = 230	?2 = 240	?3 = 120
?1 = 11	?2 = 12	?3 = 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.]

expected value	ft3
variance	ft6

(b) Would your calculations necessarily be correct if the X_i's were not independent? Explain.

The expected value would not be correct, but the variance would be correct

. Neither the expected value nor the variance would be correct.    

The expected value would be correct, but the variance would not be correct.

Both the expected value and the variance would be correct.

Answer 1

Concepts and reason

The expected value is the average value for a set of observations. This is computed by taking the sum of the values, each multiplied with their respective probabilities. The variance is the distance between the sample points.

Fundamentals

Let ${X_i}$ be n independent random variables, where $\left( {i = 1,2, \ldots n} \right)$ . Then by mathematical expectation definition, the sum of n random variables is expressed as,

$E\left( {{X_1} + {X_2} + \ldots + {X_n}} \right) = E\left( {{X_1}} \right) + E\left( {{X_2}} \right) + \ldots + E\left( {{X_n}} \right)$

And $E\left( {aX} \right) = aE\left( X \right)$

The variance is expressed as,

$\begin{array}{c}\\V\left( {{X_1} + {X_2} + \ldots + {X_n}} \right) = V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + \ldots + V\left( {{X_n}} \right)\\\\ + 2{\mathop{\rm cov}} \left( {{X_1}{X_2}} \right) + \ldots + 2{\mathop{\rm cov}} \left( {{X_1}{X_n}} \right)\\\end{array}$

When ${X_1},{X_2} \ldots {X_n}$ independent, then a covariance term becomes zero. Therefore,

$V\left( {{X_1},{X_2} \ldots {X_n}} \right) = V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + \ldots + V\left( {{X_n}} \right)$

Property: $V\left( {aX} \right) = {a^2}V\left( X \right)$

$V\left( X \right) = {\left[ {SD\left( X \right)} \right]^2}$

(a)

The given information is as follows:

$E\left( {{X_1}} \right) = 230$ , $E\left( {{X_2}} \right) = 240$ , $E\left( {{X_3}} \right) = 120$

$V\left( {{X_1}} \right) = {11^2}$ , $V\left( {{X_2}} \right) = {12^2}$ , $V\left( {{X_3}} \right) = {7^2}$

The total volume is $V = 27{X_1} + 125{X_2} + 512{X_3}$ , and ${X_1},{X_2}{\rm{ and }}{X_3}$ are independent.

Compute the expected value of the total volume:

$V = 27{X_1} + 125{X_2} + 512{X_3}$

$\begin{array}{c}\\E\left( V \right) = E\left[ {27{X_1} + 125{X_2} + 512{X_3}} \right]\\\\ = 27E\left[ {{X_1}} \right] + 125E\left[ {{X_2}} \right] + 512E\left[ {{X_3}} \right]\\\\ = \left( {27 \times 230} \right) + \left( {125 \times 240} \right) + \left( {512 \times 120} \right)\\\\ = 6210 + 30000 + 61440\\\\ = 97650\\\end{array}$

Compute the variance of the total volume:

$V = 27{X_1} + 125{X_2} + 512{X_3}$

$\begin{array}{c}\\Var\left( V \right) = Var\left[ {27{X_1} + 125{X_2} + 512{X_3}} \right]\\\\ = {27^2}Var\left[ {{X_1}} \right] + {125^2}Var\left[ {{X_2}} \right] + {512^2}Var\left[ {{X_3}} \right]\\\end{array}$

Since ${X_1},{X_2} \ldots {X_n}$ are independent, the covariance term becomes zero.

$\begin{array}{c}\\ = \left( {{{27}^2} \times {{11}^2}} \right) + \left( {{{125}^2} \times {{12}^2}} \right) + \left( {{{512}^2} \times {7^2}} \right)\\\\ = 88209 + 2250000 + 12845056\\\\ = 15183265\\\end{array}$

(b)

When the variables ${X_i}$ ’s are not independent, the variance of the total volume of the container will change. Therefore, the expected value would be correct, but the variance would be incorrect.

Ans: Part a

The expected value is 97650 and variance is $15183265$ .

Part b

The expected value would be correct, but the variance would be incorrect.