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Chapter 5: Q58E (page 241)

A shipping company handles containers in three different sizes:\(\left( 1 \right)\;27f{t^3}\;\left( {3 \times 3 \times 3} \right)\)\(\left( 2 \right) 125 f{t^3}, and \left( 3 \right)\;512 f{t^3}\). Let \({X_i}\left( {i = \;1, 2, 3} \right)\)denote the number of type i containers shipped during a given week. With \({\mu _i} = E\left( {{X_i}} \right)\)and\(\sigma _i^2 = V\left( {{X_i}} \right)\), suppose that the mean values and standard deviations are as follows:

\(\begin{array}{l}{\mu _1} = 200 {\mu _2} = 250 {\mu _3} = 100 \\{\sigma _1} = 10 {\sigma _2} = \,12 {\sigma _3} = 8\end{array}\)

a. Assuming that \({X_1}, {X_2}, {X_3}\)are independent, calculate the expected value and variance of the total volume shipped. (Hint:\(Volume = 27{X_1} + 125{X_2} + 512{X_3}\).)

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

Short Answer

Expert verified

The expected value is \(87,850\)and the variance of the volume is \(19,100,116\).

Step by step solution

01

Definition of Probability

The probability of an event happening is defined by probability. Many real-life circumstances need us to forecast the outcome of an occurrence. We may be certain or uncertain about the outcome of an event. In such instances, we say that the occurrence has a chance of occurring or not occurring.

02

Calculation for the determination of expected value and variance of volume.

(a):

The expected values of the volume

\({V_o} = 27{X_1} + 125{X_2} + 512{X_3}\)

is

\(\begin{aligned}E\left( {{V_o}} \right) &= E\left( {27{X_1} + 125{X_2} + 512{X_3}} \right)\\ &= 27E\left( {{X_1}} \right) + 125E\left( {{X_2}} \right) + 512E\left( {{X_3}} \right)\\ &= 27 \cdot 200 + 125 \cdot 250 + 512 \cdot 100\\ &= 87,850\end{aligned}\)

(1): this stands for any random variables,

(2): the expectations are given in the exercise.

The variance of the volume is

\(\begin{aligned}V\left( {{V_o}} \right) &= V\left( {27{X_1} + 125{X_2} + 512{X_3}} \right)\\ &= {27^2}V\left( {{X_1}} \right) + {125^2}V\left( {{X_2}} \right) + {512^2}V\left( {{X_3}} \right)\\ &= {27^2} \cdot {10^2} + {125^2} \cdot {12^2} + {512^2} \cdot {8^2}\\ &= 19,100,116\end{aligned}\)

03

Further Calculation for the determination of expected value and variance of volume.

(3): the given random variables are independent therefore this equality stands,

(4): the standard variations of the random variables are given in the exercise.

The variance of the volume is

\(\begin{aligned}V\left( {{V_o}} \right) &= V\left( {27{X_1} + 125{X_2} + 512{X_3}} \right)\\ &= {27^2}V\left( {{X_1}} \right) + {125^2}V\left( {{X_2}} \right) + {512^2}V\left( {{X_3}} \right)\\ &= {27^2} \cdot {10^2} + {125^2} \cdot {12^2} + {512^2} \cdot {8^2}\\ &= 19,100,116\end{aligned}\)

(3): the given random variables are independent therefore this equality stands,

(4): the standard variations of the random variables are given in the exercise.

04

Explanation is given for part b.

(b):

As mentioned in (a),the expected value would stay the same, no matter the independence. However,

the variance would change

the equality (3): does not stand when the random variables are not independent (covariances should be included).

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