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Chapter 5: Problem 58

A shipping company handles containers in three different sizes: (1) \(27 \mathrm{ft}^{3}(3 \times 3 \times 3)\), (2) \(125 \mathrm{ft}^{3}\), and (3) \(512 \mathrm{ft}^{3}\). Let \(X_{i}(i=1,2,3)\) 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: $$ \begin{array}{lll} \mu_{1}=200 & \mu_{2}=250 & \mu_{3}=100 \\ \sigma_{1}=10 & \sigma_{2}=12 & \sigma_{3}=8 \end{array} $$ a. Assuming that \(X_{1}, X_{2}, X_{3}\) are independent, calculate the expected value and variance of the total volume shipped. b. Would your calculations necessarily be correct if the \(X_{i} \mathrm{~s}\) were not independent? Explain.

Short Answer

Expert verified
a. Expected value is 87850; variance is 19000116. b. Expected value is correct; variance depends on independence.

Step by step solution

01

Understand the Total Volume

The total volume shipped, denoted as \( V \), is the sum of the volumes of each type of container shipped. Mathematically, this is expressed as \( V = 27X_1 + 125X_2 + 512X_3 \), where \( X_1, X_2, \) and \( X_3 \) are the number of type 1, 2, and 3 containers, respectively.
02

Calculate Expected Value of Total Volume

Using the linearity of expectation, the expected value of \( V \) is:\[ E(V) = E(27X_1 + 125X_2 + 512X_3) = 27E(X_1) + 125E(X_2) + 512E(X_3) \]Substitute the given expected values:\[ E(V) = 27 \times 200 + 125 \times 250 + 512 \times 100 \]Solving this gives:\[ E(V) = 5400 + 31250 + 51200 = 87850 \]
03

Calculate Variance of Total Volume

Since \( X_1, X_2, \) and \( X_3 \) are independent, the variance of \( V \) is:\[ Var(V) = Var(27X_1 + 125X_2 + 512X_3) = 27^2Var(X_1) + 125^2Var(X_2) + 512^2Var(X_3) \]Substitute the variances \( Var(X_i) = \sigma_i^2 \):\[ Var(V) = 27^2 \times 100 + 125^2 \times 144 + 512^2 \times 64 \]After calculating each term, we get:\[ Var(V) = 72900 + 2250000 + 16777216 = 19000116 \]
04

Impact of Independence on Calculations

If \( X_1, X_2, \) and \( X_3 \) were not independent, the variance calculation would not necessarily be correct. The covariance terms would need to be included, and without independence, we cannot assume those covariances are zero. However, the expected value calculation remains correct because expectation does not depend on independence.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
Expected Value is a fundamental concept in probability and statistics, representing the average or mean value of a random variable you would expect if you repeated an experiment many times. To calculate it for a linear combination of random variables, we use the principle known as the linearity of expectation. This principle states that for any constants and random variables, the expected value of a sum is the sum of the expected values.

In our shipping company exercise, we calculated the expected total volume shipped by multiplying each type of container's volume with its respective expected number:
  • \(E(V) = 27E(X_1) + 125E(X_2) + 512E(X_3)\)

This calculation illustrates how expectations can be easily combined, reflective of real-world scenarios where multiple independent elements contribute to an outcome. Note that the expected value inherently accounts for possible variability but doesn鈥檛 detail how spread out or consistent outcomes are.
Variance
Variance measures how much a set of values is spread out from the expected value, giving us insights into the variability of the data. For independent random variables, the variance of their sum is the sum of their variances. This principle is only applicable due to the lack of correlation between the variables.

In solving our problem, the variance of the total volume \(V\) is calculated as:
  • \(Var(V) = 27^2Var(X_1) + 125^2Var(X_2) + 512^2Var(X_3)\)
By substituting and calculating each variable's variance, we can understand how each type's inherent variability affects the total variance. Independent variables allow us to add variances directly without worrying about interactions, unlike dependent variables where we'd have to consider covariance, making calculations more complex.
Independence in Probability
Independence in Probability is crucial when working with random variables in statistics. It implies that the occurrence of one event or the value of one variable does not influence another. In terms of mathematical calculations, especially variance and some probability functions, independence simplifies the equations significantly, allowing direct summation of individual components as seen with variance.

In our given problem, we assumed the independence among different types of containers (\(X_1, X_2, \text{ and } X_3\)). This means the calculations for expected value and variance could be done separately and summed; variance especially relied heavily on this assumption. Had the variables not been independent, covariance terms would need to be calculated and included for variance, complicating the solution. However, independence does not affect the calculation of expected value, which remains consistent regardless of dependency among the variables.

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Most popular questions from this chapter

Let \(X\) denote the number of Canon digital cameras sold during a particular week by a certain store. The pmf of \(X\) is $$ \begin{array}{l|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline p_{X}(x) & .1 & .2 & .3 & .25 & .15 \end{array} $$ Sixty percent of all customers who purchase these cameras also buy an extended warranty. Let \(Y\) denote the number of purchasers during this week who buy an extended warranty. a. What is \(P(X=4, Y=2)\) ? [Hint: This probability equals \(P(Y=2 \mid X=4) \cdot P(X=4)\); now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty.] b. Calculate \(P(X=Y)\). c. Determine the joint pmf of \(X\) and \(Y\) and then the marginal pmf of \(Y\).

There are two traffic lights on my way to work. Let \(X_{1}\) be the number of lights at which I must stop, and suppose that the distribution of \(X_{1}\) is as follows: $$ \begin{array}{l|rrrr} x_{1} & 0 & 1 & 2 & \mu=1.1, \sigma^{2}=.49 \\ \hline p\left(x_{1}\right) & .2 & .5 & .3 \end{array} $$ Let \(X_{2}\) be the number of lights at which I must stop on the way home; \(X_{2}\) is independent of \(X_{1}\). Assume that \(X_{2}\) has the same distribution as \(X_{1}\), so that \(X_{1}, X_{2}\) is a random sample of size \(n=2\). a. Let \(T_{o}=X_{1}+X_{2}\), and determine the probability distribution of \(T_{o}\) b. Calculate \(\mu_{T_{e}}\). How does it relate to \(\mu\), the population mean? c. Calculate \(\sigma_{T_{0}}^{2}\). How does it relate to \(\sigma^{2}\), the population variance?

Show that if \(X\) and \(Y\) are independent rv's, then \(E(X Y)=\) \(E(X) \cdot E(Y)\). Then apply this in Exercise 25.

a. Use the rules of expected value to show that \(\operatorname{Cov}(a X+b\), \(c Y+d)=a c \operatorname{Cov}(X, Y)\). b. Use part (a) along with the rules of variance and standard deviation to show that \(\operatorname{Corr}(a X+b, c Y+d)=\operatorname{Corr}(X, Y)\) when \(a\) and \(c\) have the same sign. c. What happens if \(a\) and \(c\) have opposite signs?

Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 15,30 , and 20 min, respectively, and the standard deviations are 1,2 , and \(1.5\) min, respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?
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