/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 His and her earnings Researchers... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

His and her earnings Researchers randomly select a married couple in which both spouses are employed. Let \(X\) be the income of the husband and \(Y\) be the income of the wife. Suppose that you know the means \(\mu_{X}\) and \(\mu_{Y}\) and the variances \(\sigma_{X}^{2}\) and \(\sigma_{Y}^{2}\) of both variables. (a) Is it reasonable to take the mean of the total income \(X+Y\) to be \(\mu_{X}+\mu_{Y} ?\) Explain your answer. (b) Is it reasonable to take the variance of the total income to be \(\sigma_{X}^{2}+\sigma_{Y}^{2}\) ? Explain your answer.

Short Answer

Expert verified
(a) Yes, the mean is \(\mu_{X} + \mu_{Y}\). (b) No, unless \(X\) and \(Y\) are independent.

Step by step solution

01

Understanding the Mean of Total Income

The mean of a sum of random variables is equal to the sum of their means, assuming linearity of expectation. In this case, \(X\) and \(Y\) are the incomes of the husband and wife, respectively. Therefore, the mean of the total income \(X+Y\) is given by \(\mu_{X+Y} = \mu_{X} + \mu_{Y}\). This assumption is reasonable because it directly follows from the property of expected values.
02

Understanding the Variance of Total Income

To determine if the variance of the total income \(X+Y\) is \(\sigma_{X}^{2}+\sigma_{Y}^{2}\), we consider the properties of variance. Variance of the sum of two independent random variables is the sum of their variances: \(\sigma_{X+Y}^{2} = \sigma_{X}^{2} + \sigma_{Y}^{2}\), if \(X\) and \(Y\) are independent. However, since it is not specified that \(X\) and \(Y\) are independent, we cannot assume this without additional information. Thus, the variance calculation for \(X+Y\) might also need to account for their covariance: \(\sigma_{X+Y}^{2} = \sigma_{X}^{2} + \sigma_{Y}^{2} + 2Cov(X,Y)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Random Variables
In statistics, a random variable is a variable that takes on values according to a certain probability distribution. This concept is foundational in probability theory and is used to model situations where outcomes are uncertain. A random variable can be either discrete or continuous.
  • Discrete random variables have specific outcomes, like rolling a die where the outcomes are 1 through 6.
  • Continuous random variables can take on any value within a certain range, like the weight of a randomly selected person.
For the exercise in question, the incomes of the husband ( X ) and the wife ( Y ) are random variables. They are considered random because their exact incomes can vary, and different couples will have different incomes. Understanding random variables is key because they allow us to use mathematical techniques to predict and analyze these incomes.
Mean and Variance
The mean and variance are two vital concepts in statistics used to describe random variables. The mean is the average expected value, providing a measure of central tendency. For random variables, the mean is denoted as \( \mu_{X} and \mu_{Y} \) for the incomes of the husband and wife, respectively. You can simply think of it as what you would "expect" the average income to be over many observations.

The variance, on the other hand, measures the spread or dispersion of a set of values. It tells us how much the incomes are expected to vary. Variance is denoted as \( \sigma_{X}^{2} and \sigma_{Y}^{2} \). If we know these, we can infer much about the regularity or variability of incomes. Calculating the variance of total income requires understanding whether the incomes are related, as this influences the calculation.
  • If random variables are independent, their variances add up.
  • If not independent, covariances influence this addition.
Independence
Independence in statistics means that the occurrence or result of one event doesn't affect the occurrence of another. Two random variables, such as the incomes of a husband and wife, are independent if knowing the income of one provides no information about the income of the other.

When dealing with means and variances, independence plays a crucial role:
  • If \( X and Y \) are independent, the mean of their sum is simply the sum of their means.
  • Likewise, their variances also sum up directly \( \sigma_{X+Y}^{2} = \sigma_{X}^{2} + \sigma_{Y}^{2} \).
However, in real-life situations like couples' incomes, complete independence might not always be realistic. There can be correlations or dependencies due to shared economic circumstances or mutual financial decisions.
Covariance
Covariance is a statistical measure that indicates the extent to which two random variables change together. In simple terms, it shows whether increases in one variable might be associated with increases or decreases in another.
  • Positive covariance: Both variables tend to increase or decrease together.
  • Negative covariance: One variable increases when the other decreases.
When calculating the variance of a sum of random variables, if they are not independent, their covariance must be accounted for:

\[\sigma_{X+Y}^{2} = \sigma_{X}^{2} + \sigma_{Y}^{2} + 2Cov(X,Y)\]This formula shows how the covariance adjusts the variance of the total income depending on how incomes of husband and wife tend to vary together.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Fire insurance Suppose a homeowner spends \(\$ 300\) for a home insurance policy that will pay out \(\$ 200,000\) if the home is destroyed by fire. Let \(Y=\) the profit made by the company on a single policy. From previous data, the probability that a home in this area will be destroyed by fire is 0.0002 . (a) Make a table that shows the probability distribution of \(Y\) (b) Compute the expected value of Y. Explain what this result means for the insurance company.

Random digit dialing When an opinion poll calls a residential telephone number at random, there is only a \(20 \%\) chance that the call reaches a live person. You watch the random digit dialing machine make 15 calls. Let \(X=\) the number of calls that reach a live person. (a) Find and interpret \(\mu_{X}\) (b) Find and interpret \(\sigma_{X}\)

Rhubarb Suppose you purchase a bundle of 10 bareroot rhubarb plants. The sales clerk tells you that \(5 \%\) of these plants will die before producing any rhubarb. Assume that the bundle is a random sample of plants and that the sales clerk's statement is accurate. Let \(Y=\) the number of plants that die before producing any rhubarb. Use the binomial probability formula to find \(P(Y=1) .\) Interpret this result in context.

determine whether the given random variable has a binomial distribution. Justify your answer. Long or short? Put the names of all the students in your class in a hat. Mix them up, and draw four names without looking. Let \(Y=\) the number whose last names have more than six letters.

\(10 \%\) condition To use a binomial distribution to approximate the count of successes in an SRS, why do we require that the sample size \(n\) be no more than \(10 \%\) of the population size \(N ?\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.