/*! 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 43 Write a general rule for \(E(X-c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write a general rule for \(E(X-c)\) where \(c\) is a constant. What happens when \(c=\mu\), the expected value of \(X\) ?

Short Answer

Expert verified
The rule is \(E(X-c) = E(X)-c\). If \(c=\mu\), then \(E(X-c) = 0\).

Step by step solution

01

Define the Expected Value

The expected value or expectation, denoted as \(E(X)\), for a random variable \(X\) is defined as \(E(X) = \sum x_i P(x_i)\) for discrete random variables or \(E(X) = \int x f(x) \, dx\) for continuous random variables, where \(f(x)\) is the probability density function of \(X\).
02

Apply Linear Transformations to Expected Value

The expected value has a linearity property: for any constant \(c\), \(E(X - c) = E(X) - E(c)\). Because \(c\) is a constant, \(E(c) = c\). Thus, the rule becomes \(E(X - c) = E(X) - c\).
03

Substitute \(c = \mu\)

If we substitute \(c = \mu\), where \(\mu\) is the expected value of \(X\), the equation becomes \(E(X - \mu) = E(X) - \mu\). Since \(E(X) = \mu\), the expression simplifies to \(E(X - \mu) = \mu - \mu = 0\).
04

Conclusion

The general rule for \(E(X-c)\) is \(E(X) - c\). When \(c = \mu\), the expected value \(E(X-c)\) becomes 0. This holds because you are essentially calculating the expected deviation of \(X\) from its mean, which is always zero.

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
When exploring probability and statistics, a key concept is the random variable. A random variable is a numerical outcome of a random phenomenon. Imagine flipping a coin or rolling a die, both are simple events that have numerical outcomes we can describe using random variables. They can be discrete, like the number of heads we get when flipping a coin multiple times, or continuous, like the height of students in a class.
  • Discrete random variables have countable outcomes, such as integers.
  • Continuous random variables can take any value within a range.
Understanding random variables is critical for calculating the expected value, which is the core of predicting outcomes in probabilistic settings.
Probability Density Function
For continuous random variables, the probability density function (PDF) is a crucial tool. It describes the likelihood of a random variable taking on a particular value. Unlike discrete cases where we use probability mass functions (PMFs), PDFs provide a continuous graph of probabilities. The probability of a random variable falling within a particular range is given by the area under the curve of its PDF.
  • PDFs define the distribution of continuous random variables.
  • The total area under a PDF curve equals 1, reflecting complete certainty that the variable will assume some value in its range.
In the case of the expected value, the PDF helps calculate crucial expectations by integrating the function over its range.
Linearity of Expectation
One fascinating property of expected values is their linearity. This means that the expected value of a sum is the sum of the expected values. Mathematically, if you have two random variables, say, \(X\) and \(Y\), their expected values will add linearly: \(E(X + Y) = E(X) + E(Y)\). This holds even if \(X\) and \(Y\) are dependent, and it's an immensely useful property.
With the linearity of expectation, even when dealing with constants, we see that \(E(X - c) = E(X) - c\) since constants are unaffected by the probabilistic nature. This principle simplifies our understanding and calculations regarding random variables and their transformations.
Expected Deviation
Expected deviation is about determining how much a random variable deviates from its mean (or expected value). The rule we looked at, \(E(X - \mu) = 0\), highlights this phenomenon. Here, \(\mu\) represents the mean, or expected value, of \(X\), so \(X - \mu\) is the deviation from this mean.
  • When the expected deviation is zero, it tells us that, on average, the variable equals its mean.
  • This forms the basis of understanding variance, which involves squaring these deviations to avoid negatives and determine spread.
By engaging with expected deviation, one understands the predictability and spread of outcomes related to the random variable, capturing both behavior and reliability.

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

Each time a component is tested, the trial is a success ( \(S\) ) or failure \((F)\). Suppose the component is tested repeatedly until a success occurs on three consecutive trials. Let \(Y\) denote the number of trials necessary to achieve this. List all outcomes corresponding to the five smallest possible values of \(Y\), and state which \(Y\) value is associated with each one.

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate \(\alpha=8\) per hour, so that the number of arrivals during a time period of \(t\) hours is a Poisson rv with parameter \(\mu=8 t\). a. What is the probability that exactly 6 small aircraft arrive during a 1-hour period? At least 6 ? At least 10 ? b. What are the expected value and standard deviation of the number of small aircraft that arrive during a 90 -min period? c. What is the probability that at least 20 small aircraft arrive during a \(2.5\)-hour period? That at most 10 arrive during this period?

The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes" (J. of Pipeline Systems Engr. and Practice, May 2012: 36-46) proposed using the Poisson distribution to model the number of failures in pipelines of various types. Suppose that for cast-iron pipe of a particular length, the expected number of failures is 1 (very close to one of the cases considered in the article). Then \(X\), the number of failures, has a Poisson distribution with \(\mu=1\). a. Obtain \(P(X \leq 5)\) by using Appendix Table A.2. b. Determine \(P(X=2)\) first from the pmf formula and then from Appendix Table A.2. c. Determine \(P(2 \leq X \leq 4)\). d. What is the probability that \(X\) exceeds its mean value by more than one standard deviation?

NBC News reported on May 2,2013 , that 1 in 20 children in the United States have a food allergy of some sort. Consider selecting a random sample of 25 children and let \(X\) be the number in the sample who have a food allergy. Then \(X \sim \operatorname{Bin}(25, .05)\). a. Determine both \(P(X \leq 3)\) and \(P(X<3)\). b. Determine \(P(X \geq 4)\). c. Determine \(P(1 \leq X \leq 3)\). d. What are \(E(X)\) and \(\sigma_{X}\) ? e. In a sample of 50 children, what is the probability that none has a food allergy?

Eighteen individuals are scheduled to take a driving test at a particular DMV office on a certain day, eight of whom will be taking the test for the first time. Suppose that six of these individuals are randomly assigned to a particular examiner, and let \(X\) be the number among the six who are taking the test for the first time. a. What kind of a distribution does \(X\) have (name and values of all parameters)? b. Compute \(P(X=2), P(X \leq 2)\), and \(P(X \geq 2)\). c. Calculate the mean value and standard deviation of \(X\).

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.