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Let \(X=\) the outcome when a fair die is rolled once. If before the die is rolled you are offered either (1/3.5) dollars or \(h(X)=1 / X\) dollars, would you accept the guaranteed amount or would you gamble? [Note: It is not generally true that \(1 / E(X)=E(1 / X)\).

Short Answer

Expert verified
Choose to gamble; \( E(1/X) \approx 0.4083 > 0.2857 \).

Step by step solution

01

Identify the Probability Distribution

When a fair die is rolled, the outcomes are 1, 2, 3, 4, 5, and 6. Each outcome has a probability of occurring equal to \( \frac{1}{6} \).
02

Calculate the Expected Value of X

The expected value \( E(X) \) for one roll of a die is calculated as follows:\[E(X) = \sum_{i=1}^{6} x_i P(x_i) = \left( 1 \times \frac{1}{6} \right) + \left( 2 \times \frac{1}{6} \right) + \left( 3 \times \frac{1}{6} \right) + \left( 4 \times \frac{1}{6} \right) + \left( 5 \times \frac{1}{6} \right) + \left( 6 \times \frac{1}{6} \right)\]\[E(X) = 3.5\]
03

Calculate the Expected Value of h(X)

We calculate \( E\left( \frac{1}{X} \right) \) by finding the expectation of the function \( h(X) = \frac{1}{X} \):\[E\left( \frac{1}{X} \right) = \frac{1}{6} \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right)\]\[E\left( \frac{1}{X} \right) = \frac{1}{6} \left( 1 + 0.5 + 0.333 + 0.25 + 0.2 + 0.167 \right)\]\[E\left( \frac{1}{X} \right) \approx \frac{1}{6} \times 2.45 = 0.4083\]
04

Compare Expected Values

We compare the two expected values: the guaranteed amount \( \frac{1}{3.5} \approx 0.2857 \) and the expected value of \( h(X) = \frac{1}{X} \) which is approximately \( 0.4083 \).
05

Decide Whether to Accept or Gamble

Since the expected value of earning \( 1/X \) dollars (\( 0.4083 \)) is greater than the guaranteed amount (\( 0.2857 \)), it is statistically better to gamble and choose \( h(X) = \frac{1}{X} \) dollars.

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Key Concepts

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

Expected Value
The expected value, often represented as \( E(X) \), provides the average outcome of a probability distribution if you were to repeat an experiment infinite times. In simpler terms, it gives you a measure of the center of the distribution – a value you might "expect" to get in each trial as an average in the long run.

In probability theory, calculating the expected value involves multiplying each possible outcome by the probability of that outcome and then summing all of these values. In the context of a fair die, each possible outcome (1 through 6) has an equal probability of \( \frac{1}{6} \). Therefore, the expected value \( E(X) \) of rolling a die is:

\[ E(X) = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{1}{6} + 4 \times \frac{1}{6} + 5 \times \frac{1}{6} + 6 \times \frac{1}{6} = 3.5 \]

This means, on average, you can expect to roll a 3.5 with a fair six-sided die if rolled indefinitely.
Probability Theory
Probability theory is a branch of mathematics that deals with the likelihood of different outcomes. It is foundational for understanding how events occur and is pivotal in calculating probabilities for any random event that might take place.

When dealing with probability distributions like rolling a die, every outcome has a certain likelihood expressed as a probability between 0 and 1. The sum of the probabilities for all possible outcomes must equal 1 to ensure comprehensiveness.

In our die example, each face of the die is equally likely to land face up, with a probability of \( \frac{1}{6} \). This is an example of a uniform distribution—a core concept in probability theory where all outcomes have the same probability.
  • Every possible result has an assigned probability.
  • The total probability must account for 100% of possible outcomes, i.e., add to 1.
  • The theory aids in analyzing random events by quantifying uncertainty.
This comprehensive framework helps in modeling real-world random processes and understanding the underlying randomness.
Random Variables
A random variable is a fundamental concept in probability theory. It represents a numerical outcome of a random phenomenon, essentially a variable that can take on different values due to some random chance.

In the die-roll scenario, the random variable \( X \) represents the outcome of rolling the die. It can take one of the values {1, 2, 3, 4, 5, 6} corresponding to the sides of the die. Each specific value \( X \) can assume has an associated probability based on the mechanics of the experiment.
  • Random variables can be discrete or continuous.
  • Discrete random variables, like our die example, take on countable values.
  • Continuous random variables may take on any value within a range.
By understanding and manipulating these random variables, we can make informed predictions and understand the likelihood of different outcomes in randomized scenarios.

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Most popular questions from this chapter

Suppose that \(30 \%\) of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other \(70 \%\) want a used copy. Consider randomly selecting 25 purchasers. a. What are the mean value and standard deviation of the number who want a new copy of the book? b. What is the probability that the number who want new copies is more than two standard deviations away from the mean value? c. The bookstore has 15 new copies and 15 used copies in stock. If 25 people come in one by one to purchase this text, what is the probability that all 25 will get the type of book they want from current stock? [Hint: Let \(X=\) the number who want a new copy. For what values of \(X\) will all 15 get what they want?] d. Suppose that new copies cost \(\$ 100\) and used copies cost \(\$ 70\). Assume the bookstore currently has 50 new copies and 50 used copies. What is the expected value of total revenue from the sale of the next 25 copies purchased? Be sure to indicate what rule of expected value you are using.

Suppose that you read through this year's issues of the \(N e w\) York Times and record each number that appears in a news article-the income of a CEO, the number of cases of wine produced by a winery, the total charitable contribution of a politician during the previous tax year, the age of a celebrity, and so on. Now focus on the leading digit of each number, which could be \(1,2, \ldots, 8\), or 9 . Your first thought might be that the leading digit \(X\) of a randomly selected number would be equally likely to be one of the nine possibilities (a discrete uniform distribution). However, much empirical evidence as well as some theoretical arguments suggest an alternative probability distribution called Benford's law: \(p(x)=P(1\) st digit is \(x)=\log _{10}(1+1 / x) x=1,2, \ldots, 9\) a. Compute the individual probabilities and compare to the corresponding discrete uniform distribution. b. Obtain the cdf of \(X\). c. Using the cdf, what is the probability that the leading digit is at most 3 ? At least 5 ? [Note: Benford's law is the basis for some auditing procedures used to detect fraud in financial reporting-for example, by the Internal Revenue Service.]

An individual who has automobile insurance from a certain company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pmf of \(Y\) is $$ \begin{array}{l|cccc} y & 0 & 1 & 2 & 3 \\ \hline p(y) & .60 & .25 & .10 & .05 \end{array} $$ a. Compute \(E(Y)\). b. Suppose an individual with \(Y\) violations incurs a surcharge of \(\$ 100 Y^{2}\). Calculate the expected amount of the surcharge.

A consumer organization that evaluates new automobiles customarily reports the number of major defects in each car examined. Let \(X\) denote the number of major defects in a randomly selected car of a certain type. The cdf of \(X\) is as follows: $$ F(x)= \begin{cases}0 & x<0 \\ .06 & 0 \leq x<1 \\ .19 & 1 \leq x<2 \\ .39 & 2 \leq x<3 \\ .67 & 3 \leq x<4 \\ .92 & 4 \leq x<5 \\ .97 & 5 \leq x<6 \\\ 1 & 6 \leq x\end{cases} $$ Calculate the following probabilities directly from the cdf: a. \(p(2)\), that is, \(P(X=2)\) b. \(P(X>3)\) c. \(P(2 \leq X \leq 5)\) d. \(P(2

a. Show that \(b(x ; n, 1-p)=b(n-x ; n, p)\). b. Show that \(B(x ; n, 1-p)=1-B(n-x-1 ; n, p)\). [Hint: At most \(x S\) 's is equivalent to at least \((n-x) F\) 's.] c. What do parts (a) and (b) imply about the necessity of including values of \(p\) greater than \(.5\) in Appendix Table A.1?

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