/*! 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 36 An investment will be worth \(\$... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An investment will be worth \(\$ 1,000, \$ 2,000,\) or \(\$ 5,000\) at the end of the year. The probabilities of these values are \(.25, .60,\) and \(.15,\) respectively. Determine the mean and variance of the worth of the investment.

Short Answer

Expert verified
Mean: 2200, Variance: 1236000

Step by step solution

01

Understand the Problem

We are given three potential future values of an investment, each with its own probability. Our task is to find both the mean (expected value) and variance of these possible outcomes.
02

Identify the Random Variable

Let's denote the random variable representing the worth of the investment as \( X \). The possible values of \( X \) are \( 1000, 2000, \) and \( 5000 \), with probabilities \( 0.25, 0.60, \) and \( 0.15 \) respectively.
03

Calculate the Mean (Expected Value)

The mean or expected value is calculated using the formula \( E(X) = \sum_{i} (x_i \times P(x_i)) \). Compute the expected value: \[ E(X) = 1000 \times 0.25 + 2000 \times 0.60 + 5000 \times 0.15 \].Calculating each part: \( 1000 \times 0.25 = 250 \), \( 2000 \times 0.60 = 1200 \), \( 5000 \times 0.15 = 750 \).Now, sum them up: \( 250 + 1200 + 750 = 2200 \). So, \( E(X) = 2200 \).
04

Calculate Each Value's Contribution to Variance

The contribution to variance for each outcome is calculated using \( (x_i - E(X))^2 \times P(x_i) \). Calculate each part:For \( x_1 = 1000 \): \[ (1000 - 2200)^2 \times 0.25 = 144000 \times 0.25 = 36000 \].For \( x_2 = 2000 \): \[ (2000 - 2200)^2 \times 0.60 = 40000 \times 0.60 = 24000 \].For \( x_3 = 5000 \): \[ (5000 - 2200)^2 \times 0.15 = 7840000 \times 0.15 = 1176000 \].
05

Compute the Variance

Sum up all the values calculated in the previous step:\[ 36000 + 24000 + 1176000 = 1236000 \].Therefore, the variance of the investment's worth is \( 1236000 \).

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 Variable
A random variable is like a bucket that holds all possible outcomes of a certain event or process. In our example of the investment, the random variable is represented by the symbol \( X \). This random variable accounts for the possible future values of the investment at the end of the year, which can be worth \( \\( 1,000 \), \( \\) 2,000 \), or \( \$ 5,000 \). Each of these outcomes has a probability assigned to it, indicating how likely each scenario is to happen.

To put it simply, a random variable helps us to describe and organize the potential numerical outcomes of a random phenomenon. In this investment scenario, it structures our thinking so we can calculate mean and variance effectively. Without defining \( X \) as our random variable, we would lack a systematic way to evaluate the investment's possible future values.
Expected Value
Expected value, or mean, is akin to finding the average of all potential outcomes, considering their probabilities. In probability theory, it provides a summary of possible values a random variable can take. It's similar to asking "What's the average outcome we could expect based on probabilities?"

To calculate it, we multiply each possible outcome by its probability and sum the results. For the investment example, these calculations would be:
  • \( 1000 \times 0.25 \) for the \( \\( 1,000 \) outcome
  • \( 2000 \times 0.60 \) for the \( \\) 2,000 \) outcome
  • \( 5000 \times 0.15 \) for the \( \\( 5,000 \) outcome
By summing these values, we found the expected value to be \( \\) 2200 \). It indicates that, on average, we expect the investment to be worth \( \\( 2200 \). However, remember this doesn't mean the investment will actually be worth \( \\) 2200 \); rather, it's a theoretical average over many similar scenarios.
Probability Distribution
A probability distribution is a crucial concept that outlines how probabilities are distributed over different possible outcomes for a random variable. For this investment case, it shows exactly how likely it is to end up with \( \\( 1,000 \), \( \\) 2,000 \), or \( \\( 5,000 \) at the year's end.

The probabilities assigned were as follows:
  • \( \\) 1,000 \) with a probability of \( 0.25 \)
  • \( \\( 2,000 \) with a probability of \( 0.60 \)
  • \( \\) 5,000 \) with a probability of \( 0.15 \)
These probabilities must add up to \( 1 \), making them valid. This distribution helps us to understand how likely each outcome is and is the foundation for our mean and variance calculations.

By organizing probabilities this way, we can make informed decisions and analyses, such as calculating expected values or assessing risk, using the likelihood of each possible outcome. It’s like drawing a clear map of potential future scenarios with their respective chances.

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

Dr. Richmond, a psychologist, is studying the daytime television viewing habits of college students. She believes 45 percent of college students watch soap operas during the afternoon. To further investigate, she selects a sample of \(10 .\) a. Develop a probability distribution for the number of students in the sample who watch soap operas. b. Find the mean and the standard deviation of this distribution. c. What is the probability of finding exactly four watch soap operas? d. What is the probability less than half of the students selected watch soap operas?

A recent study conducted by Penn, Shone, and Borland, on behalf of LastMinute.com, revealed that 52 percent of business travelers plan their trips less than two weeks before departure. The study is to be replicated in the tri-state area with a sample of 12 frequent business travelers. a. Develop a probability distribution for the number of travelers who plan their trips within two weeks of departure. b. Find the mean and the standard deviation of this distribution. c. What is the probability exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure? d. What is the probability 5 or fewer of the 12 selected business travelers plan their trips within two weeks of departure?

In a Poisson distribution \(\mu=4\). a. What is the probability that \(x=2 ?\) b. What is the probability that \(x \leq 2 ?\) c. What is the probability that \(x>2 ?\)

A federal study reported that 7.5 percent of the U.S. workforce has a drug problem. A drug enforcement official for the State of Indiana wished to investigate this statement. In her sample of 20 employed workers: a. How many would you expect to have a drug problem? What is the standard deviation? b. What is the likelihood that none of the workers sampled has a drug problem?

Can you tell the difference between Coke and Pepsi in a blind taste test? Most people say they can and have a preference for one brand or the other. However, research suggests that people can correctly identify a sample of one of these products only about 60 percent of the time. Suppose we decide to investigate this question and select a sample of 15 college students. a. How many of the 15 students would you expect to correctly identify Coke or Pepsi? b. What is the probability exactly 10 of the students surveyed will correctly identify Coke or Pepsi? C. What is the probability at least 10 of the students will correctly identify Coke or Pepsi?

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.