Problem 36 - Basic Statistics for Business and Economics with Student

Question

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.

Step-by-Step Solution

Answer

Mean: 2200, Variance: 1236000

Step 1: 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.

Step 2: 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.

Step 3: 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, \( E(X) = 2200 \).

Step 4: 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 \( x_3 = 5000 \): \[ (5000 - 2200)^2 \times 0.15 = 7840000 \times 0.15 = 1176000 \].

Step 5: 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 \( 1236000 \).

Key Concepts

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.