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Chapter 6: Problem 36

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
The mean is \$2200, and the variance is 1560000.

Step by step solution

01

Identify the Values and Probabilities

First, list the possible outcomes of the investment and their corresponding probabilities: - The investment could be worth \(\\(1,000\) with probability \(0.25\).- The investment could be worth \(\\)2,000\) with probability \(0.60\).- The investment could be worth \(\$5,000\) with probability \(0.15\).
02

Calculate the Expected Value (Mean)

The expected value (or mean) of a random variable is calculated by multiplying each possible value by its probability and summing all of these products:\[E(X) = \sum (x_i \cdot P(x_i)) = (1000 \times 0.25) + (2000 \times 0.60) + (5000 \times 0.15) \]Calculate each term: - \(1000 \times 0.25 = 250\)- \(2000 \times 0.60 = 1200\)- \(5000 \times 0.15 = 750\)Summing these gives the expected value:\[E(X) = 250 + 1200 + 750 = 2200\]
03

Calculate the Variance

To calculate the variance, use the formula \(\text{Var}(X) = \sum ((x_i - E(X))^2 \cdot P(x_i))\), where \(E(X) = 2200\) from Step 2. Compute each term individually:\[(1000 - 2200)^2 \cdot 0.25 = 1440000 \cdot 0.25 = 360000\]\[(2000 - 2200)^2 \cdot 0.60 = 40000 \cdot 0.60 = 24000\]\[(5000 - 2200)^2 \cdot 0.15 = 7840000 \cdot 0.15 = 1176000\]Finally, sum these to find the variance:\[\text{Var}(X) = 360000 + 24000 + 1176000 = 1560000\]

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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 called the mean or average, is a key concept in probability and investment statistics. It provides a single number that summarizes the center or "average" of a set of possible outcomes. To find the expected value of an investment, one must weigh each potential outcome by its probability of occurrence.

Here's how we do it:
  • Each possible investment outcome is multiplied by its probability.
  • All the products are summed to get the expected value.
For example, consider an investment with possible outcomes of $1,000, $2,000, and $5,000, having probabilities 0.25, 0.60, and 0.15, respectively. The expected value can be calculated as follows:
  • Multiply $1,000 by 0.25, which equals $250.
  • Multiply $2,000 by 0.60, which equals $1,200.
  • Multiply $5,000 by 0.15, which equals $750.
Adding these results gives $2,200, the expected value of the investment. This figure represents the average outcome if you could repeat the investment multiple times under the same conditions.
Variance Calculation
Variance is a measure of how much the outcomes of a random variable, like an investment, deviate from the expected value. It provides insights into the risk or volatility associated with the potential outcomes.

To calculate variance:
  • Subtract the expected value from each outcome to find the deviation from the mean.
  • Square each of these deviations (to eliminate negative values).
  • Multiply each squared deviation by the probability of the corresponding outcome.
  • Sum these values to get the variance.
In the investment case:
  • For $1,000, the deviation from the mean ($2,200) is -$1,200. Squaring gives $1,440,000, which when multiplied by the probability (0.25) gives $360,000.
  • For $2,000, the deviation is -$200. Squaring gives $40,000, and when multiplied by 0.60, gives $24,000.
  • For $5,000, the deviation is $2,800. Squaring gives $7,840,000, and when multiplied by 0.15, gives $1,176,000.
Adding these provides a total variance of $1,560,000. A higher variance indicates more spread out potential outcomes, hence greater risk.
Random Variables
Random variables are essential in probability and statistics, representing outcomes of random events or processes. They assign numerical values to each event in a sample space.

In the context of investments, random variables can represent the value of an investment at the end of a period. The randomness comes from various factors, such as market fluctuations and unforeseen events.

Characteristics of random variables:
  • They can be discrete or continuous. In our example, the investment outcomes are discrete, taking only specific values ($1,000, $2,000, $5,000).
  • Each possible value of a random variable is associated with a probability, representing its chance of occurring.
  • The probabilities of all possible values must sum up to 1.
Understanding random variables enables investors to model potential outcomes and make informed decisions based on probabilistic assessments of different scenarios.

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Most popular questions from this chapter

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports they can resolve customer problems the same day they are reported in 70 percent of the cases. Suppose the 15 cases reported today are representative of all complaints. a. How many of the problems would you expect to be resolved today? What is the standard deviation? b. What is the probability 10 of the problems can be resolved today? c. What is the probability 10 or 11 of the problems can be resolved today? d. What is the probability more than 10 of the problems can be resolved today?

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Dan Woodward is the owner and manager of Dan's Truck Stop. Dan offers free refills on all coffee orders. He gathered the following information on coffee refills. Compute the mean, variance, and standard deviation for the distribution of number of refills. $$\begin{array}{|cc|}\hline \text { Refills } & \text { Percent } \\\\\hline 0 & 30 \\\1 & 40 \\\2 & 20 \\\3 & 10 \\\\\hline\end{array}$$

Under what conditions will the binomial and the Poisson distributions give roughly the same results?

Steele Electronics, Inc. sells expensive brands of stereo equipment in several shopping malls throughout the northwest section of the United States. The Marketing Research Department of Steele reports that 30 percent of the customers entering the store that indicate they are browsing will, in the end, make a purchase. Let the last 20 customers who enter the store be a sample. a. How many of these customers would you expect to make a purchase? b. What is the probability that exactly five of these customers make a purchase? c. What is the probability ten or more make a purchase? d. Does it seem likely at least one will make a purchase?
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