91Ó°ÊÓ

Exercises 57 and 58 refer to the following setting. In Exercises 14 and 18 of Section \(6.1,\) we examined the probability distribution of the random variable \(X=\) the amount a life insurance company earns on a 5 -year term life policy. Calculations reveal that \(\mu_{X}=\$ 303.35\) and \(\sigma_{X}=\$ 9707.57\) Life insurance The risk of insuring one person's life is reduced if we insure many people. Suppose that we insure two 21 -year-old males, and that their ages at death are independent. If \(X_{1}\) and \(X_{2}\) are the insurer's income from the two insurance policies, the insurer's average income \(W\) on the two policies is $$W=\frac{X_{1}+X_{2}}{2}=0.5 X_{1}+0.5 X_{2}$$ Find the mean and standard deviation of W. (You see that the mean income is the same as for a single policy but the standard deviation is less.)

Step by step solution

01

Identify Mean of Average Income

The average income from two independent policies is calculated as \( W = 0.5X_1 + 0.5X_2 \). By linearity of expectation, the mean of the average income is the same as the mean of each individual policy. Hence, \( \mu_W = 0.5 \mu_{X_1} + 0.5 \mu_{X_2} = \mu_X = \$303.35 \).

02

Calculate Variance of Average Income

Since \( X_1 \) and \( X_2 \) are independent, the variance of \( W \) is calculated using the formula \( \sigma_W^2 = (0.5)^2 \sigma_{X_1}^2 + (0.5)^2 \sigma_{X_2}^2 \). Thus, \( \sigma_W^2 = 0.25 \sigma_X^2 + 0.25 \sigma_X^2 = 0.5 \sigma_X^2 \).

03

Calculate Standard Deviation of Average Income

With the variance of the average income calculated, the standard deviation is obtained by taking the square root: \( \sigma_W = \sqrt{0.5} \times \sigma_X = \sqrt{0.5} \times 9707.57 \approx 6863.14 \).

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Key Concepts

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

Random Variable

A random variable is a fundamental concept in statistics and probability, representing a numerical outcome of a random phenomenon. In other words, it assigns numbers to each outcome in a sample space. There are two types of random variables: discrete and continuous. - **Discrete random variables** take on a finite or countable number of values. For example, the roll of a die can take on any of the discrete values 1 through 6.- **Continuous random variables** can take on an infinite number of values within a given range. An example would be the height of students in a class.In the context of our exercise, the random variable is the income, denoted by \( X \), which the insurance company earns from a life policy. Since this income can vary, it is considered a random variable.

Standard Deviation

Standard deviation is a measure that quantifies the amount of variation or dispersion of a set of data points. In a probability distribution, it describes how much the values of the random variable deviate from the mean. A small standard deviation indicates that data points are close to the mean, while a large standard deviation signifies that they are spread out over a wider range of values.In the problem scenario, we found the standard deviation of the insurance company's income \( W \) from insuring two individuals. It shows us how much the average income from these policies is expected to fluctuate. The calculation, which involves taking the square root of the variance, results in \( \sigma_W \approx 6863.14 \). This figure means that the actual average income could reasonably vary by about $6863.14 around the mean.

Mean Income

Mean income, often referred to as the expected value, is a crucial statistic that provides an average of a probability distribution. It is the sum of all possible values each weighted by their probability of occurrence.In our exercise, the mean income \( \mu_W \) from the two life insurance policies is calculated using the linearity of expectation. The mean of the combined income from two policies equals the mean income from a single policy, \( \mu_X = 303.35 \). This shows that when insuring many people, even though the income from each individual policy might result in different values, the average is consistent over many policies.

Independent Events

In probability theory, two events are independent if the occurrence of one does not affect the occurrence of the other. When dealing with random variables, independence can significantly simplify calculations, particularly when calculating variance and expected value.In our scenario, the ages at death of the two insured individuals are independent from each other. This independence allowed us to use certain properties, such as the linear combination of random variables, to determine the variance of the income \( W \) from the insurance policies. Independence in this context ensures that the insurance company can predict overall outcomes more reliably by averaging the income from multiple independent policies, reducing risk through diversification.