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Chapter 6: Problem 58

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 If four 21 -year-old men are insured, the insurer's average income is $$V=\frac{X_{1}+X_{2}+X_{3}+X_{4}}{4}=0.25 X_{1}+0.25 X_{2}+0.25 X_{3}+0.25 X_{4}$$ where \(X_{i}\) is the income from insuring one man. Assuming that the amount of income earned on individual policies is independent, find the mean and standard deviation of \(V\) . (If you compare with the results of Exercise 57 , you should see that averaging over more insured individuals reduces risk.)

Short Answer

Expert verified
Mean = $303.35, Standard Deviation = $4853.79

Step by step solution

01

Understand the Variables

You are given the random variable \(X\), which represents the amount a life insurance company earns from a single life insurance policy. You need to find the mean and standard deviation of the average income \(V\), which is composed of four independent variables: \(X_1, X_2, X_3,\) and \(X_4\).
02

Determine the Mean of V

The mean of a random variable \(V\) which is a weighted sum of random variables is given by:\[\mu_V = 0.25 \mu_{X_1} + 0.25 \mu_{X_2} + 0.25 \mu_{X_3} + 0.25 \mu_{X_4}\]Since \(X_1, X_2, X_3,\) and \(X_4\) are identical and have the same mean, \(\mu_{X_1} = \mu_{X_2} = \mu_{X_3} = \mu_{X_4} = \mu_X\).Therefore,\[\mu_V = 0.25 \cdot 4 \cdot \mu_X = \mu_X = 303.35\]
03

Determine the Standard Deviation of V

The standard deviation of a random variable \(V\) which is a weighted sum of independent random variables is given by:\[\sigma_V = \sqrt{(0.25^2 \sigma_{X_1}^2 + 0.25^2 \sigma_{X_2}^2 + 0.25^2 \sigma_{X_3}^2 + 0.25^2 \sigma_{X_4}^2)}\]Simplifying gives:\[\sigma_V = \sqrt{(4 \cdot 0.25^2 \cdot \sigma_X^2)} = \sqrt{(0.25 \cdot \sigma_X^2)} = \frac{\sigma_X}{2} = \frac{9707.57}{2} = 4853.785\]
04

Conclusion

The mean income \(\mu_V\) from insuring four men is \(303.35\), and the standard deviation \(\sigma_V\) is \(4853.785\). This indicates a reduction in risk compared to a single policy.

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Key Concepts

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

Mean and Standard Deviation
When we talk about mean and standard deviation in the context of probability distributions, we are really discussing two key concepts that help us understand and interpret data. The mean, often represented by the symbol \( \mu \), is the average value of a set of data, giving us an idea of what to expect from any individual data point in a random variable over time.
For the exercise, the mean of the random variable \( V \), which represents the average income the insurer earns from four policies, is calculated by averaging the means of each single policy. Since each policy (\( X_1 \), \( X_2 \), \( X_3 \), \( X_4 \)) has the same mean \( \mu_X = 303.35 \), this means the average remains \( 303.35 \) when distributing the risk over four policies.
Standard deviation, symbolized as \( \sigma \), measures how spread out the numbers are around the mean. A smaller standard deviation means data points tend to be closer to the mean, reducing variability. In this exercise, calculating \( \sigma_V \) reveals how grouping multiple policies lowers risk, since \( \sigma_V = 4853.785 \), which is half the standard deviation of a single policy. This shows us that embarking on more policies results in less risk per individual policy.
Random Variables
A random variable is essentially a numerical description of the outcomes of a random phenomenon. In probability, a random variable has different possible values, each with an associated probability.
In this problem, the random variable \( X \) represents the income from one life insurance policy. It can have several possible outcomes depending on various real-life factors like the policyholder's health and life circumstances. The mean and standard deviation provide key insights into the expected value and the expected variability of these outcomes.
In calculating \( V \), we treat \( X_1, X_2, X_3, \) and \( X_4 \) as independent random variables, each representing the income from four distinct policies. By considering \( V \) as a random variable itself, we can determine the overall expected income, which helps the insurance company in assessing potential earnings and risks.
Independent Events
In probability, independent events are those whose outcomes do not affect each other. This means that the occurrence of one event does not influence the likelihood of the other.
For this exercise, the income from the four life insurance policies \( X_1, X_2, X_3, \) and \( X_4 \) are assumed to be independent. This assumption simplifies the process of calculating the mean and standard deviation of \( V \), because it allows us to use the properties of independent random variables.
When dealing with independent random variables, their variances can be added directly. This property is key in understanding why the standard deviation \( \sigma_V \) is lower for the average income of four policies compared to a single policy; the variance of \( V \) is the sum of the variances of each \( X_i \) divided by the number of policies squared, thus reducing the overall risk by distributing it across multiple independent events.

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