/*! 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 57 Exercises 57 and 58 refer to the... [FREE SOLUTION] | 91Ó°ÊÓ

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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 randomly chosen 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 \(\bar{X}_{2}\) are the insurer's income from the two insurance policies, the insurer's average income \(W\) on the two policies is $$ W=\frac{X_{1}+X_{2}}{2} $$ 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.)

Short Answer

Expert verified
Mean of \(W\) is \$303.35; Standard deviation of \(W\) is \$6861.97.

Step by step solution

01

Determine the Mean of W

The random variable \(W\) represents the average income from two insurance policies, defined by \(W = \frac{X_1 + X_2}{2}\) where \(X_1\) and \(X_2\) are independent random variables each representing the income from a single policy. The expected value of \(W\) can be determined from the linearity of expectation: \[E(W) = E\left(\frac{X_1 + X_2}{2}\right) = \frac{1}{2}E(X_1) + \frac{1}{2}E(X_2)\]. Because \(E(X_1) = \mu_X\) and \(E(X_2) = \mu_X\), this simplifies to: \[E(W) = \frac{1}{2}(\mu_X + \mu_X) = \mu_X = \$303.35\].
02

Determine the Variance of W

The variance of \(W\) can be computed using the formula for the variance of the sum of independent random variables. Since \(W = \frac{X_1 + X_2}{2}\), we have: \[\text{Var}(W) = \text{Var}\left( \frac{X_1 + X_2}{2} \right) = \left(\frac{1}{2}\right)^2 [\text{Var}(X_1) + \text{Var}(X_2)]\]. Given that \(\text{Var}(X_1) = \sigma_X^2\) and \(\text{Var}(X_2) = \sigma_X^2\), we find: \[\text{Var}(W) = \frac{1}{4}(\sigma_X^2 + \sigma_X^2) = \frac{1}{4}(2\sigma_X^2) = \frac{1}{2}\sigma_X^2 \].
03

Determine the Standard Deviation of W

To find the standard deviation of \(W\), take the square root of the variance of \(W\): \[\sigma_W = \sqrt{\text{Var}(W)} = \sqrt{\frac{1}{2} \sigma_X^2} = \frac{\sigma_X}{\sqrt{2}}\]. Using the given standard deviation \(\sigma_X = \\(9707.57\), we calculate: \[\sigma_W = \frac{9707.57}{\sqrt{2}} \approx \\)6861.97\]. This means that \(W\) has a smaller standard deviation than \(X\), which is consistent with the central limit theorem for large samples.

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Key Concepts

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

Random Variable
In the realm of probability and statistics, a **Random Variable** is a fundamental concept. It is essentially a function that assigns a numerical value to each possible outcome of a random event. Imagine rolling a die: each face has an equal chance of appearing, and we can associate a number (1 through 6) with each outcome, turning our die into a random variable.

There are two types of random variables: continuous and discrete. Discrete random variables have a countable number of possibilities, like the roll of a die or the flip of a coin. On the other hand, continuous random variables can take on any value within a given range, such as temperatures or heights. For insurance policies, the random variable represents income, which can be considered continuous because it can take any value within a certain financial range.

When dealing with problems involving random variables, as in the case of insurance income analysis, it is crucial to understand their distributions. These distributions help us ascertain probabilities and perform analyses such as finding expected values and variances, critical for risk assessment and decision-making.
Mean and Standard Deviation
The **Mean** of a random variable is a measure of its central tendency. It is often denoted as \( \mu \, and it signifies the 'average' result if you could run the same process an infinite number of times. For our insurance example, where we find the mean income from two policies, the calculation demonstrates linearity of expectation: \[ E(W) = \frac{1}{2}(\mu_X + \mu_X) = \mu_X = \\(303.35 \]. This result showcases that regardless of how many independent terms we average, the mean remains unchanged at \(\\)303.35\).

**Standard Deviation**, denoted as \( \sigma \, measures the amount of variation or dispersion from the mean. A low standard deviation means that data points tend to be close to the mean, while a high standard deviation indicates that data points are spread out over a wider range of values. In our scenario, the calculation of the standard deviation for two policies, \(\frac{9707.57}{\sqrt{2}} \approx 6861.97\), confirms that spreading risk over multiple policies reduces volatility, a key insight from the statistical principle of variance reduction.
  • Mean (\mu\): Average of all possible outcomes.
  • Standard Deviation (\sigma\): Measure of the spread of those outcomes.
Central Limit Theorem
The **Central Limit Theorem (CLT)** is one of the cornerstones of probability theory and statistics. It postulates that the distribution of sample means approximates a normal distribution (Gaussian) as the sample size becomes large, regardless of the original distribution shape.

This theorem is paramount when dealing with the mean of large numbers of independent random variables. It allows statisticians and mathematicians to make inferences about population parameters even when only sample data are available. In our insurance example, the mean income from two policies has reduced variance, showcasing a small-scale application of the CLT.

The beauty of the CLT lies in its applicability, providing a reliable path from sample data to understanding population parameters. Even with a relatively small number of policies like two, the impact of CLT can be seen as the variability diminishes, underscoring the strength of aggregating independent observations in practice.

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Most popular questions from this chapter

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