International Journal of Statistical Distributions (Vol. 1, 2015) study of a variable life insurance policy, Exercise 4.97 (p. 239). Recall that a ratio (x) of the rates of return on the investment for two consecutive years was shown to have a normal distribution, with μ = 1.5 and o= .2. Consider a random sample of 100 variable life insurance policies and let represent the mean ratio for the sample. a. Find E(T) and interpret its value.

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## Statistical Analysis of Variable Life Insurance Return Rates

### Study Overview

This exercise references a study from the *International Journal of Statistical Distributions* (Vol. 1, 2015) which focuses on variable life insurance policies. The task is derived from Exercise 4.97 (p. 239).

### Problem Summary

We are examining the ratio (\(x\)) of the rates of return on investments for two consecutive years. Previous studies indicate that this ratio follows a normal distribution with:

- Mean (\(\mu\)) = 1.5
- Standard deviation (\(\sigma\)) = 0.2

A random sample of 100 variable life insurance policies is considered, with \(\bar{x}\) representing the mean ratio for the sample.

### Tasks

a. **Find \(E(\bar{x})\) and Interpret**

b. **Find Var(\(\bar{x}\))**

c. **Describe the Shape of the Sampling Distribution of \(\bar{x}\)**

d. **Calculate the z-score for \(\bar{x} = 1.52\)**

e. **Determine \(P(\bar{x} > 1.52)\)**

f. **Assess Impact of Non-Normal Distribution on Previous Answers**

### Problem Analysis

1. **Expectation of \(\bar{x}\) (\(E(\bar{x})\)):**
   - Since the sample mean is an unbiased estimator of the population mean, \(E(\bar{x}) = \mu = 1.5\).
   
2. **Variance of \(\bar{x}\) (Var(\(\bar{x}\)):**
   - The variance of the sample mean is given by Var(\(\bar{x}\)) = \(\sigma^2/n\) = \(0.2^2/100 = 0.0004\).
   
3. **Shape of the Sampling Distribution:**
   - The Central Limit Theorem states that for a sufficiently large sample size, the sampling distribution of the sample mean \(\bar{x}\) will be approximately normal. Given the sample size of 100, the distribution of \(\bar{x}\) is normal.

4. **Z-score Calculation:**
   - Z = \((1.52 - 1.5) / (0.2/\sqrt{100}) = 1\).

5. **Probability \(P(\bar{x} > 1.52)\):**
Transcribed Image Text:## Statistical Analysis of Variable Life Insurance Return Rates ### Study Overview This exercise references a study from the *International Journal of Statistical Distributions* (Vol. 1, 2015) which focuses on variable life insurance policies. The task is derived from Exercise 4.97 (p. 239). ### Problem Summary We are examining the ratio (\(x\)) of the rates of return on investments for two consecutive years. Previous studies indicate that this ratio follows a normal distribution with: - Mean (\(\mu\)) = 1.5 - Standard deviation (\(\sigma\)) = 0.2 A random sample of 100 variable life insurance policies is considered, with \(\bar{x}\) representing the mean ratio for the sample. ### Tasks a. **Find \(E(\bar{x})\) and Interpret** b. **Find Var(\(\bar{x}\))** c. **Describe the Shape of the Sampling Distribution of \(\bar{x}\)** d. **Calculate the z-score for \(\bar{x} = 1.52\)** e. **Determine \(P(\bar{x} > 1.52)\)** f. **Assess Impact of Non-Normal Distribution on Previous Answers** ### Problem Analysis 1. **Expectation of \(\bar{x}\) (\(E(\bar{x})\)):** - Since the sample mean is an unbiased estimator of the population mean, \(E(\bar{x}) = \mu = 1.5\). 2. **Variance of \(\bar{x}\) (Var(\(\bar{x}\)):** - The variance of the sample mean is given by Var(\(\bar{x}\)) = \(\sigma^2/n\) = \(0.2^2/100 = 0.0004\). 3. **Shape of the Sampling Distribution:** - The Central Limit Theorem states that for a sufficiently large sample size, the sampling distribution of the sample mean \(\bar{x}\) will be approximately normal. Given the sample size of 100, the distribution of \(\bar{x}\) is normal. 4. **Z-score Calculation:** - Z = \((1.52 - 1.5) / (0.2/\sqrt{100}) = 1\). 5. **Probability \(P(\bar{x} > 1.52)\):**
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