Problem 34 - Basic Statistics for Business and Economics with Student

Question

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is \(\$ 110,000 .\) This distribution follows the normal distribution with a standard deviation of \(\$ 40,000 .\) a. If we select a random sample of 50 households, what is the standard error of the mean? b. What is the expected shape of the distribution of the sample mean? c. What is the likelihood of selecting a sample with a mean of at least \(\$ 112,000 ?\) d. What is the likelihood of selecting a sample with a mean of more than \(\$ 100,000 ?\) e. Find the likelihood of selecting a sample with a mean of more than \(\$ 100,000\) but less than \(\$ 112,000\).

Step-by-Step Solution

Answer

a) 5,656.85; b) Approximately normal; c) 0.363; d) 0.9615; e) 0.5985.

Step 1: Finding the Standard Error of the Mean

To find the standard error of the mean, use the formula: \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 40,000 \) is the standard deviation, and \( n = 50 \) is the sample size. Plug in the values:\[ SE = \frac{40,000}{\sqrt{50}} \approx 5,656.85 \]

Step 2: Determining the Shape of the Distribution of the Sample Mean

According to the Central Limit Theorem, the distribution of the sample mean will be approximately normal if the sample size is sufficiently large (\( n \geq 30 \) is typically enough). Since our sample size is 50, the distribution of the sample mean is approximately normal.

Step 3: Calculating the Probability for the Sample Mean of \( \$112,000 \) or More

To find this probability, first compute the z-score:\[ z = \frac{112,000 - 110,000}{5,656.85} \approx 0.353 \]Use the standard normal distribution table to find the probability corresponding to \( z = 0.353 \) which is approximately 0.637. Therefore, the probability of \( z \geq 0.353 \) is \( 1 - 0.637 = 0.363 \).

Step 4: Finding the Probability for the Sample Mean More Than \( \$100,000 \)

Calculate the z-score for \( \\(100,000 \):\[ z = \frac{100,000 - 110,000}{5,656.85} \approx -1.768 \]Consult the z-table for \( z = -1.768 \), which is approximately 0.0385. The probability of the sample mean being more than \( \\)100,000 \) is \( 1 - 0.0385 = 0.9615 \).

Step 5: Probability of More Than \( \$100,000 \) but Less Than \( \$112,000 \)

This is calculated by taking the probability of \( \\(112,000 \) or less and subtracting the probability of \( \\)100,000 \) or less:- Probability of \( \leq 112,000 \) is about 0.637 (from c).- Probability of \( > 100,000 \) calculated in d is approximately 0.9615.Therefore, the probability is \( 0.637 - (1 - 0.9615) \approx 0.637 - 0.0385 \approx 0.5985 \).

Key Concepts

Normal Distribution

The normal distribution is a fundamental concept in statistics. It's often referred to as the "bell curve" due to its characteristic shape. When data is distributed normally, it forms a symmetrical curve with most values clustering around the central peak, and the probabilities tapering off equally on both sides. The mean, median, and mode of a normally distributed dataset are all equal.

In the context of the exercise, we see that the amount of life insurance per household follows a normal distribution, with a mean of $110,000 and a standard deviation of $40,000. This means that most households have life insurance amounts close to $110,000, with fewer households having significantly more or less.

The normal distribution is essential for statistical analysis, particularly in the central limit theorem. This theorem explains that distributions of sample means will tend to be normal, even if the original data is not, as long as the sample size is large enough (generally, n ≥ 30).

Standard Error

The standard error (SE) is a measure of how much the sample mean is expected to fluctuate from the true population mean. It gives us an idea of the precision of our sample mean estimate. The formula to calculate SE is \[ SE = \frac{\sigma}{\sqrt{n}} \]where \( \sigma \) is the population standard deviation, and \( n \) is the sample size.

In the exercise, with a population standard deviation of \(40,000 and a sample size of 50, we calculated a standard error of approximately \)5,656.85. This value informs us that, on average, the sample mean of household insurance amounts will deviate by \(5,656.85 from the population mean of \)110,000.

Using a smaller standard error implies more reliability in your sample mean, assuming it's an accurate reflection of the population mean.

Z-Score

A z-score, also known as a standard score, shows how many standard deviations an element is from the mean. It's a way to compare points from different normal distributions or to find probabilities using standard normal distribution tables. The formula for a z-score is:\[ z = \frac{X - \mu}{SE} \]where \( X \) is the value from the sample, \( \mu \) is the population mean, and \( SE \) is the standard error.

In our scenario, to calculate the probability of a sample mean of at least \(112,000, we found a z-score of approximately 0.353. This means \)112,000 is 0.353 standard errors above the mean of $110,000. Z-scores allow us to understand how unusual a particular sample mean is by comparing it to the standard normal distribution.

Understanding z-scores helps in determining the likelihood of certain values and forming conclusions regarding data observations.

Probability Calculation

Probability calculation is a crucial skill to interpret statistical results. After determining z-scores, it's possible to find probabilities from standard normal distribution tables, which helps assess the likelihood of different outcomes.

In the exercise, we looked at different probabilities:

• For the likelihood of a sample mean of at least $112,000, we found a probability of 0.363.
• The probability of a sample mean being more than $100,000 was 0.9615.
• For a sample mean more than $100,000 but less than $112,000, the probability calculated was about 0.5985.

Understanding probability calculations help predict outcomes and make informed decisions based on statistical data.

These calculations rely heavily on knowing how to look up z-scores in a standard normal distribution table and interpreting those values in the context of real-world data.