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Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is \(\$ 165,000 .\) 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 \(\$ 167,000 ?\) d. What is the likelihood of selecting a sample with a mean of more than \(\$ 155,000 ?\) e. Find the likelihood of selecting a sample with a mean of more than \(\$ 155,000\) but less than \(\$ 167,000\).

Short Answer

Expert verified
a. 5656.85; b. Approximately normal; c. 0.3618; d. 0.9615; e. 0.5997.

Step by step solution

01

Compute the Standard Error of the Mean

The standard error of the mean (SEM) is calculated using the formula \( \text{SEM} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation of the population and \( n \) is the sample size. Here, \( \sigma = 40,000 \) and \( n = 50 \).So, \( \text{SEM} = \frac{40,000}{\sqrt{50}} \approx 5,656.85 \).
02

Determine the Shape of the Sample Mean Distribution

According to the Central Limit Theorem, the distribution of the sample mean will be approximately normally distributed if the sample size is large enough (typically \( n > 30 \) is considered sufficient). Since we have \( n = 50 \), the sample mean distribution will be approximately normal.
03

Calculate the Probability of the Sample Mean at Least $167,000

To find this probability, compute the Z-score for \( \$167,000 \) using the formula \( Z = \frac{\bar{x} - \mu}{\text{SEM}} \), where \( \bar{x} = 167,000 \) and \( \mu = 165,000 \). So, \( Z = \frac{167,000 - 165,000}{5,656.85} \approx 0.3536 \). Using the Z-table, find the probability \( P(Z > 0.3536) \), which is \( 1 - 0.6382 = 0.3618 \).
04

Calculate the Probability of the Sample Mean More Than $155,000

Again, compute the Z-score for \( \$155,000 \):\( Z = \frac{155,000 - 165,000}{5,656.85} \approx -1.7678 \).Now, find \( P(Z > -1.7678) \), where from the Z-table, \( P(Z < -1.7678) \approx 0.0385 \), so \( P(Z > -1.7678) = 1 - 0.0385 = 0.9615 \).
05

Compute the Probability of Sample Mean Between $155,000 and $167,000

Here, calculate the probability using the results from Steps 3 and 4.The probability \( P(155,000 < \bar{x} < 167,000) \) is given by the difference between the probabilities calculated:\( 0.9615 - 0.3618 = 0.5997 \).

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Key Concepts

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

Standard Error
The standard error (SE) is a crucial component when dealing with sample data. It provides an estimate of how much the sample mean is expected to vary from the actual population mean, given the size of the sample.

To calculate the standard error of the mean (SEM), you use the formula:

\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \]

Here, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. When you have a larger sample, the SEM decreases, indicating your sample mean is likely to be closer to the actual population mean.

In the exercise provided, with a population standard deviation \(\sigma = 40,000\) and a sample size \(n = 50\), the standard error is calculated as:

\[ \text{SEM} = \frac{40,000}{\sqrt{50}} \approx 5,656.85 \]

This means that the sample means will typically vary by about \(5,656.85\) from the population mean.
Normal Distribution
Normal distribution is a foundational concept in statistics, often referred to as the "bell curve" due to its shape. It describes how values of a variable are distributed.

Key characteristics of a normal distribution include:
  • Symmetry around the mean: This means the left and right sides of the graph are mirror images.
  • Mean, median, and mode all coincide at the center.
  • A predictable proportion of data falls within certain standard deviation ranges—about 68% within one, 95% within two, and 99.7% within three standard deviations.


In the context of sample means, according to the Central Limit Theorem, if your sample size is sufficiently large (usually considered more than 30), the distribution of the sample mean will approximately follow a normal distribution, regardless of the shape of the population distribution.

With a sample size of \(n = 50\), as in this example, we expect the sample mean to follow a normal distribution, centered around the population mean of \(165,000\).
Z-score
A Z-score is a measure of how many standard deviations an element is from the mean. It is instrumental in determining probabilities in a normal distribution.

The formula to calculate a Z-score is:

\[ Z = \frac{\bar{x} - \mu}{\text{SEM}} \]

Where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, and \( \text{SEM} \) is the standard error of the mean.

In practical applications:
  • A Z-score tells you exactly how much and in what direction, your data point deviates from the mean.
  • Positive Z-scores indicate values above the mean, while negative scores indicate below.


Using a Z-score, you can look up probabilities in a Z-table that correspond to the area under the normal distribution curve—helping you understand the likelihood of observing a value at least as extreme as the one in your study.

For example, with the sample mean \(167,000\), the Z-score calculation would be:

\[ Z = \frac{167,000 - 165,000}{5,656.85} \approx 0.3536 \]

This indicates that \(167,000\) is about 0.3536 standard deviations above the mean of \(165,000\).
Probability
Probability is fundamentally about assessing the likelihood of an event occurring. In statistics, we often calculate the probability of certain outcomes based on a normal distribution.

Using Z-scores, we can determine these probabilities. Here’s how we relate Z-scores to probabilities:
  • The Z-score points to a specific spot on the normal distribution curve.
  • This location correlates to the probability of the data point being more or less extreme.
  • Z-tables or statistical software can provide these probabilities directly.


Let's consider two scenarios:

1. **Probability of the sample mean being at least \(167,000\):** Using the Z-score of about 0.3536, locate it on a Z-table to find the corresponding probability. The area to the right of this Z-score gives the probability of seeing a sample mean that large or larger.

2. **Probability of the sample mean being more than \(155,000\):** Calculate its Z-score, find the corresponding probability, and focus on the tail of the curve that represents values greater than this mean.

This probability provides insight into how often a result may naturally occur, informing decisions and hypothesis testing.

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Most popular questions from this chapter

Over the past decade, the mean number of hacking attacks experienced by members of the Information Systems Security Association is 510 per year with a standard deviation of 14.28 attacks. The number of attacks per year is normally distributed. Suppose nothing in this environment changes. a. What is the likelihood this group will suffer an average of more than 600 attacks in the next 10 years? b. Compute the probability the mean number of attacks over the next 10 years is between 500 and 600 . c. What is the possibility they will experience an average of less than 500 attacks over the next 10 years?

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