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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 \(\$ 110,000 .\) This distribution is positively skewed. The standard deviation of the population is not known. a. A random sample of 50 households revealed a mean of \(\$ 112,000\) and a standard deviation of \(\$ 40,000 .\) What is the standard error of the mean? b. Suppose that you selected 50 samples of households. 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 \(\$ 112,000\)

Short Answer

Expert verified
a) SE = 5656.85; b) Approximately normal; c) \( P(z \geq 0.3536) \); d) \( P(z > -1.7678) \); e) Difference of both probabilities.

Step by step solution

01

Calculate the Standard Error of the Mean

To calculate the standard error of the mean, we use the formula \( SE = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation and \( n \) is the sample size. Here, \( s = 40000 \) and \( n = 50 \). Thus, \( SE = \frac{40000}{\sqrt{50}} \).
02

Compute Value of Standard Error

Calculate the standard error: \( SE = \frac{40000}{\sqrt{50}} \approx 5656.85 \).
03

Understanding the Distribution Shape

According to the Central Limit Theorem, the distribution of the sample mean will be approximately normal regardless of the shape of the population distribution if the sample size is sufficiently large (typically \( n \geq 30 \)). Here, the sample size is 50.
04

Find Likelihood for Mean \( \geq \$112,000 \)

To find this probability, we first calculate the z-score using \( z = \frac{\bar{x} - \mu}{SE} \), where \( \bar{x} = 112,000 \) and \( \mu = 110,000 \). So, \( z = \frac{112000 - 110000}{5656.85} \approx 0.3536 \). Using standard normal distribution tables or a calculator, find \( P(z \geq 0.3536) \).
05

Compute Likelihood of Mean \( > \$100,000 \)

Similarly, calculate the z-score for \( \bar{x} = 100,000 \) with \( z = \frac{100000 - 110000}{5656.85} \approx -1.7678 \). Thus, \( P(z > -1.7678) \) is found using z-tables or a calculator.
06

Calculate the Probability Between Two Means

To find the likelihood of a mean between \( 100,000 \) and \( 112,000 \), subtract the cumulative probability at \( z = -1.7678 \) from that at \( z = 0.3536 \), corresponding to the earlier calculations.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics that makes it easier to understand the behavior of the mean of a sample. It indicates that, irrespective of the original distribution of the population, the distribution of the sample means will approach a normal distribution as long as the sample size is sufficiently large. Typically, a sample size of 30 or more is considered adequate.

This is crucial because it allows for the use of z-scores and other statistical measures, which are mostly based on normal distributions, even when the population distribution itself is not normal. This theorem gives us the flexibility to make probabilistic estimates about the population mean based on sample data.

In our exercise, we're examining the average life insurance in U.S. households. Although the distribution was described as positively skewed, taking a sample of 50 households is enough for the Central Limit Theorem to be applied, allowing the sample mean to approximate a normal distribution.
z-score calculation
A z-score is a statistical measurement that describes a value's relation to the mean of a group of values, measured in terms of standard deviations. A z-score can report whether a value is above or below the mean and by how many standard deviations.

The formula for calculating the z-score is:
  • \[z = \frac{\bar{x} - \mu}{SE}\]
Where:
  • \( \bar{x} \) = sample mean
  • \( \mu \) = population mean
  • \( SE \) = standard error of the mean
In our step-by-step solution, to find the probability of selecting a sample with a particular mean, we first calculated the z-score. For instance, for a sample mean of \( \\(112,000 \) with a population mean of \( \\)110,000 \) and a standard error of approximately \( 5656.85 \), the z-score is calculated as \( \approx 0.3536 \). This calculation tells us how many standard deviations away our sample mean is from the population mean.
sampling distribution
Sampling distribution refers to the probability distribution of a statistic (like a mean) obtained from a large number of samples drawn from a specific population. By understanding this concept, statisticians can make inferences about the population based on sample data.

The distribution of sample means will center around the population mean, and the spread will be determined by the standard error of the mean. With the Central Limit Theorem, we understand that the shape of the sampling distribution of the mean will be approximately normal for large sample sizes. This allows us to use techniques like z-score calculations to find probabilities related to sample means.

In exercises like ours, we often use the central concept that the sampling distribution of the sample mean becomes more normal as sample size increases, even if the population distribution is skewed. This pivotal understanding permits us to make educated guesses about the population mean.
probability calculation
Calculating probabilities using a normal distribution is a standard task in statistics. After calculating a z-score, it can be translated to a probability. This essentially tells you how likely a certain sample mean is to occur, given the population statistics.

For example, in our exercise, we wanted to determine the probability of selecting a sample with a mean of at least \( \\(112,000 \). After calculating the z-score (\( 0.3536 \)), we refer to a standard normal distribution table to find this value's corresponding probability.

For z-score calculations such as z = 0.3536, we discover the probability of the z-value being greater than the calculated value. Similarly, probabilities are calculated for mean values of over \( \\)100,000 \) and for those values falling between \( \\(100,000 \) and \( \\)112,000 \). This enables us to make informed decisions and interpretations about data behavior in real-life scenarios.

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Most popular questions from this chapter

A population consists of the following four values: \(12,12,14,\) and 16 . a. List all samples of size \(2,\) and compute the mean of each sample. b. Compute the mean of the distribution of the sample mean and the population mean. Compare the two values. c. Compare the dispersion in the population with that of the sample mean.

A population of unknown shape has a mean of \(75 .\) You select a sample of \(40 .\) The standard deviation of the sample is \(5 .\) Compute the probability the sample mean is: a. Less than 74 b. Between 74 and 76 . c. Between 76 and 77 . d. Greater than 77 .

The Crossett Trucking Company claims that the mean weight of their delivery trucks when they are fully loaded is 6,000 pounds and the standard deviation is 150 pounds. Assume that the population follows the normal distribution. Forty trucks are randomly selected and weiqhed. Within what limits will 95 percent of the sample means occur?

Appendix \(E\) is a table of random numbers. Hence, each digit from 0 to 9 has the same likelihood of occurrence. a. Draw a graph showing the population distribution. What is the population mean? Is this an example of the uniform distribution? Why? b. Following are the first 10 rows of five digits from Appendix E. Assume that these are 10 random samples of five values each. Determine the mean of each sample and plot the means on a chart similar to Chart 8-3. Compare the mean of the sampling distribution of the sample means with the population mean. $$ \begin{array}{|ccccc|} \hline 0 & 2 & 7 & 1 & 1 \\ 9 & 4 & 8 & 7 & 3 \\ 5 & 4 & 9 & 2 & 1 \\ 7 & 7 & 6 & 4 & 0 \\ 6 & 1 & 5 & 4 & 5 \\ 1 & 7 & 1 & 4 & 7 \\ 1 & 3 & 7 & 4 & 8 \\ 8 & 7 & 4 & 5 & 5 \\ 0 & 8 & 9 & 9 & 9 \\ 7 & 8 & 8 & 0 & 4 \\ \hline \end{array} $$

A normal population has a mean of 60 and a standard deviation of \(12 .\) You select a random sample of \(9 .\) Compute the probability the sample mean is: a. Greater than \(63 .\) b. Less than 56 . c. Between 56 and 63 .

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