Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $120,000. This distribution follows the normal distribution with a standard deviation of $38,000.

(A) If we select a random sample of 58 households, what is the standard error of the mean? (Round your answer to the nearest whole number.)

Standard error of the mean=?

(B) What is the expected shape of the distribution of the sample mean?

Not normal, the standard deviation is unknown.
Unknown.
Uniform
Normal.

(C) What is the likelihood of selecting a sample with a mean of at least $124,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability=?

(D) What is the likelihood of selecting a sample with a mean of more than $112,000? (Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability=?

(E) Find the likelihood of selecting a sample with a mean of more than $112,000 but less than $124,000. (Round z value to 2 decimal places and final answer to 4 decimal places.)

Probability=?

(A) SEm = SD/√n

(B) Normal

(C, D, E) Z = (score-mean)/SEm

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.

500

(A) The standard error of the mean can be calculated using the formula:

Standard error of the mean = (standard deviation of the population) / √(sample size)

Given:
Standard deviation of the population (σ) = $38,000
Sample size (n) = 58

Using the formula, we can calculate the standard error of the mean:

Standard error of the mean = $38,000 / √58 ≈ $4,992

Therefore, the standard error of the mean is approximately $4,992.

(B) The expected shape of the distribution of the sample mean is Normal. This is because, in large samples, the sample mean tends to follow a normal distribution, regardless of the shape of the population distribution, due to the Central Limit Theorem. Since we have a sample size of 58, which is relatively large, the distribution of the sample mean is expected to be normal.

(C) To find the likelihood of selecting a sample with a mean of at least $124,000, we need to calculate the z-value for this sample mean and then find the corresponding probability in the standard normal distribution table.

Z-value can be calculated using the formula:

Z = (sample mean - population mean) / (standard deviation / √sample size)

Given:
Population mean (μ) = $120,000
Standard deviation (σ) = $38,000
Sample mean = $124,000
Sample size (n) = 58

Using the formula, we can calculate the z-value:

Z = ($124,000 - $120,000) / ($38,000 / √58) ≈ 1.230

Using the z-value of 1.230, we can find the probability from the standard normal distribution table. The probability of selecting a sample with a mean of at least $124,000 is approximately 0.1094.

Therefore, the probability is approximately 0.1094.

(D) To find the likelihood of selecting a sample with a mean of more than $112,000, we follow the same process as in (C).

Given:
Sample mean = $112,000

Using the formula, we can calculate the z-value:

Z = ($112,000 - $120,000) / ($38,000 / √58) ≈ -2.105

Using the z-value of -2.105, we can find the probability from the standard normal distribution table. The probability of selecting a sample with a mean of more than $112,000 is approximately 0.9825.

Therefore, the probability is approximately 0.9825.

(E) To find the likelihood of selecting a sample with a mean of more than $112,000 but less than $124,000, we need to calculate the z-values for both sample means and then find the corresponding probabilities in the standard normal distribution table.

Z-value for the mean of $112,000 (similar to part D):
Z1 = ($112,000 - $120,000) / ($38,000 / √58) ≈ -2.105

Z-value for the mean of $124,000 (similar to part C):
Z2 = ($124,000 - $120,000) / ($38,000 / √58) ≈ 1.230

The probability of selecting a sample with a mean between $112,000 and $124,000 can be calculated by finding the area under the curve between these two z-values. This can be done by finding the difference between the probabilities corresponding to the z-values from the standard normal distribution table.

Using the standard normal distribution table, we find:
Probability(Z ≤ -2.105) ≈ 0.0175
Probability(Z ≤ 1.230) ≈ 0.8907

Therefore, the probability of selecting a sample with a mean of more than $112,000 but less than $124,000 is approximately 0.8907 - 0.0175 ≈ 0.8732.

Therefore, the probability is approximately 0.8732.

(A) To find the standard error of the mean, we need to divide the standard deviation by the square root of the sample size.

The formula for the standard error of the mean is:

Standard error = (standard deviation) / sqrt(sample size)

Given that the standard deviation is $38,000 and the sample size is 58, we can calculate the standard error as follows:

Standard error = $38,000 / sqrt(58)

Calculating sqrt(58) ≈ 7.61577 and rounding the standard error to the nearest whole number, we get:

Standard error = $38,000 / 7.61577 ≈ $4980

Therefore, the standard error of the mean is approximately $4980.

(B) The expected shape of the distribution of the sample mean can be determined by considering the sample size and the Central Limit Theorem. According to the Central Limit Theorem, as the sample size increases, the distribution of the sample mean tends to approach a normal distribution.

In this case, since we have a sample size of 58, which is larger than 30, we can expect the distribution of the sample mean to be approximately normal.

Therefore, the expected shape of the distribution of the sample mean is normal.

(C) To find the likelihood (probability) of selecting a sample with a mean of at least $124,000, we need to calculate the corresponding z-score and use the standard normal distribution table.

The formula to calculate the z-score is:

z = (sample mean - population mean) / (standard deviation / sqrt(sample size))

Given that the population mean is $120,000, the standard deviation is $38,000, the sample mean is $124,000, and the sample size is 58, we can calculate the z-score as follows:

z = ($124,000 - $120,000) / ($38,000 / sqrt(58))

Calculating sqrt(58) ≈ 7.61577, we can proceed with the calculation:

z = $4,000 / ($38,000 / 7.61577)

z ≈ 0.299

Using the standard normal distribution table or a calculator, we can find the probability associated with a z-score of 0.299. The probability is approximately 0.6179.

Therefore, the likelihood of selecting a sample with a mean of at least $124,000 is approximately 0.6179.

(D) Similarly, to find the likelihood (probability) of selecting a sample with a mean of more than $112,000, we need to calculate the corresponding z-score and use the standard normal distribution table.

The formula remains the same:

z = (sample mean - population mean) / (standard deviation / sqrt(sample size))

Given that the population mean is $120,000, the standard deviation is $38,000, the sample mean is $112,000, and the sample size is 58, we can calculate the z-score as follows:

z = ($112,000 - $120,000) / ($38,000 / sqrt(58))

Calculating sqrt(58) ≈ 7.61577, we can proceed with the calculation:

z = -$8,000 / ($38,000 / 7.61577)

z ≈ -0.391

Using the standard normal distribution table or a calculator, we can find the probability associated with a z-score of -0.391. The probability is approximately 0.3477.

Therefore, the likelihood of selecting a sample with a mean of more than $112,000 is approximately 0.3477.

(E) To find the likelihood (probability) of selecting a sample with a mean of more than $112,000 but less than $124,000, we need to calculate the corresponding z-scores and use the standard normal distribution table.

First, let's calculate the z-score for a mean of $112,000:

z1 = ($112,000 - $120,000) / ($38,000 / sqrt(58))

Calculating sqrt(58) ≈ 7.61577, we can proceed with the calculation:

z1 = -$8,000 / ($38,000 / 7.61577)

z1 ≈ -0.391

Next, let's calculate the z-score for a mean of $124,000:

z2 = ($124,000 - $120,000) / ($38,000 / sqrt(58))

Calculating sqrt(58) ≈ 7.61577, we can proceed with the calculation:

z2 = $4,000 / ($38,000 / 7.61577)

z2 ≈ 0.299

To find the probability between these two z-scores, we subtract the probability associated with the smaller z-score from the probability associated with the larger z-score:

Probability = P(z2) - P(z1)

Using the standard normal distribution table or a calculator, we can find the probabilities:

P(z2) ≈ 0.6179

P(z1) ≈ 0.3477

Therefore, the likelihood of selecting a sample with a mean of more than $112,000 but less than $124,000 is approximately 0.6179 - 0.3477 ≈ 0.2702.