In Exercises 55–60, find the indicated probabilities and interpret the results.


The mean annual salary for Level 1 actuaries in the United States is about \$72,000. A random sample of 45 Level 1 actuaries is selected. What is the probability that the mean annual salary of the sample is (b) more than \$68,000? Assume sigma = \$11,000.

Verified step by step guidance

1. Step 1: Identify the key information provided in the problem. The population mean (μ) is \$72,000, the population standard deviation (σ) is \$11,000, the sample size (n) is 45, and we are tasked with finding the probability that the sample mean (x̄) is greater than \$68,000.
2. Step 2: Recognize that the sampling distribution of the sample mean follows a normal distribution because the sample size is sufficiently large (n > 30). The mean of the sampling distribution is equal to the population mean (μ = \$72,000), and the standard error of the mean (SE) is calculated as SE = σ / √n.
3. Step 3: Calculate the standard error of the mean using the formula SE = σ / √n. Substitute the values: σ = \$11,000 and n = 45. This will give you the standard error, which represents the spread of the sampling distribution.
4. Step 4: Convert the sample mean (\$68,000) to a z-score using the formula z = (x̄ - μ) / SE. Substitute the values: x̄ = \$68,000, μ = \$72,000, and the SE calculated in Step 3. The z-score represents how many standard errors the sample mean is below the population mean.
5. Step 5: Use the z-score obtained in Step 4 to find the corresponding probability from the standard normal distribution table or a statistical software. Since the problem asks for the probability that the sample mean is more than \$68,000, calculate 1 - P(Z ≤ z), where P(Z ≤ z) is the cumulative probability up to the z-score. Interpret the result in the context of the problem.

Key Concepts

Central Limit Theorem

The Central Limit Theorem states that the distribution of the sample mean will approach a normal distribution as the sample size increases, regardless of the population's distribution, provided the sample size is sufficiently large (typically n > 30). This theorem is crucial for calculating probabilities related to sample means, especially when dealing with large samples.

Standard Error

The Standard Error (SE) measures the dispersion of sample means around the population mean. It is calculated as the population standard deviation divided by the square root of the sample size (SE = sigma / √n). In this case, with a population standard deviation of $11,000 and a sample size of 45, the SE helps determine how much the sample mean is expected to vary from the true mean.

Z-Score

A Z-score indicates how many standard deviations an element is from the mean. It is calculated using the formula Z = (X - μ) / SE, where X is the value of interest, μ is the population mean, and SE is the standard error. In this scenario, calculating the Z-score for a sample mean of $68,000 allows us to find the corresponding probability using the standard normal distribution.