An April 15, 2002 report in Time Magazine stated that the average age for women to marry in the United States is now 24.5 years of age. If the standard deviation is assumed to be 4.1 years, find the probability that a random sample of 32 U.S. women would show a mean age at marriage of less than or equal to 22.7 years.
To find the probability, we need to use the standard deviation and the sample size to calculate the standard error. Then we can use the standard error and the given mean to calculate the z-score. Finally, we can use the z-score to find the probability using a standard normal distribution table or a calculator.
Here are the steps to calculate the probability:
Step 1: Calculate the standard error (SE) using the formula SE = σ / √n, where σ is the standard deviation and n is the sample size.
Given: σ = 4.1 and n = 32
SE = 4.1 / √32 ≈ 0.725
Step 2: Calculate the z-score using the sample mean (x̄), population mean (μ = 24.5), and standard error (SE) using the formula z = (x̄ - μ) / SE.
Given: x̄ = 22.7, μ = 24.5, and SE = 0.725
z = (22.7 - 24.5) / 0.725 ≈ -2.483
Step 3: Find the probability using the z-score. We want to find the probability that the mean age at marriage is less than or equal to 22.7 years, so we need to find the area to the left of the z-score (-2.483).
Using a standard normal distribution table or a calculator, we find that the probability is approximately 0.006.
Therefore, the probability that a random sample of 32 U.S. women would show a mean age at marriage of less than or equal to 22.7 years is approximately 0.006, or 0.6%.