the mean age at which men in the united states marry for the first time follows the normal distribution with a mean of 24.5 years. the standard deviation is 2.8 years. for a rabdom sample of 59 men, what is the likelihood that the age they were married were less than 25

z = (x-u)/sigma = (25 -24.5)/2.8 = .18

F(.18) from normal table is about .57

You're wrong

To find the likelihood that the age at which the men were married is less than 25, we need to calculate the probability of a random sample of 59 men having a mean age less than 25.

Given:
Mean (μ) = 24.5 years
Standard deviation (σ) = 2.8 years
Sample size (n) = 59

Step 1: Convert the problem to the standard normal distribution.
To do this, we need to calculate the z-score, which represents the number of standard deviations a given value is from the mean.

z = (x - μ) / σ

where:
x = desired value (25 in this case)

Step 2: Calculate the z-score.
z = (25 - 24.5) / 2.8 = 0.1786

Step 3: Look up the z-score in the standard normal distribution table.
The standard normal distribution table provides the cumulative probability up to a given z-score. In this case, we want to find the probability that the z-score is less than 0.1786.

Using the standard normal distribution table (or a calculator), we find that the cumulative probability corresponding to a z-score of 0.1786 is approximately 0.5714.

Step 4: Interpret the result.
The probability that a random sample of 59 men in the United States will have a mean age at first marriage less than 25 years is approximately 0.5714, or 57.14%.