An April 15, 2002 report in Time Magazine stated that the average age for women to marry in the United States is now 24.9 years of age. If the standard deviation is assumed to be 4.9 years, find the probability that a random sample of 33 U.S. women would show a mean age at marriage of less than or equal to 23.2 years. (Give your answer correct to four decimal places.)

Find the area under the standard normal curve to the left of z = 1.88, P(z < 1.88). (Give your answer correct to four decimal places.)

Z = (score-mean)/SEm

SEm = SD/√n

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.

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To find the probability that a random sample of 33 U.S. women would show a mean age at marriage of less than or equal to 23.2 years, we can use the concept of the sampling distribution of the mean.

The sampling distribution of the mean is the distribution of the sample means obtained from all possible samples of a specific size taken from a population. The sampling distribution of the mean follows a normal distribution when the sample size is large enough (central limit theorem).

In this case, the population mean is given as 24.9 years, and the standard deviation is given as 4.9 years. Since we have a large enough sample size of 33, we can assume that the sampling distribution of the mean follows a normal distribution.

To calculate the probability of obtaining a mean age at marriage of less than or equal to 23.2 years, we need to standardize the sample mean using the z-score formula.

The z-score formula is:
z = (x - μ) / (σ / √n)

Where:
- z is the standardized score
- x is the value we are interested in (23.2 years)
- μ is the population mean (24.9 years)
- σ is the standard deviation (4.9 years)
- n is the sample size (33)

Substituting the given values into the formula, we get:
z = (23.2 - 24.9) / (4.9 / √33)

Calculating this, we get:
z ≈ -1.893

Now, we can calculate the probability using the standard normal distribution table or calculator. The probability is the area under the curve to the left of the z-score.

Using the standard normal distribution table or calculator, we find that the probability corresponding to a z-score of -1.893 is approximately 0.0293.

Therefore, the probability that a random sample of 33 U.S. women would show a mean age at marriage of less than or equal to 23.2 years is approximately 0.0293 (or 0.0293 to four decimal places).