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.)
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To solve this problem, we need to use the concept of the sampling distribution of the mean. Given the average age of women to marry in the United States and the standard deviation, we can determine the probability of a random sample of women having a mean age at marriage less than or equal to 23.2 years.
First, we calculate the standard error of the mean (SEM) using the formula:
SEM = standard deviation / √sample size
SEM = 4.9 / √33 ≈ 0.854
Next, we will use the normal distribution to find the Z-score associated with the given mean age at marriage (23.2 years) using the formula:
Z = (X - μ) / SEM
Z = (23.2 - 24.9) / 0.854 ≈ -1.988
Now, we can use a Z-table or a statistical calculator to find the cumulative probability associated with the Z-score. The cumulative probability represents the probability of getting a value less than or equal to the given mean age at marriage.
Looking up the Z-score of -1.988 in the Z-table, we find that the cumulative probability is approximately 0.0235.
Finally, we round the probability to four decimal places, giving us:
Probability = 0.0235
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 0.0235.