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.)
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/probability related to the Z score.
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 Central Limit Theorem and convert it into a z-score problem.
1. First, convert the given mean and standard deviation into the corresponding standard normal distribution with a mean of 0 and a standard deviation of 1. We can use the formula:
z = (x - μ) / σ
Where z is the z-score, x is the value we want to convert, μ is the mean, and σ is the standard deviation.
In this case, the given mean (μ) is 24.9 and the standard deviation (σ) is 4.9. So, we can calculate the z-score as:
z = (23.2 - 24.9) / 4.9
2. Calculate the z-score using the formula and find the corresponding area under the standard normal curve. We can use a standard normal distribution table or a calculator to find this value. Let's denote this probability as P(Z ≤ z).
3. Given that the sample size is 33, we can use the Central Limit Theorem to assume that the sampling distribution of the sample mean follows a normal distribution with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
So, the mean of the sampling distribution (μ) is still 24.9, and the standard deviation (σ) becomes 4.9 / sqrt(33).
4. Now, we need to find the z-score for the sample mean using the same formula as before:
z = (x - μ) / (σ / sqrt(n))
Where x is the value of the sample mean, μ is the mean of the sampling distribution, σ is the standard deviation of the sampling distribution, and n is the sample size.
In this case, the sample mean (x) is 23.2, the mean of the sampling distribution (μ) is 24.9, and the standard deviation of the sampling distribution (σ) is 4.9 / sqrt(33). So, we can calculate the z-score as:
z = (23.2 - 24.9) / (4.9 / sqrt(33))
5. Calculate the z-score for the sample mean using the formula and find the corresponding area under the standard normal curve. We can use a standard normal distribution table or a calculator to find this value. Let's denote this probability as P(Z ≤ z_sample_mean).
7. Finally, we compare the probabilities obtained in step 2 (P(Z ≤ z)) and step 5 (P(Z ≤ z_sample_mean)) to determine 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.
The probability can be calculated as:
Probability = P(Z ≤ z_sample_mean) - P(Z ≤ z)
Remember to round your answer to four decimal places.
Hope this helps!