Question 5, 8.1.19
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The length of human pregnancies is approximately normal with mean $\mu=266$ days and standard deviation $\sigma=16$ days. Complete parts (a) through (f).
Click here to viow the slandard normal distribution table (page 1)
Click here to view the standard normald sistibution table (page 2).
A. If 100 independent random samples of size $n=14$ pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of 258 days or more.
B. If 100 independent random samples of size $n=14$ pregnancies were obtained from this population, we would expect sample(s) to have a sample mean of exactly 258 days
If 100 independent random samples of size $n=14$ pregnancies were obtained from this population, we would expect 3 sample(s) to have a sample mean of 258 days or less.
(d) What is the probability that a random sample of 29 pregnancies has a mean gestation period of 258 days or less?
The probability that the mean of a random sample of 29 pregnancies is less than 258 days is (Round to four decimal places as needed.)
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Final Answer: The probability that the mean of a random sample of 29 pregnancies is less than 258 days is \(\boxed{0.0035}\).
Step 1 ：We are given that the length of human pregnancies is approximately normal with mean \(\mu=266\) days and standard deviation \(\sigma=16\) days. We are asked to find the probability that the mean of a random sample of 29 pregnancies is less than 258 days.
Step 2 ：This is a problem of normal distribution. We can use the formula for the z-score to find the probability. The z-score is calculated as \((X - \mu) / (\sigma / \sqrt{n})\), where X is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and n is the sample size.
Step 3 ：Substitute the given values into the z-score formula: X = 258, \(\mu\) = 266, \(\sigma\) = 16, and n = 29. The z-score is calculated as \((-2.692582403567252)\).
Step 4 ：After calculating the z-score, we can use a z-table to find the probability. The probability corresponding to the z-score is \(0.003545050611907765\).
Step 5 ：Rounding to four decimal places, the final answer is \(0.0035\).
Step 6 ：Final Answer: The probability that the mean of a random sample of 29 pregnancies is less than 258 days is \(\boxed{0.0035}\).
