(c) What is the probability that a random sample of 19 pregnancies has a mean gestation period of 215 days or less? Notice that the probability in this case is for the mean of a sample of the population. This means that the sampling distribution of the mean should be used and not the population distribution. In part (b). µ was calculated to be 222 and was calculated to be 4.358899. Therefore, the sampling distribution of x for a sample size of 19 is approximately normal with a mean μ- = 222 and standard deviation o-=4.358899. While either technology or a standard normal distribution table could be used to calculate the desired probability, in thi problem, use technology.

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Mean is 222

Standard Dev is 19

Question is asking me to use technology to find the answer to this problem. Can you explain how to do this such as with either the Graphing calculator and Excel?

(c) What is the probability that a random sample of 19 pregnancies has a mean gestation period of 215 days or less?

Notice that the probability in this case is for the mean of a sample of the population. This means that the sampling distribution of the mean should be used and not the population distribution.

In part (b), the sample mean (\( \mu_{\bar{x}} \)) was calculated to be 222 and the standard deviation (\( \sigma_{\bar{x}} \)) was calculated to be 4.358899. Therefore, the sampling distribution of \( \bar{x} \) for a sample size of 19 is approximately normal with a mean \( \mu_{\bar{x}} = 222 \) and a standard deviation \( \sigma_{\bar{x}} = 4.358899 \). While either technology or a standard normal distribution table could be used to calculate the desired probability, in this problem, use technology.
Transcribed Image Text:(c) What is the probability that a random sample of 19 pregnancies has a mean gestation period of 215 days or less? Notice that the probability in this case is for the mean of a sample of the population. This means that the sampling distribution of the mean should be used and not the population distribution. In part (b), the sample mean (\( \mu_{\bar{x}} \)) was calculated to be 222 and the standard deviation (\( \sigma_{\bar{x}} \)) was calculated to be 4.358899. Therefore, the sampling distribution of \( \bar{x} \) for a sample size of 19 is approximately normal with a mean \( \mu_{\bar{x}} = 222 \) and a standard deviation \( \sigma_{\bar{x}} = 4.358899 \). While either technology or a standard normal distribution table could be used to calculate the desired probability, in this problem, use technology.
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