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In: Statistics and Probability

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean...

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean

mu equals μ=197 daysand standard deviation sigma equals sσ=14 days. Complete parts​ (a) through​ (f) below.

​(a) What is the probability that a randomly selected pregnancy lasts less than 192 ​days?

The probability that a randomly selected pregnancy lasts less than 192 days is approximately 0.3604 ​(Round to four decimal places as​ needed.)

Interpret this probability. Select the correct choice below and fill in the answer box within your choice. ​(Round to the nearest integer as​ needed.)

B.If 100 pregnant individuals were selected independently from this​ population, we would expect 36 pregnancies to last less than

192 days.

Suppose a random sample of 20 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies.

The sampling distribution of x overbarx is normal with mu Subscript x overbar μx equals=197 and sigma Subscript x overbarσxequals=3.1305 ​(Round to four decimal places as​ needed.)

What is the probability that a random sample of 20 pregnancies has a mean gestation period of 192 days or​ less?

The probability that the mean of a random sample of 20 pregnancies is less than 192 days is approximately 0.0551.

​(Round to four decimal places as​ needed.)

Interpret this probability. Select the correct choice below and fill in the answer box within your choice. ​(Round to the nearest integer as​ needed.)

A.If 100 independent random samples of size nequals=20 pregnancies were obtained from this​ population, we would expect 6 ​sample(s) to have a sample mean of 192 days or less.

d) What is the probability that a random sample of 48 pregnancies has a mean gestation period of 192 days or​ less?

The probability that the mean of a random sample of 48 pregnancies is less than 192 days is approximately ______. ​(Round to four decimal places as​ needed.)

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