Question

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 127 days and standard deviation sigma equals 12 days. Complete parts​ (a) through​ (f) below. ​(a) What is the probability that a randomly selected pregnancy lasts less than 123 ​days? The probability that a randomly selected pregnancy lasts less than 123 days is approximately 0.3694. ​(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 pregnant individuals were selected independently from this​ population, we would expect 37 pregnancies to last less than 123 days. B. If 100 pregnant individuals were selected independently from this​ population, we would expect nothing pregnancies to last more than 123 days. C. If 100 pregnant individuals were selected independently from this​ population, we would expect nothing pregnancies to last exactly 123 days. ​(b) 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 overbar is normal with mu Subscript x overbarequals 127 and sigma Subscript x overbarequals 2.9104. ​(Round to four decimal places as​ needed.) ​(c) What is the probability that a random sample of 20 pregnancies has a mean gestation period of 123 days or​ less? The probability that the mean of a random sample of 20 pregnancies is less than 123 days is approximately nothing. ​(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 nequals20 pregnancies were obtained from this​ population, we would expect nothing ​sample(s) to have a sample mean of 123 days or less. B. If 100 independent random samples of size nequals20 pregnancies were obtained from this​ population, we would expect nothing ​sample(s) to have a sample mean of 123 days or more. C. If 100 independent random samples of size nequals20 pregnancies were obtained from this​ population, we would expect nothing ​sample(s) to have a sample mean of exactly 123 days. ​(d) What is the probability that a random sample of 39 pregnancies has a mean gestation period of 123 days or​ less? The probability that the mean of a random sample of 39 pregnancies is less than 123 days is approximately nothing. ​(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 nequals39 pregnancies were obtained from this​ population, we would expect nothing ​sample(s) to have a sample mean of 123 days or less. B. If 100 independent random samples of size nequals39 pregnancies were obtained from this​ population, we would expect nothing ​sample(s) to have a sample mean of exactly 123 days. C. If 100 independent random samples of size nequals39 pregnancies were obtained from this​ population, we would expect nothing ​sample(s) to have a sample mean of 123 days or more. ​(e) What might you conclude if a random sample of 39 pregnancies resulted in a mean gestation period of 123 days or​ less? This result would be ▼ expected, unusual, so the sample likely came from a population whose mean gestation period is ▼ less than equal to greater than 127 days. ​(f) What is the probability a random sample of size 15 will have a mean gestation period within 12 days of the​ mean? The probability that a random sample of size 15 will have a mean gestation period within 12 days of the mean is nothing. ​(Round to four decimal places as​ needed.)

Answer 1

a) P(X < 123)

= P((X - )/ < (123 - )/)

= P(Z < (123 - 127)/12)

= P(Z < -0.33)

= 0.3707

Expected value = 100 * 0.3707 = 37.07 = 37

Option - A) If 100 pregnantndividuals were selected independently from this population, we would expect 37 pregnancies to last less than 123 days.

b) = 127

=

     = 12/ = 2.6833

The samplinf distribution of is normal is normal with = 127 and = 2.6833

P(< 123)

= P(( - )/( ) < (123 - )/( ))

= P(Z < (123 - 127)/2.6833)

= P(Z < -1.49)

= 0.0681

Expected value = 0.0681 * 100 = 6.81 = 7

If 100 independent random samples of size n = 20 pregnancies were obtained from this population , we would expect 7 samples to have a sample mean of 123 days or less.

d) P(< 123)

= P(( - )/( ) < (123 - )/( ))

= P(Z < (123 - 127)/(12/))

= P(Z < -2.08)

= 0.0188

Expected value = 0.0188 * 100 = 1.88 = 2

If 100 independent random samples of size n = 39 pregnancies were obtained from this population , we would expect 2 samples to have a sample mean of 123 days or less.

e) This result would be unusual, so the sample likely came from a population whose mean gestation period is equal to 127 days.

f) P(115 < < 139)

= P((115 - )/( ) < ( - )/( ) < (139 - )/( ))

= P((115 - 127)/(12/) < Z < (139 - 127)/(12/))

= P(-3.87 < Z < 3.87)

= P(Z < 3.87) - P(Z < -3.87)

= 1 - 0 = 1.0000