Question

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

mu equals 142 daysμ=142 days

and standard deviation

sigma equals 9 daysσ=9 days.

Complete parts​ (a) through​ (f) below.

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

139139

​days?The probability that a randomly selected pregnancy lasts less than

139139

days is approximately

0.36940.3694.

​(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

nothing

pregnancies to last exactly

139139

days.Your answer is not correct.

B.If 100 pregnant individuals were selected independently from this​ population, we would expect

3737

pregnancies to last less than

139139

days.

C.If 100 pregnant individuals were selected independently from this​ population, we would expect

nothing

pregnancies to last more than

139139

days.​(b) Suppose a random sample of

1818

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 overbarμxequals=142142

and

sigma Subscript x overbarσxequals=2.12132.1213.

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

​(c) What is the probability that a random sample of

1818

pregnancies has a mean gestation period of

139139

days or​ less?The probability that the mean of a random sample of

1818

pregnancies is less than

139139

days is approximately

nothing.

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

Answer 1

c) P(Mean days < 139)

139-142 = P(z < 9/V18

= P(z < -1.41)

= 0.0786