Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 171 days and standard deviation sigma equals 17 days.

Question

f. What is the probability a random sample of size 18 will have a mean gestation period within 12 days of the​ mean?

Answer 1

The sample means are distributed normally with a mean of 171 and a standard deviation

sigma / N ^ 0.5 = 17 days / 18 ^ 0.5 = 17 days / 4.2426 = 4.0 days

The question calls for the probability that the sample mean x bar deviates from 171 by no more than 12 days. Because 12 days / 4 days = 3, we want the probability that z lies between -3 and +3. From a normal distribution table,

P(-3 < z < +3) = P(z < 3) - P(z < -3) = 0.99865 - 0.00135 = 0.9975, a virtual certainty.

Answer 2

Within 12 days of the mean gestational period equates to a probability that is between 159 days and 183 days. The probability could be written as P( 159 days < or = 171 < or = 183 days). Using the CENTRAL LIMIT THEOREM, the STANDARD DEVIATION of the RANDOM sample of 18 is 17/SQRT 18. The STANDARD DEVIATION of the random sample of 18 is 4.007. There is a 99.7% probability that the MEAN GESTATIONAL PERIOD will fall within 12 days of the mean of 171 days.