Question

An expert witness for a paternity lawsuit testifies that the length of a pregnancy is normally distributed with a mean of 280 days and a standard deviation of 13 days. An alleged father was out of the country from 246 to 306 days before the birth of the child, so the pregnancy would have been less than 246 days or more than 306 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. What is the probability that he was NOT the father? Calculate the z-scores first, and then use those to calculate the probability. (Round your answer to four decimal places.)


What is the probability that he could be the father? (Round your answer to four decimal places.)

Answer 1

We are given the length of a pregnancy is normally distributed with a mean of 280 days and std. Deviation of 13 day.

If a person is the father then pregnancy would have been less than 246 days or more than 306 days long

Hence we need to find the Probability

P*( 246>X>306) = P(X<246) +P(X>306)

P(X<246)

P(X>306)

P*( 246>X>306) = P(X<246) +P(X>306) = 0.0045+0.0228=0.0273

Hence the probability that the man is the father is 0.0273

Not the father.

The man was away from 246 to 306 days, Hence the probability of the pregnancy between these days is