An expert witness for a paternity lawsuit testifies that the length of a pregnan

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 249 to 307 days before the birth of the child, so the pregnancy would have been less than 249 days or more than 307 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention.

A) 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.)

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

Explanation / Answer

If pregnancy is between 249 and 307, we shall assume that he is not the father

P(he is not the father) = P(249 < X < 307) = P(X<307) - P(X<249)

P(X<x) = P(Z<(x-mean)/standard deviation)

So, P(249 < X < 307) = P(Z < (307-280)/13) - P(Z < (249-280)/13)

= P(Z < 2.08) - P(Z < -2.38)

= 0.9812 - 0.0087

= 0.9725

A) P(he is not the father) = 0.9725

B) P(he could be the father) = 1 - 0.9725 = 0.0275

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