Problem 84

Given information

Given information is that 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 240 to 306 days before the birth of the child, so the pregnancy would have been less than 240 days or more than 306 days long if he was the father. The birth was uncomplicated, and the child needed no medical intervention. Then we have to find out the probability that he was NOT the father and the probability that he could be the father.

Step-by-step explanation

According to the given information, be the random variable that represents the length of the pregnancy. Then 

z-score can be calculated using the following formula:

where is the raw score,  is the mean and  is the standard deviation.

Now we have to calculate z-score for and 

 z-score for , 

 z-score for , 

From the standard normal table, and