Finding Probability (How would you state the following in a rare event rule fram

Question

Finding Probability (How would you state the following in a rare event rule framework)

In general, length of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.

One classic use of the normal distribution is inspired by a letter to “Dear Abby” in which a wife claimed to have given birth 306 days after a brief visit from her husband, who was serving in the Navy. Given this information, find the probability of a pregnancy lasting 306 days or longer. What does the result suggest?

Z= 306-268/15

=.0057 or .57%

Explanation / Answer

Normal Distribution
Mean ( u ) = 268
Standard Deviation ( sd )=15
Normal Distribution = Z= X- u / sd ~ N(0,1)  
P(X < 306) = (306-268)/15
= 38/15= 2.5333
= P ( Z <2.5333) From Standard Normal Table
= 0.9944                  
              
P(X > = 306) = (1 - P(X < 306)
= 1 - 0.9944 = 0.0057 = 0.57%  

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