The lengths of pregnancies are normally distributed with a mean of 268 days and

Question

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. a. A wife claimed to have given birth 308 days after a brief visit from her husband, who was away serving in the Navy. Find the probability of a pregnancy lasting 308 days or longer. What does the result suggest? b. If we stipulate that a baby is premature if the length of pregnancy is in the lowest 3%, find the length that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care.

Explanation / Answer

Normal Distribution
Mean ( u ) =268
Standard Deviation ( sd )=15
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P(X < 308) = (308-268)/15
= 40/15= 2.6667
= P ( Z <2.6667) From Standard Normal Table
= 0.9962                  
P(X > = 308) = (1 - P(X < 308)
= 1 - 0.9962 = 0.0038                  

b.
P ( Z < x ) = 0.03
Value of z to the cumulative probability of 0.03 from normal table is -1.881
P( x-u/s.d < x - 268/15 ) = 0.03
That is, ( x - 268/15 ) = -1.88
--> x = -1.88 * 15 + 268 = 239.785  

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