separates the smallest 99% from the largest 1%. ancies The lengths of pregnancie
ID: 2949078 • Letter: S
Question
separates the smallest 99% from the largest 1%. ancies The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days 3 One classical use of the normal distribution is inspired by a letter to "Dear Abby"' in which a ife dlaimed to have given birth 308 days after a brief visit from her husband, who was serving h the Navy. Given this information, find the probability of a pregnancy lasting 308 days or onger. What does the result suggest? If we stipulate that a baby is premature if the length of pregnancy is in the lowest 4%, 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 or that care.Explanation / Answer
a) as for normal distribution z score =(X-mean)/std deviaiton
theefore P(pregnancy last 308 days or longer)=P(X>308)=P(Z>(308-268)/15)=P(Z>2.67)=0.0038
as probability of happening is rare ; the result suggest the length of pregnancy was unusually long
b)
for 4th percentile ; z score =-1.75
therefore corresponding period =mean+z*std deviation =268-1.75*15=241.75 days
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