Question 1109338: The lengths of pregnancies are normally distributed with a mean of 270 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 309 days or longer. b. If the length of pregnancy is in the lowest 4%, then the baby is premature. Find the length that separates premature babies from those who are not premature.
I have a TI-84 but im not sure how to solve
Answer by stanbon(75887)   (Show Source):
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The lengths of pregnancies are normally distributed with a mean of 270 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 309 days or longer. b. If the length of pregnancy is in the lowest 4%, then the baby is premature. Find the length that separates premature babies from those who are not premature.
I have a TI-84 but im not sure how to solve
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a) z(309) = (309-270)/15 = 2.6
P(x > 309) = P(z > 2.6) = normalcdf(2.6,100) = 0.0047
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b) Find the z-value with a left tail of 0.04
invNorm(0.04) = -1.7507
Find the corresponding # of days::
d = z*s+u
d = -1.7507*15+270 = 243.74 days
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Cheers,
Stan H.
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