document.write( "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.
\n" ); document.write( " I have a TI-84 but im not sure how to solve
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Algebra.Com's Answer #724407 by stanbon(75887)\"\" \"About 
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\n" ); document.write( " 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.
\n" ); document.write( "I have a TI-84 but im not sure how to solve
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\n" ); document.write( "a) z(309) = (309-270)/15 = 2.6
\n" ); document.write( "P(x > 309) = P(z > 2.6) = normalcdf(2.6,100) = 0.0047
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\n" ); document.write( "b) Find the z-value with a left tail of 0.04
\n" ); document.write( "invNorm(0.04) = -1.7507
\n" ); document.write( "Find the corresponding # of days::
\n" ); document.write( "d = z*s+u
\n" ); document.write( "d = -1.7507*15+270 = 243.74 days
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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