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.
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document.write( " I have a TI-84 but im not sure how to solve \n" );
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Algebra.Com's Answer #724407 by stanbon(75887) ![]() You can put this solution on YOUR website! \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 \n" ); document.write( "------------------- \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 \n" ); document.write( "------ \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 \n" ); document.write( "------------------ \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "---------- \n" ); document.write( " |