The lengths of pregnancies are normally distributed with a mean of 268 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.
Accepted Solution
A:
Answer:probability of a pregnancy lasting 309 days is 0.0032the length of pregnancy is in the lowest 4% is 241.74 daysStep-by-step explanation:Given data mean = 268standard deviation = 15to find out probability of a pregnancy lasting 309 days or longer and the length of pregnancy is in the lowest 4%solutionwe know mean (M) is 268 and standard deviation (SD) is 15 so probability will be in 1st part where pregnancy lasting 309 daysP(X > 309 ) = P( X− mean > 309 − mean )we know Z = (309 − mean) / standard deviation , it will be Z = 309−268/15 = 2.73 we can say these both are equalP(X > 309 ) = P( Z > 2.73)now we use the standard normal z-table i.e.P( Z > 2.73 ) = 0.0032so P( X > 309 ) is 0.0032probability of a pregnancy lasting 309 days is 0.0032and in 2nd part z value with 4% .i.e 0.04z = -1.7507 from the standard table so days = z × standard deviation + mean days = -1.7507 × 15 + 268days = 241.74the length of pregnancy is in the lowest 4% is 241.74 days