Time Re : 02:5318 Submit Qui This Question: 1 pt 13 of 15 (0 This Quiz: 15 pts T

Question

Time Re : 02:5318 Submit Qui This Question: 1 pt 13 of 15 (0 This Quiz: 15 pts 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 lasing 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 p er ature babies from those who are not premature a. The probability that a pregnancy wil last 309 days or longer is (Round to four decimal places as needed.) (Round to the nearest integer as needed.) 0 of 4 120min

Explanation / Answer

Mean = 270 days

Standard deviation = 15 days

P(X < A) = P(Z < (A - mean)/standard deviation)

a) P(pregnancy will last 309 days) = P(X > 309)

= 1 - P(X < 309)

= 1 - P(Z < (309 - 270)/15)

= 1 - P(Z < 2.6)

= 1 - 0.9953

= 0.0047

b) Let P denote the maximum pregnancy period that will be considered premature.

P(X < P) = 0.04

P(Z < (P - 270)/15) = 0.04

From standard normal distribution table,

(P - 270)/15 = -1.75

P = 243.75

Babies who are born on or before 243 days are considered premature

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