a. The probability that a pregnancy will last 309 days or longer is   ​(Round to four decimal places as​ needed.)   b. Babies who are born on or before____days are considered premature. ​   (Round to the nearest integer as​ needed.)

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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 2​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.
 
a. The probability that a pregnancy will last 309 days or longer is
 
​(Round to four decimal places as​ needed.)
 
b. Babies who are born on or before____days are considered premature.
 
(Round to the nearest integer as​ needed.)

 

Expert Solution
Step 1
a. The probability that a pregnancy will last 309 days or longer is
 
​(Round to four decimal places as​ needed.)

The following information has been provided

μ=268, σ=15\mu = 268, \text{ } \sigma = 15

We need to compute Pr(X309)\Pr(X \ge 309)

 The corresponding z-value needed to be computed is:

Z=Xμσ=30926815=2.7333Z = \frac{X - \mu}{\sigma} = \frac{ 309 - 268}{ 15} = 2.7333

Therefore, we get that

Pr(X309)=Pr(Z30926815)=Pr(Z2.7333)\Pr(X \geq 309) = \Pr\left(Z \ge \frac{ 309 - 268}{ 15}\right) = \Pr(Z \ge 2.7333)

=10.9969=0.0031 = 1 - 0.9969 = 0.0031

The probability that a pregnancy will last 309 days or longer = 0.0031

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