Question

The lengths of pregnancies are normally distributed with a mean of

266 days and a standard deviation of 15 days. a. Find the probability of a pregnancy lasting 307 days or longer. b. If the length of pregnancy is in the lowest 3%, then the baby is premature. Find the length that separates premature babies from those who are not premature.Click to view page 1 of the table.

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a. The probability that a pregnancy will last

307307

days or longer is

nothing.

​(Round to four decimal places as​ needed.)

b. Babies who are born on or before

nothing

days are considered premature.

​(Round to the nearest integer as​ needed.)

Answer 1

Solution :

Given that ,

mean = = 266

standard deviation = = 15

a) P(x > 307) = 1 - p( x< 307)

=1- p P[(x - ) / < (307 - 266) / 15 ]

=1- P(z < 2.73 )

Using z table,

= 1 - 0.9968

= 0.0032

b) Using standard normal table,

P(Z < z) = 3%

= P(Z < z ) = 0.03

= P(Z < -1.88 ) = 0.03  

z = -1.88

Using z-score formula,

x = z * +

x = -1.88 * 15 +266

x = 237.8

238 days are considered premature