Question

The lengths of pregnancies are normally distributed with a mean of

267267

days and a standard deviation of

1515

days. a. Find the probability of a pregnancy lasting

308308

days or longer. b. If the length of pregnancy is in the lowest

22​%,

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

308308

days or longer is

Answer 1

Solution :

Given that ,

a) P(x > 308) = 1 - p( x< 308)

=1- p P[(x - ) / < (308 - 267) / 15 ]

=1- P(z < 2.73)

= 1 - 0.9968

= 0.0032

b) Using standard normal table,

P(Z < z) = 2%

= P(Z < z ) = 0.02

= P(Z < -2.05 ) = 0.02  

z = -2.05

Using z-score formula,

x = z * +

x = -2.05 * 15 + 267

x = 236.25 days