Question

he lengths of pregnancies are normally distributed with a mean of

267

days and a standard deviation of

1515

days. a. Find the probability of a pregnancy lasting

309309

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

3 percent

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

308

days or longer is

nothing.

Answer 1

Solution :

Given that ,

a) P(x 309 ) = 1 - P(x   309)

= 1 - P[(x - ) / (309 - 267) / 15]

= 1 -  P(z 2.80)

= 1 - 0.9974

= 0.0026

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 + 267

x = 238.8 days