Question

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 308 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.

a. The probability that a pregnancy will last 308 days or longer is ?

b. Babies who are born on or before ____ days are considered premature.

Answer #1

Solution :

Given that ,

mean = = 268

standard deviation = = 15

(a)

P(x 308) = 1 - P(x 308)

= 1 - P((x - ) / (308 - 268) / 15)

= 1 - P(z 2.67)

= 1 - 0.9962

= 0.0038

Probability = 0.0038

(b)

P(Z < z) = 3%

P(Z < -1.88) = 0.03

z = -1.88

Using z-score formula,

x = z * +

x = -1.88 * 15 + 268 = 239.8

Babies who are born on or before 239.8 days are considered premature .

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