Question

The lengths of pregnancies are normally distributed with a mean of 270 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 22%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

Answer #1

We know that:

mean= 270 days

standard deviation = 15 days

**a.**

P(pregnancy>309)= 1- P(pregnancy<309)

= 1- P(Z< (309-270)/15)

= 1- P(Z<2.6) [This value can be found from a z distribution table]

= 1- 0.9953

= 0.0047

**b.**

We know that the pregnancy is premature in the lowest 22% of the z distribution. Thus, the z value will have a probability of 22% or 0.22 to its left.

Thus, the z value is [This value can be found from a z distribution table] -0.7768.

Thus,

=> (x-270)/15 = -0.7768

=> x-270 = -11.652

=> x= 258.348

Thus, the length is 258.348 days ~ 259 days.

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