The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probability of a pregnancy lasting days or longer. b. If the length of pregnancy is in the lowest ​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of , in which is the mean and is the standard deviation.

b) We have to find X when Z has a p-value of , and X is given by: , in which is the mean and is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean , standard deviation

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of , in which is the mean and is the standard deviation.

b. If the length of pregnancy is in the lowest a​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of , and X is given by: , in which is the mean and is the standard deviation.

answer: the answer is 30 miles per hour.

ft/s -> mphmph = ft/s 1.466667mph = 44 divided by 1.466667mph = 30.00000

- r3ktforgood ☕