Answer:
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 ![\sigma](https://tex.z-dn.net/?f=%5Csigma)
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.