Respuesta :
Answer:
a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of [tex]Z = \frac{X - \mu}{\sigma}[/tex], in which [tex]\mu[/tex] is the mean and [tex]\sigma[/tex] is the standard deviation.
b) We have to find X when Z has a p-value of [tex]\frac{a}{100}[/tex], and X is given by: [tex]X = \mu - Z\sigma[/tex], in which [tex]\mu[/tex] is the mean and [tex]\sigma[/tex] 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex], standard deviation [tex]\sigma[/tex]
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 [tex]Z = \frac{X - \mu}{\sigma}[/tex], in which [tex]\mu[/tex] is the mean and [tex]\sigma[/tex] 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 [tex]\frac{a}{100}[/tex], and X is given by: [tex]X = \mu - Z\sigma[/tex], in which [tex]\mu[/tex] is the mean and [tex]\sigma[/tex] is the standard deviation.
