Respuesta :
Answer:
a) 0.005% probability of a pregnancy lasting 308 days or longer
b) The pregnancy length that separates premature babies from those who are not premature is 229 days.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 250, \sigma = 15[/tex]
a. Find the probability of a pregnancy lasting 308 days or longer?
This is 1 subtracted by the pvalue of Z when X = 308. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{308 - 250}{15}[/tex]
[tex]Z = 3.87[/tex]
[tex]Z = 3.87[/tex] has a pvalue of 0.99995
1 - 0.99995 - 0.00005
0.005% probability of a pregnancy lasting 308 days or longer
b. If we stipulate that a baby is premature if the length of the pregnancy is in the lowest 8% (8th percentile), find the length that separates premature babies from those who are not premature.
The 8th percentile is X when Z has a pvalue of 0.08. So it is X when Z = -1.405.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.405 = \frac{X - 250}{15}[/tex]
[tex]X - 250 = -1.405*15[/tex]
[tex]X = -1.405*15 + 250[/tex]
[tex]X = 229[/tex]
The pregnancy length that separates premature babies from those who are not premature is 229 days.
