The lengths of pregnancies are normally distributed

Ответ:

jennifercastill3

28.01.2022

Mathematics

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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 250, \sigma = 15$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{308 - 250}{15}$

Z = 3.87

Z = 3.87 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.

$Z = \frac{X - \mu}{\sigma}$

$-1.405 = \frac{X - 250}{15}$