Answer:
a) The probability that a pregnancy will last 308 days or longer is 0.0038
b) Babies who are born on or before 256 days are considered prematures.
Step-by-step explanation:
Let X be the random variable that represents the length of a pregnancy. Then, X is normally distributed with a mean of 268 days and a standard deviation of 15 days.
a) The z-score related to 308 days is z = (308-268)/15 = 2.6667, so, the probability of a pregnancy lasting 308 days or longer is P(Z > 2.6667) = 0.0038
b) We are looking for a value q such that P(X < q) = 0.22, i.e., P((X-268)/15 < (q-268)/15) = 0.22, here, Z = (X-268)/15 is a standard normal random variable and z = (q-268)/15 is the 22nd quantile of the standard normal distribution, i.e., z = -0.7722 = (q-268)/15 and (-0.7722)(15) + 268 = q, i.e., q = 256.417, so, babies who are born on or before 256 days are considered premature.
Answer:
a) P(X>308) = 0.00383
b) 256.42 days
Step-by-step explanation:
Population mean (μ) = 268 days
Standard deviation (σ) = 15 days
a) P(X>308)?
The z-score for any length of pregnancy 'X' is given by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
The z-score for X=308 is:
[tex]z=\frac{308-268}{15}\\z=2.6667[/tex]
A z-score of 2.6667 is equivalent to the 99.617-th percentile of a normal distribution. Thus, the probability of a pregnancy lasting 208 days or longer is:
[tex]P(X>308)=1-0.99617\\P(X>308) = 0.00383[/tex]
b) X at which P(X) < 0.22?
At the 22-nd percentile, a normal distribution has an equivalent z-score of -0.772. Therefore, the length of pregnancy, X, that separates premature babies from those who are not premature is:
[tex]-0.772=\frac{X-268}{15}\\X= 256.42[/tex]
