Answer:
The answer is below
Step-by-step explanation:
Given that mean (μ) = 266 days, standard deviation (σ) = 16 days, sample size (n) = 16 women.
a) The mean of the sampling distribution of Xbar ([tex]\mu_x[/tex]) is given as:
[tex]\mu_x=\mu=266\ days[/tex]
The standard deviation of the sampling distribution of Xbar ([tex]\sigma_x[/tex]) is given as:
[tex]\sigma_x=\frac{\sigma}{\sqrt{n} } =\frac{16}{\sqrt{16} }=4[/tex]
b) The z score is a measure in statistics used to determine by how much the raw score is above or below the mean. It is given by:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} }[/tex]
For x > 270 days:
[tex]z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{270-266}{\frac{16}{\sqrt{4} } }=1[/tex]
The probability the average pregnancy length exceed 270 days = P(x > 270) = P(z > 1) = 1 - P(z < 1) = 1 - 0.8413 = 0.1587 = 15.87%
