Question

Systolic blood pressure for women between the ages of 18 and 24 is normally distributed with a mean of 114.8 (millimeters of mercury) and a standard deviation of 13.1.
A. What is the likelihood that an individual woman has a blood pressure abouve 125?
B. Suppose a random sample of n=300 women is selected and the mean blood pressure for tyhe sample is computed. What is the likelihood that the mean blood pressure for the sample will be above 125?
C. Suppose a random sample of n=300 women is selected and the mean blood pressure for the sample is computed. What is the likelihood that the mean blood pressure for the sample will be bleow. 114?

Answer 1

A. Z = (score-mean)/SD (for distribution of scores)

B & C. Z = (mean1 - mean2)/standard error (SE) of difference between means (for distribution of means)

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Since only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to these Z scores.