Solve the problem.
-For women aged 18-24, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1. If 36 women are selected at random from a population of 300 women aged 18-24,
Find the probability that their mean systolic blood pressure will be less than 110 mm Hg. Assume that the
Sampling is done without replacement and use a finite population correction factor with N = 300.

Question

Quizplus · Accepted Answer

The correct answer is A.The finite population correction factor is:$$fpc = \sqrt{\frac{N-n}{N-1}} = \sqrt{\frac{300-36}{300-1}} = 0.968$$The standard error of the mean is:$$SE = \frac{\sigma}{\sqrt{n}} * fpc = \frac{13.1}{\sqrt{36}} * 0.968 = 2.15$$The z-score for a mean systolic blood pressure of 110 mm Hg is:$$z = \frac{110 - 114.8}{2.15} = -2.23$$The probability that the mean systolic blood pressure will be less than 110 mm Hg is:$$P(Z < -2.23) = 0.0125$$Using the finite population correction factor, the probability becomes:$$P(Z < -2.23) * fpc = 0.0125 * 0.968 = 0.0121$$Therefore, the probability that the mean systolic blood pressure will be less than 110 mm Hg is 0.0121, which is closest to choice A, 0.0096.

Solve the Problem. -For Women Aged 18-24, Systolic Blood Pressures (In Mm Hg)