In clinical nursing care facilities, there are about twice as many women as men,

Question

In clinical nursing care facilities, there are about twice as many women as men, since the average lifespan of women is considerably longer than for men. A medical research project is underway to study geriatric problems. A group of 100 patients is selected at random for the study. What is the probability that 75 or more of the group would be female? What would be the smallest group of patients that should be selected in order to make the chance less that 0.05 of having more than 70% of them female?

Explanation / Answer

Since there are 2x as many females as males, p(female) = 2/3. Since the population is large, we can assume that selecting asingle person (or 100 persons) will not affect thisprobability. As a result, we can use a z-test. A) P(75 people out of 100) = P(.75 proportion) = (.75 - 2/3) /sqrt((2/3 * 1/3) / 100) = (1 / 12) / (sqrt(2) / 30) = 1.77 -> p= .9616 Since we want the probability that 75 or MORE people are females,we subtract this probability from 1 and get 1-.9616 = .0384. B) We want to find n such that the probability of having theproportion = .7 is less than 5%. First, we find theassociated z score.   p=.95 --> z = 1.645 = (.7 - 2/3)/ sqrt(2/3 * 1/3 / n). Solving for n, we get 541.205. We round up to make sure this tail is less than or equal to.05. Therefore, our final answer is 542.

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