An article in Parenting magazine reported that 60% of Americans needed a vacation after visiting their families for the holidays. Suppose this is the true proportion of Americans who feel this way. A random sample of 100 Americans is taken. Using the Normal approximation, what is the probability that less than 50% of the people in the sample feel that they need a vacation after visiting their families for the holidays?

Question

Quizplus · Accepted Answer

We can use the Normal approximation to the binomial distribution with mean np = 100(0.6) = 60 and standard deviation sqrt(npq) = sqrt(100(0.6)(0.4)) = 4.9. Then, we can standardize using the formula z = (x - np)/sqrt(npq), where x is the number of people in the sample who feel that they need a vacation. We want to find the probability that less than 50% of the people in the sample feel this way, so we want to find P(x < 50), which is equivalent to finding P(z < (50 - 60)/4.9) = P(z < -2.04). Using a standard normal distribution table or calculator, we find that P(z < -2.04) = 0.0207. Thus, the answer is C.

An Article in Parenting Magazine Reported That 60% of Americans