Based on data collected by the National Center for Health Statistics and made available to the public in the Sample Adult database (A-5), an estimate of the percentage of adults who have at some point in their life been told they have hypertension is 23.53 percent. If we select a simple random sample of 20 U. S. adults and assume that the probability that each has been told that he or she has hypertension is .24, find the probability that the number of people in the sample who have been told that they have hypertension will be:

Step-by-step explanation:

The question is incomplete. The complete question is:

Based on data collected by the National Center for Health Statistics and made available to the public in the Sample Adult database (A-5). an estimate of the percentage of adults who have at some point in their life been told they have hypertension is 23.53 percent. If we select a simple random sample of 20 U.S. adults and assume that the probability that each has been told that he or she has hypertension is .24. find the probability that the number of people in the sample who have been told that hypertension will be: a) Exactly three (b) Three or more (c) Fewer than three (d) Between three and seven, inclusive

Solution:

This is a binomial probability distribution. If an adult is selected randomly, the outcome is either he has been told that he has hypertension or he has not been told. The probability of success, p would be that a randomly selected adult, x has been told. Therefore,

p = 0.24

n = 20

a)We want to determine P(x = 3). From the binomial probability distribution calculator,

P(x = 3) = 0.15

b) P(x ≥ 3) = 0.89

c) P(x < 3) = 0.11

d) P(3 ≤ x ≤ 7)

It would be P(x ≤ 7) - P(x ≥ 3)

P(x ≤ 7) = 0.92

Therefore,

P(3 ≤ x ≤ 7) = 0.92 - 0.89 = 0.03

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!

the answer is:

mean: 6

mad: 2.14