A national survey reported that 32% of adults in a certain country have hypertension (high blood pressure). A sample of 22 adults is studied. Round the answers to four decimal places.Part 1 of 4(a) What is the probability that exactly 6 of them have hypertension?The probability that exactly 6 of them have hypertension is .Part 2 of 4(b) What is the probability that more than 8 have hypertension?The probability that more than 8 have hypertension is .Part 3 of 4(c) What is the probability that fewer than 3 have hypertension?The probability that fewer than 3 have hypertension is .Part 4 of 4(d) Would it be unusual if more than 10 of them have hypertension?It ▼(Choose one) be unusual if more than 10 of them find have hypertension since the probability is
Question
A national survey reported that 32% of adults in a certain country have hypertension (high blood pressure). A sample of 22 adults is studied. Round the answers to four decimal places.Part 1 of 4(a) What is the probability that exactly 6 of them have hypertension?The probability that exactly 6 of them have hypertension is .Part 2 of 4(b) What is the probability that more than 8 have hypertension?The probability that more than 8 have hypertension is .Part 3 of 4(c) What is the probability that fewer than 3 have hypertension?The probability that fewer than 3 have hypertension is .Part 4 of 4(d) Would it be unusual if more than 10 of them have hypertension?It ▼(Choose one) be unusual if more than 10 of them find have hypertension since the probability is
Solution
To answer these questions, we will use the binomial probability formula:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes
Part 1 of 4(a) The probability that exactly 6 of them have hypertension is calculated as follows:
P(X=6) = C(22, 6) * (0.32^6) * ((1-0.32)^(22-6))
Using a calculator, we find that P(X=6) = 0.2363.
Part 2 of 4(b) The probability that more than 8 have hypertension is the sum of the probabilities that 9, 10, 11, ..., 22 have hypertension. This is calculated as follows:
P(X>8) = 1 - P(X<=8) = 1 - [P(X=0) + P(X=1) + ... + P(X=8)]
Using a calculator, we find that P(X>8) = 0.6016.
Part 3 of 4(c) The probability that fewer than 3 have hypertension is the sum of the probabilities that 0, 1, or 2 have hypertension. This is calculated as follows:
P(X<3) = P(X=0) + P(X=1) + P(X=2)
Using a calculator, we find that P(X<3) = 0.0218.
Part 4 of 4(d) To determine if it would be unusual if more than 10 of them have hypertension, we need to calculate the probability that more than 10 have hypertension and compare it to a threshold. A common threshold is 0.05. If the probability is less than 0.05, it would be considered unusual.
P(X>10) = 1 - P(X<=10) = 1 - [P(X=0) + P(X=1) + ... + P(X=10)]
Using a calculator, we find that P(X>10) = 0.3174.
Since 0.3174 is greater than 0.05, it would not be unusual if more than 10 of them have hypertension.
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