SOLUTION: Twenty percent of Americans ages 25 to 75 have high blood pressure. Suppose a random sample of 16 Americans age 25 to 75 is selected. Find the following probabilities: a) (2 mar

Question 1170267: Twenty percent of Americans ages 25 to 75 have high blood pressure. Suppose a random sample of 16 Americans age 25 to 75 is selected. Find the following probabilities:
a) (2 marks) None will have high blood pressure.
b) (2 marks) Exactly four will have high blood pressure.
c) (2 marks) At least one will have high blood pressure.
d) (3 marks) Determine the mean and the standard deviation of the distribution
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step by step.
**Understanding the Problem**
This is a binomial probability problem because:
* There are a fixed number of trials (n = 16).
* Each trial has only two possible outcomes (high blood pressure or not).
* The probability of success (having high blood pressure) is constant (p = 0.20).
* The trials are independent.
**Given Information**
* Probability of having high blood pressure (success), p = 0.20
* Probability of not having high blood pressure (failure), q = 1 - p = 1 - 0.20 = 0.80
* Number of trials (sample size), n = 16
**Formula for Binomial Probability**
The probability of getting exactly k successes in n trials is:
P(X = k) = (nCk) * p^k * q^(n-k)
where:
* nCk = n! / (k! * (n-k)!) (the number of combinations of n items taken k at a time)
**a) None will have high blood pressure (k = 0)**
P(X = 0) = (16C0) * (0.20)^0 * (0.80)^(16-0)
P(X = 0) = 1 * 1 * (0.80)^16
P(X = 0) ≈ 0.0281
**b) Exactly four will have high blood pressure (k = 4)**
P(X = 4) = (16C4) * (0.20)^4 * (0.80)^(16-4)
P(X = 4) = (16! / (4! * 12!)) * (0.20)^4 * (0.80)^12
P(X = 4) = 1820 * 0.0016 * 0.068719476736
P(X = 4) ≈ 0.2001
**c) At least one will have high blood pressure (k ≥ 1)**
It's easier to find the probability of the complementary event (none have high blood pressure) and subtract it from 1:
P(X ≥ 1) = 1 - P(X = 0)
P(X ≥ 1) = 1 - 0.0281
P(X ≥ 1) ≈ 0.9719
**d) Mean and Standard Deviation**
For a binomial distribution:
* Mean (μ) = n * p
* Standard deviation (σ) = √(n * p * q)
Mean (μ) = 16 * 0.20 = 3.2
Standard deviation (σ) = √(16 * 0.20 * 0.80) = √(2.56) = 1.6
**Answers:**
a) Approximately 0.0281
b) Approximately 0.2001
c) Approximately 0.9719
d) Mean: 3.2, Standard deviation: 1.6

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