Answer:
a. mean = 6.96 , standard deviation = 1.71 b. 0.0641
Step-by-step explanation:
Here is the complete question
According to the Behavioural Risk Factor Surveillance System, 58% of all Americans adhere to a sedentary lifestyle (sedentary means does not exercise).
a. If you selected repeated samples of 12 from the U.S. population, what would be the mean number of individuals per sample who do not exercise regularly? What would be the standard deviation?
b. Suppose you select a sample of 12 individuals and find that 10 of them do not exercise regularly. Assuming that the Surveillance System is correct, what is the probability that you would have obtained results as bad or worse than expected?
<em>Solution</em>
a. Since we can only have two types of outcomes, that is, those who exercise and those who do not exercise, the problem follows a binomial distribution.
Since the probability of those who do not exercise, p = 58 % = 0.58,
the mean of the binomial distribution μ = np where n = sample number = 12
So, μ = 12 × 0.58 = 6.96
The standard deviation of a binomial distribution σ = √(npq) where q = probability that people exercise = 1 - p = 1 - 0.58 = 0.42
So, σ = √(12 × 0.58 × 0.42) = √2.9232 = 1.71
b. To find the worst case, we consider the probability that at best, two exercise. Since two exercise, the probability that at best two exercise is ¹²C₂q²p¹⁰ + ¹²C₁qp¹¹ + ¹²C₀q⁰p¹²
= (12 × 11/2)(0.42)²(0.58)¹⁰ + 12(0.42)(0.58)¹¹ + 1 × (0.42)⁰(0.58)¹²
= 0.05 + 0.0126 + 0.00145
= 0.06405
≅ 0.0641
