According to the Behavioral 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?

Step-by-step explanation:

Hello!

The variable is X: number of sedentary people in a sample of 12.

This is a discrete variable with a binomial distribution and the probability of success is p=0.58. To determine if a variable has or not a binomial distribution you have to see if it follows the binomial criteria:

Binomial criteria:

1. The number of observation of the trial is fixed n=12

2. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial. The activity habits on each person are independent of the others.

3. The probability of success in the same from one trial to another p=0.58

So X≈ Bi (n;ρ)

a. If you were to take a sample of n=12.

The expected value (mean) is E(X)= n*p= 0.58*12= 6.96

And the Standard deviation:

V(X)= n*p*q= 12*0.58*(1-0.58)= 2.92

√V(X)= 1.71

b. I'll calculate the probability of exactly 10 persons having sedentary habits using the table of cumulative probabilities, to do so, you have to look for the accumulated probability until 10 and subtract the accumulative probability until 9:

P(X=10) = P(X≤10) - P(X≤9)= 0.9859 - 0.9358= 0.050

I hope you have a SUPER day!

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