Question

According to the Behavioral Risk Factor Surveillance System, 58% of all Americans adhere to a sedentary lifestyle. Suppose that you select a sample of 10 individuals and find that 8 of them do not exercise regularly. Assuming that the Surveillance System is correct, what is the probability of finding 8 or more individuals who are sedentary, from a sample of 10?

Answer 1

Binomial distribution is being applied to this problem

If 'X' is the random variable representing the number of successes, the probability of getting ‘r’ successes and ‘n-r’ failures, in 'n' trails, ‘p’ probability of success ‘q’=(1-p) is given by the probability function

Given,

Percentage of Americians adhere to a sedentary life style = 58%

Probability that an inividual adhere to a sedentary life style : p = 58/100 = 0.58

q = 1-p = 1-0.58 = 0.42

n : Number individuals selected in a sample = 10

X : Number of individuals who are sedentary

X follows a Binomial distribution

Therefore , Probability of 'r' individuals from a sample of 10 individuals are sedentary where p=0.58 is the probability of an individual is sedentary

Probability of finding 8 or more individuals who are sedentary, from a sample of 10 = P(X8)

P(X8) = P(X=8) + P(X=9) + P(X=10)

P(X8) = P(X=8) + P(X=9) + P(X=10) = 0.1017+0.0312+0.0043=0.1372

Probability of finding 8 or more individuals who are sedentary, from a sample of 10 = P(X8) = 0.1372

Probability of finding 8 or more individuals who are sedentary, from a sample of 10 = 0.1372