According to a study conducted by a statistical organization

According to a study conducted by a statistical organization the proportion of Americans who are satisfied with the way things are going in their lives is 0.76. Suppose that a random sample of 110 Americans is obtained.

What is the probability that at least 80 Americans in the sample are satisfied with their lives?____

What is the probability that 77 of fewer Americans in the sample are satisfied with their lives? _____

(Round all answers four decimal places as needed).

Solution

mean= n*p=110*0.76 = 83.6

standard deviation =sqrt(n*p*(1-p))

=sqrt(110*0.76*(1-0.76))

=4.479286

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(a) So the probability that at least 80 Americans in the sample are satisfied with their lives is

P(X>=80) = P((X-mean)/s >(80-83.6)/4.479286)

=P(Z>-0.80) = 0.7881 (from standard normal table)

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(b) So the probability that 77 of fewer Americans in the sample are satisfied with their lives is

P(X<77) = P(Z<(77-83.6)/4.479286)

=P(Z<-1.47) = 0.0708 (from standard normal table)