Homework SourseAccording to a study done by the Gallup organization the pro
According to a study done by the Gallup organization, the proportion of Americans who are satisfied with the way things in their lives are going is 0.82.
a) Suppose a random sample of 100 Americans is asked \"Are you satisfied with the way things are going in your life?\" Describe the sampling of p?, the proportion of Americans who are satisfied with the way things are going in their life. Be sure to verify the model requirements.
b) In the sample obtained in part a, what is the probability the proportion who are satisfied with the way things are going in their life exceeds 0.85?
c) Would it be unusual for a survey of 100 Americans to reveal that 75 or fewer are satisfied with the way things are going in their life? Why?
Solution
a) Suppose a random sample of 100 Americans is asked \"Are you satisfied with the way things are going in your life?\" Describe the sampling of p?, the proportion of Americans who are satisfied with the way things are going in their life. Be sure to verify the model requirements.
Since the sample size is larger than 30, we can use normal distribution approximated.
It follows normal distribution with mean=n*p=100*0.82=82
and standard deviation =sqrt(n*p*(1-p))
=sqrt(100*0.82*(1-0.82)) =3.841875
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b) In the sample obtained in part a, what is the probability the proportion who are satisfied with the way things are going in their life exceeds 0.85?
P(phat>0.85) = P((phat-p)/sqrt(p*(1-p)/n) >(0.85-0.82)/sqrt(0.82*(1-0.82)/100))
=P(Z>0.78) =0.2177 (from standard normal table)
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c) Would it be unusual for a survey of 100 Americans to reveal that 75 or fewer are satisfied with the way things are going in their life? Why?
P(X<75) = P((X-mean)/s <(75-82)/3.841875)
=P(Z<-1.82) = 0.0344 (from standard normal table)
Since the probability is less than 0.05, it would be unusual for a survey of 100 Americans to reveal that 75 or fewer are satisfied with the way things are going in their life
