Question 1156611: The U.S. Citizenship and Immigration Services collects and reports information about naturalized persons in Statistical Yearbook. During one year, there were 463,204 persons who became naturalized citizens of the United States and, of those, 41.5% were originally from some country in Asia. Suppose 200 people who became naturalized citizens of the United States that year are selected at random.
(i) Which method would you use to find the probability that the number who were originally from some Asian country is between 60 and 95, inclusive? Provide the reason in your own words.
(ii) Find the probability that the number who were originally from some Asian country is between 60 and 95, inclusive.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! One might prefer to use the normal approximation given the size of the sample. But one can do binomial with a calculator but not by hand given the size of the numbers. The binomial is exact and is the second below.
mean is np=0.415*200=83
variance is np(1-p)=83*.585=48.555
sd is sqrt(V)=6.97
less than or equal to 95
z<(95.5-83)/6.97= 1.721 We use 95.5 to cover the fact that it is 95 or fewer. That z has probability 0.9633
now use z<60 (60-83)/6.97
z<-23/6.97=-3.29. That probability is 0.0005
The difference between those probabilities is 0.9628
Calculator left of 95 has probability of 0.9630
to the left of 60 has probability of 0.0005
the difference is probability of 0.9625
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