Question 1057713: Naturalization. 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. If 200 people who became naturalized citizens of the United States that year are selected at random, what is the probability that the number who were originally from some Asian country is
a. fewer than 75?
b. between 80 and 90, inclusive?
c. less than 70 or more than 90?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Because the sample size is > 30, we can assume a normal distribution
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The mean of the population is the mean of the sample, that is, 0.415
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The sample standard deviation is square root ( 0.415 * (1-0.415) / 200 ) = 0.0348
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a) 75/200 = 0.375
z-value = ( 0.375 - 0.415 ) / 0.0348 = -1.1494 approx -1.15
lookup z-value in table of z-values to find the associated probability(Pr)
Pr ( X < 75 ) = 0.1251
:
b) 80/200 = 0.4 and 90/200 = 0.45
Pr ( 80 < X < 90 ) = Pr ( X < 90 ) - Pr ( X < 80 )
z-value for 80 is ( 0.4 - 0.415 ) / 0.0348 = -0.431 approx -0.43
z-value for 90 is ( 0.45 - 0.415 ) / 0.0348 = 1.0057 approx 1.01
Pr ( 80 < X < 90 ) = 0.8438 - 0.3336 = 0.5102
:
c) 70/200 = 0.35
Pr ( X < 70 or X > 90 ) = Pr ( X < 70 ) + ( 1 - Pr ( X < 90 )
z-value for 70 is ( 0.35 - 0.415 ) / 0.0348 = -1.8678 approx -1.87
Pr ( X < 70 or X > 90 ) = 0.0307 + 0.1562 = 0.1869
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