After years of rapid growth, illegal immigration into the United States has decl
ID: 3053008 • Letter: A
Question
After years of rapid growth, illegal immigration into the United States has declined, perhaps owing to the recession and increased border enforcement by the United States (Los Angeles Times, September 1, 2010). While its share has declined, California still accounts for 29% of the nation’s estimated 11.9 million undocumented immigrants. Use Table 1.
In a sample of 40 illegal immigrants, what is the probability that more than 23% live in California?(Round “z” value to 2 decimal places, and final answer to 4 decimal places.)
In a sample of 100 illegal immigrants, what is the probability that more than 23% live in California? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.
After years of rapid growth, illegal immigration into the United States has declined, perhaps owing to the recession and increased border enforcement by the United States (Los Angeles Times, September 1, 2010). While its share has declined, California still accounts for 29% of the nation’s estimated 11.9 million undocumented immigrants. Use Table 1.
Explanation / Answer
Result:
After years of rapid growth, illegal immigration into the United States has declined, perhaps owing to the recession and increased border enforcement by the United States (Los Angeles Times, September 1, 2010). While its share has declined, California still accounts for 29% of the nation’s estimated 11.9 million undocumented immigrants. Use Table 1.
a.
In a sample of 40 illegal immigrants, what is the probability that more than 23% live in California?(Round “z” value to 2 decimal places, and final answer to 4 decimal places.)
P=0.29, n=40
Standard deviation = sqrt(0.29*(1-0.29)/40) =0.0717
Z value for p=0.23, z =(0.23-0.29)/0.0717 = -0.84
P( p>0.23) = P( z > -0.84)
=0.7995
b.
In a sample of 100 illegal immigrants, what is the probability that more than 23% live in California? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.
P=0.29, n=100
Standard deviation = sqrt(0.29*(1-0.29)/100) =0.0454
Z value for p=0.23, z =(0.23-0.29)/0.0454 = -1.32
P( p>0.23) = P( z > -1.32)
=0.9066