Question

After years of rapid growth, illegal immigration into the US declined as the country began the most serious economic slump since the Great Depression (LA Times, 9/1/2010). While its share declined, California still accounted for 27% of the nation’s estimated 14.1 million undocumented immigrants. Of a random sample of 45, what is the probability that fewer than 30% lived in California?

Answer 1

Solution:

We are given n = 45, p = 0.27, q = 1 – p = 1 – 0.27 = 0.73

We have to find probability that fewer than 30% lived in California.

30% of 45 = 13.5

That is, we have to find P(X<13.5) ≈ P(X<13) (by using continuity correction)

Here, we have to use normal approximation.

Mean = np = 45*0.27 = 12.15

SD = sqrt(npq) = sqrt(45*.27*.73) = 2.97817

Z = (X – mean) / SD

Z = (13 - 12.15) / 2.97817

Z = 0.28541

P(Z<0.28541) = P(X<13) = 0.612335

(by using z-table)

Required probability = 0.612335