Question

In 2009 the percentage of the U.S. population who was foreign-born was 12.2. Choose 60 U.S. residents at random. How many would you expect to be American-born? Find the mean, variance, and standard deviation for the number who are foreign-born.

Answer 1

Compute the number of American-born.

The percentage of the foreign-born is 12.2% and the percentage of the Americans is 87.8% (=100%–12.2%).

 

The American-born is,

nq = 60(0.878)

     = 52.68

     ≈ 53

 

Thus, approximately 53 are expected to be American-born.

 

Obtain the mean, variance, and standard deviation for the binomial distribution when n = 60, and p = 0.122.

Mean:

µ = np

   = 60(0.122)

   = 7.3

 

Variance:

σ2 = npq

     = 60(0.122)(1 – 0.122)

     = 60(0.122)(0.878)

     = 6.4

 

Standard deviation:

σ = √npq

   = √6.4

   = 2.5

Mean = 7.3

Variance = 6.4

Standard deviation = 2.5