Question

In 2014, the percentage of the U.S population who was foreign-born was 13.1. Choose 60 U.S residents at random. Find the mean, variance, and Standard Deviation for the number who are foreign-born.

Answer 1

Hence Mean ,variance ,standard deviation who are foreign born are as

8, 6.83 ,2.613 respectively

Explanation:

Given:

probability of success for foreign-born is 13.1 % p=0.131

probability of failure= q=1-0.131=0.869

n=60

To Find:

Mean ,variance and standard deviation

Solution:

We have here binomial distribution as probability for success (p)and number of trails are given (n).

So we know that formula for mean ,variance and standard deviation of binomial distribution are as follows ,

p=0.131

q=0.869

n=60

So Mean=μ= n*p

=60*0.131

=7.86

Hence mean will be 7.86 which approximately 8 persons.

Now for Variance=n*p*q

=60*0.131*0.869

=6.830

So variance will be 6.83

And for Standard deviation=Sqrt(variance)=Sqrt(n*p*q)

=Sqrt(6.83)

=2.613

Hence Mean ,variance ,standard deviation who are foreign born are as

8, 6.83 ,2.613 respectively.