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.

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

8, 6.83  ,2.613 respectively

Step-by-step 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.

my acc plz

step-by-step explanation:

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)