According to the national institute of allergy and infectious diseases about 3.7% of american adults have a food allergy. A large company plans a lunch reception for its 400 employees. Assume that employees are independent. Let the randome variable x be the number of company employeees who have a food allergy. The mean and standard deviation of the distribution of x are closest to

The binomial probability is the probability of exactly x successes on n repeated trials, and each trial can only have two outcomes, with probabilities [tex]\pi[/tex] and [tex]1 - \pi[/tex].

For an experiment with n repeated trials and a probability of success [tex]\pi[/tex], we have that:

The mean is: [tex]\mu_{x} = n*\pi[/tex]

The standard deviation is: [tex]\sigma_{x} = \sqrt{n*\pi*(1-\pi)}[/tex]

In this problem

There are 400 employees, each with a probability of 3.7% of having food allergy. So [tex]n = 400, \pi = 0.037[/tex]. So:

[tex]\mu_{x} = n*\pi = 400*(0.037) = 14.8[/tex]

The mean of x is close to 14.8.

[tex]\sigma_{x} = \sqrt{n*\pi*(1-\pi)} = \sqrt{400*(0.037)*(1-0.037)} = 3.78[/tex]

The standard deviation of x is close to 3.78.