Question

According to the National Institute of Allergy and Infectious Diseases, 6% of American adults have a food allergy. A large company plans a lunch reception for its 500 employees. Assume that employees are independent. Let the random variable X be the number of company employees who have a food allergy.
(a) What are the assumptions/requirements of a Binomial distribution? Does this situation meet all these requirements?
(b) What are the expected value and standard deviation of X (i.e., of the population)?
(c) What is the probability that none of the 500 employees has a food allergy?

Answer 1

Answer:

a) The requirements is that each trial can only have two outcomes(success/failure), and each trial has the same probability of a success, that is, they are independent of each other.

In this problem, for each person, either they are allergic to some kind of food, or they are not. The probability of a person being allergic is independent of any other people. So this situations meets all these requirements.

b) The expected value is 30 and the standard deviation is 5.31.

c) Close to 0% probability that none of the 500 employees has a food allergy

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have food allergy, or they do not. The probability of an adult having allergy is independent of other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

6% of American adults have a food allergy

This means that $p = 0.06$

500 employees.

This means that $n = 500$

(a) What are the assumptions/requirements of a Binomial distribution? Does this situation meet all these requirements?

The requirements is that each trial can only have two outcomes(success/failure), and each trial has the same probability of a success, that is, they are independent of each other.

In this problem, for each person, either they are allergic to some kind of food, or they are not. The probability of a person being allergic is independent of any other people. So this situations meets all these requirements.

(b) What are the expected value and standard deviation of X (i.e., of the population)?

$E(X) = np = 500*0.06 = 30$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{500*0.06*0.94} = 5.31$

The expected value is 30 and the standard deviation is 5.31.

(c) What is the probability that none of the 500 employees has a food allergy?

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{500,0}.(0.06)^{0}.(0.94)^{500} \cong 0$

Close to 0% probability that none of the 500 employees has a food allergy