Question

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

Answer 1

The mean of x is close to 14.8.

The standard deviation of x is close to 3.78.

Explanation:

Binomial probability distribution

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

For an experiment with n repeated trials and a probability of success
`\pi`, we have that:

The mean is:
`\mu_(x) = n*\pi`

The standard deviation is:
`\sigma_(x) = √(n*\pi*(1-\pi))`

In this problem

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

`\mu_(x) = n*\pi = 400*(0.037) = 14.8`

The mean of x is close to 14.8.

`\sigma_(x) = √(n*\pi*(1-\pi)) = √(400*(0.037)*(1-0.037)) = 3.78`

The standard deviation of x is close to 3.78.