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

Answer:

The mean of x is close to 14.8.

The standard deviation of x is close to 3.78.

Step-by-step 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: $\sigma_{x} = \sqrt{n*\pi*(1-\pi)}$

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.

$\sigma_{x} = \sqrt{n*\pi*(1-\pi)} = \sqrt{400*(0.037)*(1-0.037)} = 3.78$

The standard deviation of x is close to 3.78.