Question

According to the National Institute of Allergy and Infectious Diseases, approximately 7% of American children 4 years of age or under have a food allergy. A day care has capacity for 8 children in that age range. Assume that the children attending the daycare are independent. Let the random variable X
be the number of children in this day care who have a food allergy. The probability P(X < 2) is

Answer 1

Answer:

89.65%

Step-by-step explanation:

This can be modeled with a binomial distribution with probability of “success” (having a food allergy) equals 7% (0.07) and the probability of “failure” (not having food allergy) equals 0.93 in 8 “trials” (8 children).

P(X < 2) = P(X = 0) + P(X = 1) =

$\large \binom{8}{0}(0.07)^0(0.93)^{8}+\binom{8}{1}(0.07)^1(0.93)^{7}$

we can compute this either by hand or computer assisted and we get our probability equals

0.8965 0r 89.65%