Question

Anita's, a fast-food chain specializing in hot dogs and garlic fries, keeps track of the proportion of its customers who decide to eat in the restaurant (as opposed to ordering the food "to go"), so it can make decisions regarding the possible construction of in-store play areas, the attendance of its mascot Sammy at the franchise locations, and so on. Anita's reports that 52% of its customers order their food to go. If this proportion is correct, what is the probability that, in a random sample of 5 customers at Anita's, exactly 4 order their food to go?
Round your response to at least three decimal places

Answer 1

Using the binomial distribution, it is found that there is a 0.1755 = 17.55% probability that, in a random sample of 5 customers at Anita's, exactly 4 order their food to go.

For each customer, there are only two possible outcomes, either they order their food to go, or they do not. The orders of each customer are independent of other customers, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• 52% of its customers order their food to go, hence $p = 0.52$.

• Random sample of 5 customers, hence $n = 5$.

The probability that exactly 4 order their food to go is P(X = 4), hence:

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

$P(X = 4) = C_{5,4}.(0.52)^{4}.(0.48)^{1} = 0.1755$

0.1755 = 17.55% probability that, in a random sample of 5 customers at Anita's, exactly 4 order their food to go.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377