Question

At a Noodle & Company restaurant, the probability that a customer will order a nonalcoholic beverage is . 43. Find the probability that in a sample of 7 customers, fewer that 5 will order a nonalcoholic beverage.

Answer 1

Answer:

0.8716 = 87.16% probability that in a sample of 7 customers, fewer that 5 will order a nonalcoholic beverage.

Step-by-step explanation:

For each customer, there are only two possible outcomes. Either they order a nonalcoholic beverage, or they order an alcoholic beverage. The probability of a customer ordering a nonalcoholic beverage is independent of any other customer, which means that the binomial probability distribution is used 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.

At a Noodle & Company restaurant, the probability that a customer will order a nonalcoholic beverage is 0.43.

This means that $p = 0.43$

Find the probability that in a sample of 7 customers, fewer that 5 will order a nonalcoholic beverage.

This is:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

In which

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

$P(X = 0) = C_{7,0}.(0.43)^{0}.(0.57)^{7} = 0.0195$

$P(X = 1) = C_{7,1}.(0.43)^{1}.(0.57)^{6} = 0.1032$

$P(X = 2) = C_{7,2}.(0.43)^{2}.(0.57)^{5} = 0.2336$

$P(X = 3) = C_{7,3}.(0.43)^{3}.(0.57)^{4} = 0.2937$

$P(X = 4) = C_{7,4}.(0.43)^{4}.(0.57)^{3} = 0.2216$

Then

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0195 + 0.1032 + 0.2336 + 0.2937 + 0.2216 = 0.8716$

0.8716 = 87.16% probability that in a sample of 7 customers, fewer that 5 will order a nonalcoholic beverage.