Question

At a Noodles & Company restaurant, the probability that a customer will order a nonalcoholic beverage is .50. Find the probability that in a sample of 14 customers, at least 7 will order a nonalcoholic beverage

Answer 1

For each customer, there are only two possible outcomes. Either they will order an alcoholic beverage, or they will not. The probability of a customer ordering an alcoholic 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 Noodles & Company restaurant, the probability that a customer will order a nonalcoholic beverage is .50

This means that $p = 0.5$

Sample of 14 customers

This means that $n = 14$

Probability that at least 7 will order a nonalcoholic beverage

This is:

$P(X \geq 7) = 1 - P(X < 7)$

In which

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

Then

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

$P(X = 0) = C_{14,0}.(0.5)^{0}.(0.5)^{14} = 0.0001$

$P(X = 1) = C_{14,1}.(0.5)^{1}.(0.5)^{13} = 0.0009$

$P(X = 2) = C_{14,2}.(0.5)^{2}.(0.5)^{12} = 0.0056$

$P(X = 3) = C_{14,3}.(0.5)^{3}.(0.5)^{11} = 0.0222$

$P(X = 4) = C_{14,4}.(0.5)^{4}.(0.5)^{10} = 0.0611$

$P(X = 5) = C_{14,5}.(0.5)^{5}.(0.5)^{9} = 0.1222$

$P(X = 6) = C_{14,6}.(0.5)^{6}.(0.5)^{8} = 0.1833$

So

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0001 + 0.0009 + 0.0056 + 0.0222 + 0.0611 + 0.1222 + 0.1833 = 0.3954$

$P(X \geq 7) = 1 - P(X < 7) = 1 - 0.3954 = 0.6046$

0.6046 = 60.46% probability that at least 7 will order a nonalcoholic beverage.

For more on the binomial distribution, you can check brainly.com/question/15557838