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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 4 customers at Anita's, exactly 3 order their food to go?

Answer :

opudodennis

Answer:

[tex]P(X=3)=0.2700[/tex]

Step-by-step explanation:

#Notice this is a binomial probability distribution function with n=4,x=3 and p=52%=0.52

-The binomial probability function is expressed as:

[tex]PX=x)={n\choose x}p^x(1-p)^{n-x}[/tex]

Where:

n-is total number of events

x-total number of successful events

p-is the probability of success.

Hence, the probability that, in a random sample of 4 customers at Anita's, exactly 3 order their food to go is:

[tex]PX=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X=3)={4\choose3}0.52^30.48^1\\\\=0.2700[/tex]

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