only the answer to question D please 5. A fast-food restaurant sells hamburgers

Question

only the answer to question D please

5. A fast-food restaurant sells hamburgers and chicken sandwiches. On a typical weekday the demand for hamburgers is normally distributed with mean 313 and standard deviation 57; the demand for chicken sandwiches is normally distributed with mean 93 and standard deviation 22. a. How many hamburgers must the restaurant stock to be 98% sure of not running out on a given day? b. Answer part a for chicken sandwiches. c. If the restaurant stocks 400 hamburgers and 150 chicken sandwiches for a given day, what is the probability that it will run out of hamburgers or chicken sandwiches (or both) that day? Assume that the demand for hamburgers and the demand for chicken sandwiches are probabilistically independent.

d. The restaurant opens 7 days a week. Assume that the daily demand for both hamburgers and sandwiches are independent. In any given week, what is the probability that the restaurant will run out of hamburgers or chicken sandwiches (or both) in no more than () 1 day?

Explanation / Answer

Ans:

d)First calculate the probability of out of stock for single day

Let H = hamburger and C = chicken

P(H > 400) =P(z>1.526)= 0.0635

P(C > 150) =P(z>2.59)= 0.0048

P(Both) = 0.0635 x 0.0048 = 0.0003

So,P(H or C or both out o stock)=0.0635+0.0048+0.0003=0.0686

Now we will use Binomial distribution with n=7,p=0.0686

P(x<=1)=BINOMDIST(1,7,0.0686,TRUE)=0.9216

So,the required probability that the restaurant will run out of hamburgers or chicken sandwiches (or both) in no more than () 1 day is 0.9216

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