Sahara, Lexington\'s best restaurant, is considering an all-you-can-eat special

ID: 2506217 • Letter: S

Question

Sahara, Lexington's best restaurant, is considering an all-you-can-eat special on their famous

spicy chicken entree. Based on a similar promotion a few years back, the owners have assembled a probability

distribution for the random variable X, representing the number of servings of chicken a customer will eat:

x p(x)

1 .210

2 .535

3 .199

4 .056

a. Determine the expected value, variance, and standard deviation of X.

b. Suppose that a serving of chicken costs the restaurant $5. What is the minimum price for the all-you-

can-eat option so that to break even in expectation?

c. What is the minimum price for the all-you-can-eat option so that they break even or better with 94.4%

probability?

a) Expected Value = sum (x*p(x)) = 1*0.21+ 2*0.535 + 3*0.199 + 4 *0.056 = 2.101

E(x^2)= (1)^2*(0.210)+(2)^2*(0.535)+(3)^2*(0.199)+ (4)^2*(0.056)=5.037

Variance = E(x^2)- [E(x)]^2 = 5.037 -(2.101)^2 = 0.6228

Standard Deviation = sqrt(0.6228) =0.7892

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.