Suppose that 30% of all students who have to buy a text for a particular course

Question

Suppose that 30% of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other 70% want a used copy. Consider randomly selecting 15 purchasers.

(b)

What is the probability that the number who want new copies is more than two standard deviations away from the mean value? (Round your answer to three decimal places.)

(c)

The bookstore has 10 new copies and 10 used copies in stock. If 15 people come in one by one to purchase this text, what is the probability that all 15 will get the type of book they want from current stock? [Hint: Let X = the number who want a new copy. For what values of X will all 15 get what they want?] (Round your answer to three decimal places.)

(d)

Suppose that new copies cost $140 and used copies cost $60. Assume the bookstore currently has 50 new copies and 50 used copies. What is the expected value of total revenue from the sale of the next 15 copies purchased? [Hint: Let h(X) = the revenue when X of the 15 purchasers want new copies. Express this as a linear function.]

Explanation / Answer

Probability that students who have to buy a text for a particular course want a new copy = 0.30

(b) What is the probability that the number who want new copies is more than two standard deviations away from the mean value?

Answer :

Mean value = 15 * 0.3 = 4.5

Standard Deviation = Sqrt ( 0.3 * 0.7 * 15) = 1.7748

so Mean + 2 standard deviation = 4.5 + 2 * 1.7748 = 8.05

so We have to find the probability that more 8.05 or say 9 out of 15 people will want a new copy.

so Pr( X >= 9 ; 4.5; 1.7748) = 0.01524

but if here we have done normall approximation to the binomial so

Pr (X > + 2 ) = 1 -0.9772 = 0.0228

(c) New copies = 10

Old Copies = 10

As we have to find the probability that out of 15 people, probability that no more than 10 people will take new copies and same as no more than 10 people will take old copies.

Let X = the number who want a new copy.

so Pr(X<= 10) WHere Mean number = 4.5 and Std. deviation (X) = 1.7748

so Pr( X<= 10; 4.5 ; 1.7748) = 0.9997

but here the problem is with the people who wants a old copy.

LEts Y = People who wants a used copy where there is only 10 used Copy.

Mean number of people who wants a used copy = 15 * 0.7 = 10.5

standard deviation of such number of people = 1.775

so Pr (Y <= 10; 10.5; 1.775) =?

Z = ( 10 - 10.5)/ 1.775= -0.28

Pr (Y <= 10; 10.5; 1.775) = 0.3897

so Probability that each one will get the what he wants = Pr( X<= 10; 4.5 ; 1.7748) * Pr (Y <= 10; 10.5; 1.775)  

= 0.9997 * 0.3897 = 0.3896

d) Here there are sufficient number of new and used copies so there is no concern of anyone getting deprived of getting any copies.

Let h(X) = the revenue from New copies when X of the 15 purchasers want new copies.

g(X) = the revenue from old copies when X of the 15 purchasers want new copies.

Total revenue = h(X) + g(X)

= 140 * X + 60 * (15-X) = 900 + 80X

Total Revenue = 900 + 80X

so E(Total Revenue) = E(900 + 80X)

= 900 + 80 E(X) = 900 + 80 * 4.5 = $1260

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