3. [6 points] Suppose that 30% of all students who have to buy a text for a part

Question

3. [6 points] Suppose that 30% of all students who have to buy a text for a particular course want a used copy. Consider randomly selecting 25 want a new copy, whereas the other 70% (a) What are the mean value and standard deviation of the number who want a new copy of the book? (b) What is the probability that the number who want new copies is more than two standard deviations away from the mean value? (c) The bookstore has 15 new copies and 15 used copies in stock. If 25 people come in one by one to purchase this text, what is the probability that all 25 will get the type of book they want from current stock? Hint: Let X be the number who want a new copy. For what values of X will all 25 get what they want? (d) Suppose that new copies cost $100 and used copies cost $70. 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 25 copies purchased? Be sure to indicate what rule of expected value you are using. [Hint: Let h(X) be the revenue when X of the 25 purchasers want new copies. Express this as a linear function.]

Explanation / Answer

Let X = number of students who want to buy new copies in a sample of 25.

Then, X ~ B(25, p), where p = probability a student would want a new copy, which is estimated by the proportion of students who want to buy new copies. Thus, from the given data, p = 0.3 and so X ~ B(25, 0.3) ……………………………………………………….(A)

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then

probability mass function (pmf) of X is given by

p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n ………………………………..(1)

[The above probability can also be directly obtained using Excel Function of Binomial Distribution] ……………………………………………………………………………….(1a)

Mean (average) of X = E(X) = np…………………………………………………………..(2)

Variance of X = V(X) = np(1 – p)…………………………………………………………..(3)

Standatd Deviation of X = SD(X) = { np(1 – p)} ………………………………………...(4)

For any random variable, X, E(aX + b) = aE(X) + b, where a and b are constants. ……….(5)

Part (a)

Vide (A), (2) and (4) above,

Mean value of number of students who want to buy new copies = 25 x 0.3 = 7.5 ANSWER 1

SD of number of students who want to buy new copies = sqrt(25 x 0.3 x 0.7)

= 2.29 ANSWER 2

Part (b)

Probability that the number of students who want to buy new copies is more than 2 SD away from the mean value

= P[X > {7.5 + (2 x 2.29)}

= P(X > 12.08)

= P(X 13)

= 1- 0.98253 [using Excel Function of Binomial Distribution, n = 25, p = 0.3, x 12]

= 0.01747 ANSWER

Part (c)

Given 15 copies each of new and used copies available, all students will get what copies they want if x = 10 (1) 15, where x is as defined earlier. Hence, probability that all students will get what copies they want = P(10 X 15)

= P(X 15) - P(X 9)

= 0.999546 – 0.810564 [using Excel Function of Binomial Distribution, n = 25, p = 0.3,

x 15, x 9]

= 0.188982 ANSWER

Part (d)

Revenue, R = 100x + 70(25 - x)

= 25x + 1750

So, expected revenue = E(R) = E(25x + 1750)

= 25E(X) + 1750 [vide (5)]

= (25 x 7.5) + 1750

= $1937.50 ANSWER

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