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 25purchasers.

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=the number who want a new copy. For what values of X will all 15 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 5o 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: Leth(X)=the revenue when X of the 25 purchasers want new copies. Express this as a linear function. }

Answer 1

From the given information, \(30 \%\) of all students buy a new copy of the text book, whereas \(70 \%\) of the students buy a used copy.

a)

Compute the mean and standard deviation of the number who want a new copy.

Let \(X\) is the number of students out of 25 who want to buy a new copy of the book.

Here \(X \sim \operatorname{Binom}(n=25, p=0.3)\).

Now, the mean of \(X\) :

$$ \begin{aligned} \mu_{X} &=E(X) \\ &=n p \\ &=25(0.3) \\ &=7.5 \end{aligned} $$

The standard deviation of \(X\) :

$$ \begin{aligned} \sigma_{x} &=\sqrt{n p q} \\ &=\sqrt{25(0.3)(0.7)} \\ &=2.2913 \end{aligned} $$

b)

Compute the probability that the number who wants new copies is more than two standard deviations away from the mean. That is, \(P\left(\left|X-\mu_{X}\right|>2 \sigma_{X}\right)\).

$$ \begin{aligned} P\left(\left|X-\mu_{X}\right|>2 \sigma_{X}\right) &=P(|X-7.5|>2 \cdot 2.29) \\ &=P(X>7.5+4.58 \text { or } X<7.5-4.58) x="">12.08)+P(X<2.92) x="">12)+P(X \leq 2) \\ &=1-B(12 ; 25,0.3)+B(2 ; 25,0.3) \\ &=1-0.9825+0.0090\left(\begin{array}{l} \text { Use Excel } \\ = & \text { BINOM.DIST(12,25,0.3,TRUE) } \\ = & \text { BINOM.DIST( } 2,25,0.3, \text { TRUE }) \end{array}\right) \\ &=0.0264 \end{aligned} $$

Therefore, the probability that the number who wants new copies is more than two standard deviations away from the mean is 0.0264 .

c)

All 25 people get the type of book they want if the number of people who want a new copy is at most \(15(X \leq 15)\) and the number of people who want an old copy is at most \(15(25-X \leq 15\), or \(X \geq 10)\)

Thus, the probability that all 25 people get the type of book they want is given by,

\(P(10

$$ \begin{array}{l} =B(15 ; 25, .3)-B(9 ; 25, .3) \quad\left(\begin{array}{l} \text { use Excel } \\ =\text { BINOM.DIST(15,25,0.3,TRUE) } \\ =\text { BINOM.DIST(9,25,0.3,TRUE) } \end{array}\right) \\ =0.9995-0.8106 \\ =0.1890 \end{array} $$

Therefore, the required probability is \(\mathbf{0 . 1 8 9 0}\).

d)

The total revenue of the next 25 copies purchased is

$$ \begin{aligned} h(X) &=\$ 100 \cdot X+\$ 70 \cdot(25-X) \\ &=\$ 30 \cdot X+\$ 1750 \end{aligned} $$

Now the expected value of the total revenue is given by,

$$ \begin{aligned} E[h(X)] &=E(\$ 30 \cdot X+\$ 1750) \\ &=\$ 30 \cdot E(X)+\$ 1750\left(\begin{array}{l} \text { By using } \\ E(a X+b)=a E(X)+b \end{array}\right) \\ &=\$ 30 \cdot 7.5+\$ 1750 \\ &=\$ 1975 \end{aligned} $$

Therefore, the expected revenue is \(\$ 1975\).