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.

(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 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?]

(d) Suppose that new copies cost $150 and used copies cost $90. 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.]

Indicate what rule of expected value you are using.

$$ \begin{aligned} &E(a X+b)=a \cdot E(X) . \\ &E(a X+b)=a^{2} \cdot E(X) \\ &E(a X+b)=E(X)+b . \\ &E(a X+b)=a \cdot E(X)+b \end{aligned} $$

Answer 1

a)

mean E(x)=μ=np=4.5

standard deviation σ=√(np(1-p))=1.77

b)

2 standard deviation from mean values are μ -/+ 2*σ=(0.95,8.05)

P(more than 2 standard deviation from mean) =P(X=0 )+P(X>=9)

=P(1<=X<=8)=

1-∑x=ab     (nCx)px(1−p)(n-x)    =

1-0.9800 =0.0200

c)

P(5<=X<=10)=

∑x=ab     (nCx)px(1−p)(n-x)    =

0.484

d)

expected value =4.5*150+(15-4.5)*90 =1620