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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 $100 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.]

Indicate what rule of expected value you are using.

1. E(aX + b) = E(X) + b.
2. E(aX + b) = a · E(X) + b
3. E(aX + b) = a · E(X).
4. E(aX + b) = a2 · E(X)

Respuesta :

Answer:

Step-by-step explanation:

Given 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.

Each purchaser is independent of the other and there are only two outcomes

X - no of persons who prefer new copies is Bin (15, 0.3)

a) E(x) = np =[tex]15(0.3) =4.5[/tex]

Var(x) = npq = [tex]4.5*0.7\\=3.15[/tex]

Std dev (x) = 1.775

b)[tex]P(X>4.5+2(1.775)\\= P(X>8.050)\\= P(9)+P(10)+...+P(15)\\= 1-F(8)\\= 0.0152[/tex]

c) [tex]P(X\leq 10)\\=F(10)\\=0.9993[/tex]

d) h(X) = revenue when x of 15 purchasers want new copies

E(h(x)) = 100(E(x) = 450

Here h(x) = 100x

Hence E(100x) = 100E(X)

Using the concept of standard deviation to conclude the points;

  • The standard deviation is 1 - F(8).
  • New copy more than the mean value is 0.0152.
  • The probability that all 15 0.9993.
  • E(100x) = 100E(x)

What is a standard deviation?

It is the measure of the dispersion of statistical data. Dispersion is the extent to which the value is in a variation.

Given

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)  The mean value and standard deviation of the number who want a new copy of the book

E(x) np = 15(0.3) = 4.5

Var(x) = npq = 4.5 (0.7) = 3.15

Standard deviation (x) = 1.775

[tex]\rm = P(x>4.5+2*1.775)\\\\= P(x>8.050)\\\\= P(9)+P(10)+....+P(15)\\\\= 1-F(8)[/tex]

b) The number who want new copies is more than two standard deviations away from the mean value.

[tex]\rm = 0.0152\\\\[/tex]

c) The probability that all 15 will get the type of book they want from the current stock

[tex]\rm = P(x\leq 10)\\\\= F(10)\\\\ = 0.9993[/tex]

d) h(x) = revenue when x of 15 purchase want new copies

E(H(x)) = 100(E(x)) = 450

Here h(x) = 100x

Hence E(100x) = 100E(x)

More about the standard deviation link is given below.

https://brainly.com/question/12402189

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