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Chapter 5: Q 5.110 (page 224)

A variable y of a finite population has the following frequency distribution:

y1234
f22106

Suppose a member is selected at random from the population and let Y denote the value of the variable y for the member obtained.

a. Determine the probability distribution of the random variable Y.

b. Use random-variable notation to describe the events that Y takes on the value 3, a value less than 3, and a value of at least 3.

c. Find P(Y = 3), P(Y <3), and P(Y 鈮� 3). Interpret your results.

d. Construct a probability histogram for the random variable Y.

Short Answer

Expert verified

Part a.

Random Variable (Y)

Probability

1

2/20 = 0.1

2

2/20 = 0.1

3

10/20 = 0.5

4

6/20 = 0.3

Part b.

On the value of 3: Y = 3

A value less than 3: Y < 3

A value of at least 3 : Y鈮� 3

Part c.

P(Y=3) = 0.5

P(Y < 3) = 0.1 + 0.1 = 0.2

P(Y 鈮� 3) = 0.5 + 0.3 = 0.8

Part d.

Step by step solution

01

Part (a) Step 1. Given information

The frequency distribution of a variable x in a finite population is as follows: :

y

1

2

3

4

f

2

2

10

6

Assume a member is chosen at random from the population, and Y is the value of the variable y for that member.

02

Part (a) Step 2. Formula Used 

The experiment is conducted out assuming that the random variable (X=x) occurs n times out of a total of N times. As a result of the f/N rule:

P (X=x) = n/N

03

Part (a) Step 3. Solution

Here, N = 2+2+10+6 = 20. As a result, the probability of random variable Y is as follows:

Random Variable

Probability

1

2/20 = 0.1

2

2/20 = 0.1

3

10/20 = 0.5

4

6/20 = 0.3

04

Part (b) Step 1. Solution 

Event where Y takes the value 3, can also be written as:

Y = 3

Event where Y takes a value less than 3, can also be written as :

Y < 3

Event where Y takes a value of at least 3, can also be written as :

Y 鈮� 3

05

Part (c) Step 1. Solution

From the table,

P(Y = 3) = 0.5

P(Y < 3) = 0.1 + 0.1 = 0.2

P(Y 鈮� 3 ) = 0.5 + 0.3 = 0.8

06

Part (d) Step 1. Solution

The probability histogram for the random variable Y is as follows:

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Most popular questions from this chapter

The Hypergeometric Distribution. In this exercise, we discuss the hypergeometric distribution in more detail. When sampling is done without replacement from a finite population, the hypergeometric distribution is the exact probability distribution for the number of members sampled that have a specified attribute. The hypergeometric probability formula is

PX=x=NpxN1-pn-xNn,

where Xdenotes the number of members sampled that have the specified attribute, Nis the population size, nis the sample size, and pis the population proportion.

To illustrate, suppose that a customer purchases 4 fuses from a shipment of 250, of which 94 % are not defective. Let a success correspond to a fuse that is not defective.

(a) Determine N, n, and p.

(b) Apply the hypergeometric probability formula to determine the probability distribution of the number of nondefective fuses that the customer gets.

Key Fact 5.6 shows that a hypergeometric distribution can be approximated by a binomial distribution, provided the sample size does not exceed 5% of the population size. In particular, you can use the binomial probability formula

P(X=x)=nxpx1-pn-x

with n=4andp=0.94, to approximate the probability distribution of the number of nondefective fuses that the customer gets.

(c) Obtain the binomial distribution with parameters n=4andp=0.94.

(d) Compare the hypergeometric distribution that you obtained in part (b) with the binomial distribution that you obtained in part (c).

Craps. The game of craps is played by rolling two balanced dice. A first roll of a sum of 7 or 11 wins; and a first roll of a sum of 2,3 , or 12 loses. To win with any other first sum, that sum must be repeated before a sum of 7 is thrown. It can be shown that the probability is 0.493 that a player wins a game of craps. Suppose we consider a win by a player to be a success, s

a. Identify the success probability, p

b. Construct a table showing the possible win-lose results and their probabilities for three games of craps. Round each probability to three decimal places.

c. Draw a tree diagram for part (b).

d. List the outcomes in which the player wins exactly two out of three times.

e. Determine the probability of each of the outcomes in part (d). Explain why those probabilities are equal.

f. Find the probability that the player wins exactly two out of three times.

g. Without using the binomial probability formula, obtain the probability distribution of the random variable Y, the number of times out of three that the player wins.

h. Identify the probability distribution in part (g).

What does it mean three events to be mutually exclusive.?

In Exercises 5.16-5.26, express your probability answers as a decimal rounded to three places.

Preeclampsia. Preeclampsia is a medical condition characterized by high blood pressure and protein in the urine of a pregnant woman. It is a serious condition that can be life-threatening to the mother and child. In the article "Women's Experiences Preeclampsia: Australian Action on Preeclampsia Survey of Wom and Their Confidants" (Journal of Pregnancy,Vol. 2011, Issue 1, Article ID 375653), C. East et al. examined the experiences of 68 women with preeclampsia. The following table provides a frequency distribution of instances of prenatal or infant death for infants of women with preeclampsia.

Suppose that one of these women with preeclampsia is randomly selected. Find the probability that the child of the woman selected

(a) died.

(b). died one week to six months after birth.

(c). lived at least six weeks.

Equipment breakdowns: A factory manager collected data on the number of equipment breakdown per day. From those data she derived the probability distribution shown in the following table, Where W denote the number of breakdown on a given day.

Part (a) Determine 渭wand蟽w, Round your answer for the standard deviation to three decimal places.

Part (b) On average, how many breakdown occur per day?

Part (c) About how many breakdown are expected during a 1 year period assuming 250 work days per year?
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