chapter 4 random variables and...

CHAPTER 4 RANDOM VARIABLES AND DISTRIBUTIONS

Every baby is born either male or female. So every child is the outcome of a human binomial experiment.

HS_PS_S1_04_CO.indd 130HS_PS_S1_04_CO.indd 130 8/31/11 6:49 PM8/31/11 6:49 PM

Copyright © 2011, K12 Inc. All rights reserved. This material may not be reproduced in whole or in part, including illustrations, without the express prior written consent of K12 Inc.


In This ChapterYou will learn how random variables and probability distributions can describe problem situations. You’ll also learn how to use expected value to make decisions. Specific discrete and continuous distributions, such as binomial and normal distributions, will be covered as well as techniques for using the standard normal distribution to compare outcomes.

Topic List ► Chapter 4 Introduction

► Creating Probability Distributions

► Interpreting Probability Distributions

► Expected Value

► Binomial Distributions

► Continuous Random Variables

► The Normal Distribution

► Standardizing Data

► Comparing Scores

► The Standard Normal Curve

► Finding Standard Scores

► Chapter 4 Wrap-Up

In a family with one child, what is the chance that the child is a girl? In a family with two children, what is the chance that both of them are girls? What about the chance of having three girls, or four, or five? Random variables can help describe situations like these.

RANDOM VARIABLES AND DISTRIBUTIONS 131

HS_PS_S1_04_CO.indd 131HS_PS_S1_04_CO.indd 131 8/31/11 6:49 PM8/31/11 6:49 PM



Though individual events can be random, they can yield patterns when repeated multiple times.

Probability Distributions

It is often predicted that the probability that a newborn baby will be a boy (or girl, as the case may be) is 50%. Due to the natural gender ratio, however, the actual probability that a newborn will be a boy is not exactly 50%. In the United States, for example, the chances are slightly higher that a newborn will be a boy rather than a girl.

The relative frequency histogram shows the proportions for the number of boys in a family with 3 children, given the predicted probability of 50%.

According to this graph, there is a 37.5% chance of having exactly 1 boy out of 3 children. You can also see that there is a 25% chance that all 3 children will have the same gender (just add the heights of the first and last bar).

The graph for the actual number of boys in a family with 3 children, however, tells a slightly different story. The relative frequency histogram estimates the actual proportions considering the natural gender ratio.

Notice that, compared to the first graph, the chances are greater that a family of 3 will have mostly boys (as indicated by the higher bars for 2 and 3 boys). Although there are slightly more all-boy families than would be predicted by pure probability, the histograms are quite similar.

Chapter 4 Introduction

Probability Distributions

ItIt i iss ofoftten predicted that the probability that a newborn baby will be aboy (or girl, as the case may be) is 50%. Due to the natural gender ratio, however, the actual probability that a newborn will be a boy is not exactly 50%. In the United States, for example, the chances are slightly higher that % , p , g y g

40

35

30

25

20

15

10

5

0 1 2Number of boys

Number of Boys in Family(Predicted)

30

Pro

babi

lity

(%)

40

35

30

25

20

15

10

5

0 1 2Number of boys

Number of Boys in Family(Actual)

30

Pro

babi

lity

(%)

132 CHAPTER 4 RANDOM VARIABLES AND DISTRIBUTIONS

HS_PS_S1_04_IN_RG.indd 132HS_PS_S1_04_IN_RG.indd 132 9/17/11 2:24 AM9/17/11 2:24 AM




Applying It

In this chapter, you will learn how to create and interpret probability and binomial distributions using both tables and histograms. You will work with real data to learn about expected value and continuous random variables. You will also learn about normal distributions and how the standard normal curve is used to interpret data and make predictions.

The Normal Distribution

When a data set has a normal distribution, a histogram that represents the data set is symmetric and has a bell shape.

The symmetry of this curve demonstrates that more data are near the mean than in the outlying regions.

The normal distribution has properties that are very useful in statistics. When data are normally distributed, about 68% of the data values are within a range of 1 standard deviation above and below the mean.

For example, suppose the average length of a full-term newborn baby is 20.1 inches with a standard deviation of 0.4 inches. Because lengths of babies are normally distributed, we can assume that roughly 68% of newborn babies are between 19.7 and 20.5 inches in length. These values are obtained by subtracting and adding 0.4 from 20.1.

Other properties of the normal distribution will be explored later in the chapter.

Thhe Normal Distributionh

WWhhenen a d data set has a normal distribution, a histogram that represents theWhWhdata set is symmetric and has a bell shape.at

The symmetry of this curve demonstrates that more data are near the meanhth i th tl i ithaan in the outlying regionsa

Mean


19.3 19.7 20.1

68% of newbornbabies

20.5 20.9

CHAPTER 4 INTRODUCTION 133

HS_PS_S1_04_IN_RG.indd 133HS_PS_S1_04_IN_RG.indd 133 9/17/11 2:24 AM9/17/11 2:24 AM



Number of Children per Family

10

12

8

6

4

2

01 2 3 4 5 6

Children

Fam

ilies

Preparing for the ChapterReview the following skills to prepare for the concepts in Chapter 4.► Find probabilities of events.

► Interpret histograms.

► Determine areas of figures in the coordinate plane.

► Solve linear equations in the form c = x − a _____ b .

► Simplify the expression 1 − x for different values of x.

Problem Set

A stack of 22 cards are numbered 1 to 22. If a card is randomly selected, find the probability of the event.

1. The number 4 is selected.

2. An even number is selected.

3. An odd number less than 8 is selected.

4. A prime number or an even number is selected.

5. A number less than 11 is selected.

6. A number less than 17 or greater than 20 is selected.

The histogram shows the number of children in 32 selected families.

7. How many families have 5 children?

8. How many families have 2 or more children?

9. How many families have from 1 to 3 children?

10. What percent of the families have from 2 to 4 children?


HS_PS_S1_04_IN_PS.indd 134HS_PS_S1_04_IN_PS.indd 134 9/24/11 1:54 AM9/24/11 1:54 AM




x

1

1

0

234567

2 3 4 5 6 7y

x

1

1

0

2345

2 3 4 5 6 7 8 9y

x

1

1

0

23456

2 3 4 5 6 7 8 9y

x

1

1

0

23456

2 3 4 5 6 7 8 9 10 11y

Determine the area of the shaded figure.

Determine the value of x.

Find the value of the expression 1 − x for the given value of x.

11.

12.

13.

14.

15. 1.1 = x − 30 ______ 3

16. 2.05 = x − 1.22 ________ 0.1

17. −1.15 = x − 120 _______ 4.8

18. −2.5 = x − 450 _______ 12

19. x = 3 __ 8 21. x = 2 __ 9 23. x = 0.381

20. x = 0.92 22. x = 0.3072 24. x = 0.0011

CHAPTER 4 INTRODUCTION 135

HS_PS_S1_04_IN_PS.indd 135HS_PS_S1_04_IN_PS.indd 135 9/17/11 2:29 AM9/17/11 2:29 AM



X 0–9 10–19 20–29 30–39 40–49

f 81 73 35 47 14

X 0–9 10–19 20–29 30–39 40–49

P 0.324 0.292 0.14 0.188 0.056

REMEMBERThe letter X is used to represent a discrete random variable. For example, when a coin is tossed, the variable X would represent the two outcomes: heads and tails. These outcomes are both discrete and random.

THINK ABOUT ITThe sum of the probabilities in a probability distribution table always equals one.

Creating Probability DistributionsProbabilities of all outcomes of a discrete random variable can be summarized in a table.

Creating Probability Distribution Tables

A study of a new treatment for high cholesterol undergoes clinical testing with the results summarized in this frequency table.

In this table, X stands for the number of points that the patient’s cholesterol level decreased, and f stands for the number of patients. For instance, 73 patients had their cholesterol decrease between 10 and 19 points after the new treatment.

A patient wants to know the probability that the new treatment will lower his cholesterol from 20 to 29 points. Here, a probability distribution table would be useful because it shows the probabilities for all of the outcomes.

For the frequency table describing the clinical testing, the total number of frequencies is 81 + 73 + 35 + 47 + 14 = 250. So 250 patients took part in the study. Here is the complete probability distribution table.

According to this table, the probability that the new treatment will lower a patient’s cholesterol between 20 and 29 points is 0.14 or 14%.

To create a probability distribution table from a frequency table

Step 1 Find the sum of the frequencies.

Step 2 Divide each frequency by the sum found in Step 1.

Step 3 Replace each frequency in the table with the ratio found in Step 2.

Step 4 Change the label f to P.

HOW TO CREATE A PROBABILITY DISTRIBUTION TABLE


HS_PS_S1_04_T01_RG.indd 136HS_PS_S1_04_T01_RG.indd 136 9/17/11 2:33 AM9/17/11 2:33 AM



3

4

2 2

1

1

THINK ABOUT ITEvery probability distribution for a discrete random variable must satisfy both properties.

X 1 2 3 4

P 1 __ 3 1 __ 3 1 __ 6 1 __ 6

X 1 2 3 4 5

P 0.5 0.1 0.3 0.4−0.3

X 5 10 15 20 25 30

P 0.34 0.5 0.101 0.32 0.34 0.122

Every probability distribution for a discrete random variable has the following properties:• Each probability is a number between 0 and 1.• The sum of all probabilities is equal to 1.

PROPERTIES OF PROBABILITY DISTRIBUTIONS

In a different study, patients are separated into four groups according to a spinner.

For the spinner, there are four different outcomes: Group 1, Group 2, Group 3, and Group 4. There are 6 sections overall.

Notice that some outcomes are more likely than others. For instance, there are two sections for groups 1 and 2, but only one section for groups 3 and 4.

The probability distribution table shows that the chances of being in Group 1 or Group 2 are higher than the chances of being in Group 3 or Group 4.

This probability distribution table violates the first property because each probability is not a number between 0 and 1. So it is not a probability distribution.

This probability distribution table violates the second property because the sum of all probabilities is not equal to 1. So it is not a probability distribution.

Properties of Probability Distributions

There are two properties that every probability distribution table has.

CREATING PROBABILITY DISTRIBUTIONS 137





Problem Set

A survey asked a group of people the size of their households. The results are in the table.

1. How many people were surveyed?

2. How many people reported a household size of 4?

3. Complete the probability distribution table.

Match the frequency table with the probability distribution table.

A. B. C. D.

4. 5. 6. 7.

Household Size 1 2 3 4 5

ƒ 4 12 6 2 1

Household Size 1 2 3 4 5

P 0.08

X ƒ

0 5

1 10

2 15

3 5

4 15

X ƒ

0 10

1 12

2 5

3 13

4 10

X ƒ

0 10

1 10

2 10

3 10

4 10

X ƒ

0 5

1 6

2 16

3 8

4 15

X P

0 0.1

1 0.12

2 0.32

3 0.16

4 0.3

X P

0 0.1

1 0.2

2 0.3

3 0.1

4 0.3

X P

0 0.2

1 0.24

2 0.1

3 0.26

4 0.2

X P

0 0.2

1 0.2

2 0.2

3 0.2

4 0.2

HS_PS_S1_04_T01_PS.indd 138HS_PS_S1_04_T01_PS.indd 138 9/17/11 2:45 AM9/17/11 2:45 AM



CREATING PROBABILITY DISTRIBUTIONS 139

Create a probability distribution table.

10. In a medical study, the heights of a group of men were measured to the nearest inch. There were 120 men with heights of 60–64 inches, 280 men with heights of 65–69 inches, 312 men with heights of 70–74 inches, and 162 men with heights of 75–79 inches.

11. The Mosquitoes are playing the Gnats in a three-game series. The probability is 0.25 that the Gnats will win all 3 games, 0.4 that they will win 2 of the games, 0.2 that they will win 1 of the games, and 0.15 that they will not win any of the games.

12. What is the probability of rolling a 1, 2, 3, 4, 5, or 6 when rolling a 6-sided number cube?

13. There is a 40% chance it will rain tomorrow, a 30% chance it will snow, and a 30% chance it will be partly cloudy.

8.

9.

Explain why the table is not a probability distribution.

14. 15.

Find the value of n using the probability distribution table.

16. 17.

Solve.

18. Challenge Create a probability distribution for rolling two 6-sided number cubes, where X is the sum of the numbers on the number cube, and P is the probability of obtaining the sum.

0

1

2

3

4

5

6

Yellow Red Orange GreenBlue

0

1

2

3

4

5

6

7

8

9

10

Tue Wed Fri Sat SunThuMon

Result P

0 0.25

1 0.45

2 0.0

3 0.35

Length P

5–14 0.25

15–24-0.3

25–34 0.4

35–44 0.5

45–54 0.15

Number of Guests per Room

0 1 2 3 4

P 0.42 0.31 0.16 n 0.09

Color red blue gold silver

P 1 __ 3 1 __ 8 n 1 __ 6




X 10 11 12 13 14

P 0.15 0.3 0.2 0.25 0.1

Interpreting Probability DistributionsThe likelihood of many events can be summarized by a table or histogram.

Interpreting Probability Distribution Tables

A summer camp program director is planning activities for the summer. The probability distribution table shows the probability P that the age of a randomly chosen camper is X.

For the first activity, the program director selects a camper at random to read announcements. What is P (X = 13): the probability that the camper selected is 13 years old?

According to the probability distribution table, when X = 13, the probability is P = 0.25. So, P(X = 13) = 0.25.

In other words, the probability of selecting a 13-year-old camper at random is 0.25, or 25%.

Suppose the program director wants to determine P(X ≥ 13), the probability that a randomly selected camper is at least 13 years old.

This can be answered by adding individual probabilities.

P (X ≥ 13) = P (X = 13) + P (X = 14)

= 0.25 + 0.1

= 0.35

There is a 35% probability of selecting a camper who is at least 13 years old.





0.1

0.2

0.3

0.4

0.5

60 12 18 24

Ages of Staff Members

30 36 42 48 54X

P

Interpreting Probability Distribution Histograms

The program director now has to decide which staff members to assign to which activities. For this, she is using this probability distribution histogram.

In this histogram, X stands for the ages of individual staff members. P is the probability that if a single staff member is chosen at random, then the staff member will be that age.

For the first activity, the program director selects a staff member at random to lead the songs at dinner. What is P (18 ≤ X < 24): the probability that the staff member selected is at least 18 years old and less than 24 years old?

The first bar of the histogram represents 18 ≤ X < 24. The height of this bar is at the value P = 0.2. So the probability of selecting a staff member who is at least 18 years old and less than 24 years old is 0.2%, or 20%.

Probability distribution histograms can help determine more than just single entries. Suppose the program director wants to determine P (X ≥ 30), the probability that a randomly selected staff member is at least 30 years old.

This can be answered by adding individual probabilities.

P (X ≥ 30) = P (30 ≤ X < 36) + P (36 ≤ X < 42)

= 0.3 + 0.1

= 0.4

There is a 40% probability of selecting a staff member who is at least 30 years old.

REMEMBERA histogram is a graph showing the frequency distribution of data.

INTERPRETING PROBABILITY DISTRIBUTIONS 141





The probability distribution histogram shows the weight distribution of hamadryas baboons that live in a certain region. Find the probability of the event.

17. P (5 ≤ X < 20)

18. P (25 ≤ X < 35)

19. P (X < 15 or X ≥ 25)

20. P (X < 30)

21. P (X < 15)

22. P (X < 20 or X ≥ 30)

Problem Set

Use the probability distribution table to find the probability of the event.

1. P (X = 3)

2. P (X = 2 or X = 0)

3. P (1 < X ≤ 3)

4. P (0 < X < 3)

5. P (X > 2 or X = 0)

6. P (X ≠ 3)

7. P (X < 4)

8. P (X ≥ 4)

Use the probability histogram to find the probability of the event.

9. P (X = 6)

10. P (X = 1 or X = 3)

11. P (1 < X ≤ 6)

12. P (0 < X < 4)

13. P (X = 1 or X > 3)

14. P (X ≠ 6)

15. P (X > 4)

16. P (X ≤ 5)

1 2 3 4X

P

5 6 7

0.25

0.2

0.15

0.1

0.05

0

X 0 1 2 3 4 5

P 0.3 0.05 0.1 0.15 0.15 0.25

0.25

0.3

0.2

0.15

0.1

0.05

5 10 15 20 25Weight (lb)

Weights of Hamadryas Baboons

30 35 40 45

P




INTERPRETING PROBABILITY DISTRIBUTIONS 143

Use the Weights of Hamadryas Baboons probability histogram to find the probability of the event.

29. Challenge Two hamadryas baboons from the region are randomly selected. What is the probability that both weigh less than 30 pounds?

23. Create a probability distribution table for this situation. What is the probability that he does not visit the library on Friday?

24. Let X be the number of times he visits the library on Friday for the next two weeks. Create a probability distribution table for X. Find P (X > 0).

Each week, Evan randomly picks a weekday (Monday through Friday) to visit the library.

A message center receives calls from 6 p.m. until 6 a.m. Typically, 50% of the calls come in between 6 p.m. and 9 p.m.; 25% of the calls come in between 9 p.m. and midnight; 20% of the calls come in between midnight and 3 a.m.; and 5% of the calls come in between 3 a.m. and 6 a.m.

25. What is the probability that a call will come in between 6 p.m. and 9 p.m.?

26. What is the probability that a call will come in between 6 p.m. and midnight?

27. What is the probability that a call will come in between midnight and 6 a.m.?

28. What is the probability that a call will come in between 9 p.m. and 6 a.m.?




Expected ValueProbability distributions can tell us what to expect when all outcomes are considered.

Using Probability for Estimation

Each week the manager at a large restaurant orders eggs that are used for breakfast items at the restaurant. The probability distribution table shows the probability P that a customer will order a breakfast item with X number of eggs.

The food manager expects about 800 customers to have breakfast each week at the restaurant. How many of these customers are expected to order breakfast items with two eggs?

According to the probability distribution table, the probability that a customer will order a breakfast with two eggs is 0.42, or 42%.

Out of 800 patrons, 42% are expected to order breakfast items with two eggs. So about 0.42 · 800 = 336 customers are expected to order breakfast items with two eggs.

Finding Expected Value

The prediction above is based on a single outcome—that a customer will order a breakfast item that has two eggs.

What if the restaurant manager wants to know the average number of eggs for all breakfast items served at the restaurant?

The manager needs to know the expected value—the expected number of eggs in a typical breakfast served at the restaurant.

Suppose each outcome of a random variable X is x1, x2, . . . ., xn and has probability p1, p2, . . . . , pn, respectively. The average, or expected value, of X is

E(X ) = x1 · p1 + x2 · p2 + . . . + xn · pn.

EXPECTED VALUE

REMEMBERIn a probability distribution table, the sum of all probabilities is equal to 1, and each individual probability is a number between 0 and 1.

X P

0 0.29

1 0.18

2 0.42

3 0.11





The expected value is the sum of the products of the individual probabilities and their respective probabilities.

E(X ) = x1 · p1 + x2 · p2 + x3 · p3 + x4 · p4

= 0 · (0.29) + 1 · (0.18) + 2 · (0.42) + 3 · (0.11)

= 0 + 0.18 + 0.84 + 0.33

= 1.35

The expected value of 1.35 means that the food manager can expect that each breakfast ordered will have 1.35 eggs on average.

Solving Problems with Expected Value

The recommended amount of riboflavin per day is 1.7 milligrams. Suppose there is a 30% chance that a person will take 0 milligrams of riboflavin per day; a 40% chance that a person will take 1 milligram of riboflavin per day; a 20% chance that a person will take 2 milligrams of riboflavin per day; a 5% chance that a person will take 3 milligrams of riboflavin per day; and a 5% chance that a person will take 4 milligrams of riboflavin per day. What is the expected value of riboflavin intake per day?

E(X ) = x1 · p1 + x2 · p2 + x3 · p3 + x4 · p4

= 0 · (0.3) + 1 · (0.4) + 2 · (0.2) + 3 · (0.05) + 4 · (0.05)

= 0 + 0.4 + 0.4 + 0.15 + 0.2

= 1.15

So the expected value of riboflavin intake for a single person is about 1.15 milligrams per day. This value is below the recommended amount of 1.7 milligrams per day.

THINK ABOUT ITWith an expected value of 1.35 eggs per breakfast, the manager can predict that about 1.35 · (800) = 1080 eggs will be consumed in a week.

BY THE WAYRibofl avin is also known as vitamin B2.

EXPECTED VALUE 145





Problem Set

The table shows the distribution of household sizes in a certain city.

1. If there were 6000 households in the city, how many households would you expect to be of size 1?

2. If there were 4000 households in the city, how many households would you expect to be of size 4?

3. What is the expected value of household size in this city?

Determine the expected value of X.

Household Size P

1 0.16

2 0.48

3 0.24

4 0.08

5 0.04

4. 6. 8.

5. 7. 9.

X P

0 1 __ 3

1 1 __ 3

2 1 __ 6

3 1 __ 6

X P

0 1 __ 3

6 1 __ 3

12 1 __ 6

18 1 __ 6

X P

0 0.1

1 0.1

2 0.1

3 0.1

4 0.1

5 0.5

X P

0 0.5

1 0.1

2 0.1

3 0.1

4 0.1

5 0.1

X P

1998 0.42

1999 0.2

2000 0.15

2001 0.1

2002 0.13

X P

1998 0.2

2000 0.2

2002 0.2

2004 0.2

2006 0.2




EXPECTED VALUE 147

15. Challenge If the expected value of this distribution is 8.5, fi nd the value of m and n.

Solve.

The Columbus Crew of Major League Soccer had the following results last year.

Use the probability distribution table.

10. A clothing store has a section of jeans on sale. There is a 30% chance of picking a pair with a 30-inch inseam; a 25% chance of picking a pair with a 32-inch inseam; a 10% chance of picking a pair with a 34-inch inseam; a 10% chance of picking a pair with a 36-inch inseam; a 10% chance of picking a pair with a 38-inch inseam; and a 15% chance of picking a pair with a 40-inch inseam. Create a probability distribution table and fi nd the expected value of the inseam length of the jeans.

11. Raina is entering a 64-player cribbage tournament. The probability of her playing exactly 1 match is 1 __ 2 ; exactly 2 matches is 1 __ 4 ; exactly 3 matches is 1 __ 8 ; exactly 4 matches is 1 ___ 16 ; exactly 5 matches is 1 ___ 32 ; and exactly 6 matches is 1 ___ 32 . Create a probability distribution table for this situation, and fi nd the expected value of the number of matches Raina will play.

12. What is the expected value of the number of goals the Columbus Crew will score in a match?

13. If the Columbus Crew played 30 matches, convert the probability distribution table into a frequency table.

14. Determine the mean number of goals the Columbus Crew scored in a match. Explain why this is greater than, less than, or equal to the expected value.

Number of Goals 0 1 2 3

P 1 __ 3 1 __ 5 3 ___ 10 1 __ 6

X P

6 m

8 1 __ 2

n 1 __ 4




Binomial DistributionsA table or graph can be used to represent outcomes of a variable that has a binomial distribution.

Determining Whether a Variable Has a Binomial Distribution

A doctor knows that for a certain type of adult migraine headache, Medicine Q works about 70% of the time. If the medicine is tried with two different patients with this condition, the resulting probability distribution is a binomial distribution.

A variable has a binomial distribution if• There are a fi xed number of independent trials.• The outcome of each trial is success or failure.• The probability of success for each trial is the same.

BINOMIAL DISTRIBUTION

This situation meets all of the conditions for a binomial distribution.

• There are a fixed number of independent trials; there are two trials, and each patient taking Medicine Q is independent of the other.

• The outcome of each trail is a success or failure; the medicine will help the migraine (success) or it will not (failure).

• The probability of success for each trial is the same; there is a 0.7 probability of success when a patient takes Medicine Q.

Creating Binomial Distributions

Suppose later in the day, the doctor is seeing two patients who have this type of adult migraine headache. We can create a binomial distribution table or graph to represent outcomes for these two patients.

The variable X can take on three different values: Medicine Q will work for none of the patients, one of the patients, or both patients. So the variable X is equal to 0, 1, or 2.

THINK ABOUT ITTwo or more coin tosses have a binomial distribution. The probability of success and failure is the same—50%.





X 0 1 2

P 0.09 0.42 0.49 0.55

0.50.45

0.35

0.25

0.15

0.05

0.4

0.3

0.2

0

0.1

1 20 X

P

When X = 0, it means that Medicine Q failed to work for both patients. Since P (success) = 0.7, P (failure) = 0.3.

P(X = 0) = P (failure) · P (failure)

= 0.3 · 0.3

= 0.09

When X = 1, it means that the medicine worked for one of the patients, but not both.

P(X = 1) = P (success) · P (failure) + P (failure) · P (success)

= 0.7 · 0.3 + 0.3 · 0.7

= 0.21 + 0.21

= 0.42

When X = 2, it means that the medicine worked for both patients.

P(X = 0) = P (success) · P (success)

= 0.7 · 0.7

= 0.49

We can now use the probabilities that we calculated for each outcome to create a binomial distribution table and a binomial distribution histogram.

Interpreting Binomial Distributions

The doctor would like to be successful with at least one of the two patients. In the language of probability, the question becomes what is P (X ≥ 1)?

The answer to this question is found by adding the appropriate probabilities in the table.

P(X ≥ 1) = P (X = 1) + P (X = 2)

= 0.42 + 0.49

= 0.91

There is a 91% probability that Medicine Q will be successful with at least one of the patients.

Binomial Distribution Table Binomial Distribution Histogram

REMEMBERIf A and B are independent events, then P(A and B) = P(A) · P(B).

BINOMIAL DISTRIBUTIONS 149





Determine whether the variable X has a binomial distribution. If it has a binomial distribution, then give P (success) and the number of trials. If X does not have a binomial distribution, then explain why.

1. When a coin is tossed, X is the number of tosses it takes to get tails.

2. Over the next 4 days, X is the number of days it will rain.

3. When a coin is tossed 4 times, X is the number of tails.

4. When a 6-sided number cube is rolled 3 times, X is the number of 2s rolled.

5. A bag contains a red, a yellow, and a blue marble. When a marble is selected 2 times in a row without replacement, X is the number of yellow marbles.

6. A bag contains a red, a yellow, and a blue marble. When a marble is selected from the bag 3 times in a row with replacement, X is the number of yellow marbles.

A. P (success) = 0.8

B. P (success) = 0.7

C. P (success) = 0.2

D. P (success) = 0.3

Match the probability histogram with the description of the binomial distribution. Assume that there are 2 trials.

Problem Set

7.

0.5

0.6

0.4

0.3

0.2

0.1

00 1 2

X

P

9.

0.5

0.6

0.7

0.4

0.3

0.2

0.1

00 1 2

X

P

8.

0.5

0.6

0.4

0.3

0.2

0.1

00 1 2

X

P

10.

0.5

0.6

0.7

0.4

0.3

0.2

0.1

00 1 2

X

P




BINOMIAL DISTRIBUTIONS 151

Suppose the dial on the spinner is spun 2 times in a row. Create a probability histogram for the variable X.

11. X is the number of times the dial lands on section A.

12. X is the number of times the dial lands on section C.

13. X is the number of times the dial lands on section A or C.

14. X is the number of times the dial lands on section B or A.

15. When a coin is tossed twice, X is the number of tails. Find P (X < 2).

16. A basketball player’s probability of making a free throw is 9 ___ 10 . When the player shoots 2 free throws in a row, X is the number of free throws made. Find P (X = 2).

17. When a coin is tossed 3 times, X is the number of tails. Find P (X > 1).

18. A basketball player’s probability of making a free throw is 9 ___ 10 . When the player shoots 3 free throws in a row, X is the number of free throws made. Find P (X = 2).

19. A basketball player’s probability of making a free throw is 0.6. When the player shoots 2 free throws in a row, X is the number of free throws made. Find P (X = 1).

20. A basketball player’s probability of making a free throw is 0.6. When the player shoots 3 free throws in a row, X is the number of free throws made. Find P (1 ≤ X < 3).

21. Challenge Suppose the dial on the spinner is spun 3 times in a row. Find the probability that the dial will land 2 or more times on section B.

Create a probability distribution table for the variable X. Then use the table to find the indicated probability.

Use the spinner to solve.

C B

AD

C

BA




Uniform ProbabilityDistribution

0.5

0.4

0.3

0.2

10099

0.1

101 102 103 104 1050 X

P

Continuous Random VariablesAreas within probability distributions can be used to compute probability.

Exploring Continuous Random Variables

An incubator is being used to hatch chickens. The incubator temperatures X, in degrees Fahrenheit, are distributed according to this probability distribution.

When all values of a random variable X are equally likely to occur, the resulting probability distribution is called a uniform probability distribution.

For the incubator, all of the values of X, from 100°F to 104°F are equally likely to occur. Because any temperature between these values, such as 101.5°F or 103.9°F, may occur, X, in this case, is considered to be a continuous random variable.

Using Area to Determine Probability

When X is a continuous random variable, the resulting probability distribution is a geometric shape with a total area of 1. The probability that a ≤ X ≤ b will be the area of the shape bordered by the values a and b.

For instance, suppose the temperature of the incubator is taken at 10 a.m. What is the probability that the temperature is between 101°F and 103°F?

Q & A

Q How is a discrete random variable diff erent from a continuous random variable?

A A discrete random variable has a limited number of values, while a continuous random variable has an infi nite number of values.

A variable X is a continuous random variable if it can take on any value in the possible range of the random variable.

CONTINUOUS RANDOM VARIABLE





0.5

0.4

0.3

0.2

10099

0.1

101 102 103 104 1050 X

P

Rectangularregion

Nonuniform Probability Distribution

0 1

0.2

0.4

0.6

2 3 4 5 6X

P

Trapezoidalregion

0 1

0.2

0.4

0.6

2 3 4 5 6X

P

To answer this question, you can compute the area of the rectangular region where 101 ≤ X ≤ 103, which is shown here.

The area of the rectangular region is the product of the base and the height: 2 · 0.25 = 0.5. There is a 50% probability that the temperature will be between 101°F and 103°F degrees at 10 a.m.

Using Continuous Nonuniform Probability Distributions

Here is another continuous probability distribution that is not uniform and is in the shape of a triangle.

Suppose we want to know P(1 ≤ X ≤ 4), the probability that X is between 1 and 4.

The region bounded by X = 1 and X = 4 is a trapezoid with base lengths 0.1 and 0.4. The height of the trapezoid is 3.

Area = 1 __ 2 h ( base1 + base2)

= 1 __ 2 · 3 (0.1 + 0.4)

= 0.75

So the probability that X is between 3 and 4 is 0.75, or 75%.

REMEMBERThe area of the graph of a probability distribution is equal to 1.

CONTINUOUS RANDOM VARIABLES 153





Problem Set

Use the uniform probability distribution to solve.

1. What is the area of the distribution?

2. What is the height of the distribution?

3. What is the area of the part of the region where X ≤ 3?

4. What is the area of the part of the region where 1.5 ≤ X ≤ 3?

5. Determine P (X ≤ 3).

6. Determine P (X < 3).

7. Determine P (1.5 ≤ X ≤ 3).

8. Determine P (1.5 < X ≤ 3).

9. If P (X < a) = 0.5, what is the value of a?

10. If P (X ≤ a) = 0.5, what is the value of a?

11. If P (X < a) = 2 __ 3 , what is the value of a?

12. If P (X ≤ a) = 1 __ 6 , what is the value of a?

13. Suppose the animal is weighed at 11 a.m. What is probability that the animal’s weight will be less than 150 pounds?

14. Suppose the animal is weighed at 2 p.m. What is probability that the animal’s weight will be less than 151 pounds?

15. Suppose the animal is weighed in the morning. What is probability that the animal’s weight will be within 1 __ 2 pound of 150 pounds?

16. Suppose the animal is weighed in the afternoon. What is probability that the animal’s weight will be within 3 __ 4 pound of 149 pounds?

The weights X of an animal are distributed according to the probability distribution shown.

0 1 2 3 4X

P

147 148 149 150 151 152 153X




CONTINUOUS RANDOM VARIABLES 155

Use the probability distribution to solve.




20. Determine P (X < 3 or X > 5).

21. Determine P (X ≤ 3 or X > 6).

22. Determine P (3 < X < 4).

23. Determine P (1.5 < X ≤ 6.5).


25. If P (X > a) = 0.5, what is the value of a?




A random number generator generates real numbers from 0 to 3, according to this probability distribution.

29. What is the probability that a number between 1 and 2 is generated?

30. What is the probability that a number between 2 and 3 is generated?

31. What is the probability that a number less than 2 is generated?

32. What is the probability that a number between 0.5 and 2.5 is generated?

33. What is the probability that a number less than 1 or greater than 1.5 is generated?

34. Challenge Suppose the probability is 0.4 that a number less than a is generated. What is the value of a?

0.3

0.4

0.5

0.6

0.1

0

0.2

1 2 3 4X

P

0.1

0.2

0.3

0 1 2 3 4 5 6 7 8X

P




The Normal DistributionThe bell-shaped curve is often used for data analysis and probability.

Using the Normal Distribution

When a data set has a normal distribution, a histogram that represents the data set is symmetric and has a bell shape.

When a curve is drawn through the tops of the bars, the curve is in the shape of a bell.

Normal distributions have important properties that help us study data sets that are otherwise too large to study.

For example, the mean is always at the center of the distribution, so about half of the data set is above the mean and half of the data set is below the mean.

The standard deviation of a normal distribution also gives us useful information about the data set.

FACTIn a normal distribution, the mean, median, and mode all have approximately the same value.

If a data set has a normal distribution, then about• 68% of the data is within one standard deviation of the mean.• 95% of the data is within two standard deviations of the mean.• 99.7% of the data is within three standard deviations of the mean.

PROPERTIES OF A NORMAL DISTRIBUTION

Mean






The properties involving standard deviation are illustrated in more detail in this diagram.

For example, we know that about 68% of the data is within one standard deviation of the mean. This means that about 34% is one standard deviation above the mean, and 34% of the data is one standard deviation below the mean.

Using Normal Distribution Properties

A pediatrician read an article that states that the length of the flu incubation period is distributed normally with a mean length of 7 days and a standard deviation of 2 days. Afterward, he wonders what the probability is of the flu incubating for at least 9 days.

Assuming that flu incubation periods are normally distributed, a normal curve can be drawn to represent this situation.

We know that 9 days is one standard deviation above the mean of 7 days. So, 34% of the data is between 7 days and 9 days. Also, 50% of the data is below the mean.

This means that 50% + 34% = 84% of flu cases have an incubation period of at most 9 days. By subtracting 84% from 100%, we can obtain the percent of cases that have an incubation period of at least 9 days.

Therefore, the probability is about 16% that a flu case will have an incubation period of at least 9 days.

REMEMBERThe Greek letters μ and σ are used to represent the mean and standard deviation of a data set, respectively.

REMEMBERIn a normal distribution, about 50% of the data is above the mean, and about 50% of the data is below the mean.

Flu Incubation Period

Number of days

50% 34%

3 5 7 9 11

Flu Incubation Period

Number of days

16%

3 5 7 9 11

μ - 3σ μ - 2σ μ - 1σ μ μ + 1σ μ + 2σ μ + 3σ2.35% 2.35%13.5%13.5% 34%

68%

95%

99.7%


34%

THE NORMAL DISTRIBUTION 157





Problem Set

Suppose normal human body temperatures are normally distributed with a mean of 37°C and a standard deviation of 0.2°C.

1. What temperature is one standard deviation below the mean temperature?

2. What temperature is three standard deviations above the mean?

3. What percent of humans have a temperature above 37°C?

4. What percent of humans have a temperature below 37.2°C?

5. What percent of humans have a temperature below 36.8°C?

6. What percent of humans have a temperature between 36.8°C and 37.2°C?

7. What percent of humans have a temperature between 36.6°C and 37.4°C?

8. What percent of humans have a temperature between 37°C and 37.4°C?

Suppose that the weights of 5400 registered female Labrador retrievers in the United States are distributed normally with a mean of 62.5 pounds and a standard deviation of 2.5 lb.

9. What percent of the Labrador retrievers weigh less than 62.5 pounds?

10. What percent of the Labrador retrievers weigh less than 65 pounds?

11. What percent of the Labrador retrievers weigh between 60 pounds and 65 pounds?

12. What percent of the Labrador retrievers weigh between 60 lb and 62.5 pounds?

13. Approximately how many of the Labrador retrievers weigh less than 62.5 pounds?

14. Approximately how many of the Labrador retrievers weigh less than 65 pounds?

15. Approximately how many of the Labrador retrievers weigh between 60 pounds and 65 pounds?

16. Approximately how many of the Labrador retrievers weigh between 60 pounds and 62.5 pounds?




THE NORMAL DISTRIBUTION 159

Suppose that the number of petals on 2400 fully developed roses is normally distributed with a mean of 32 petals and a standard deviation of 2 petals.

17. Approximately how many fl owers have fewer than 32 petals?

18. Approximately how many fl owers have between 30 and 34 petals?



21. Approximately how many fl owers have 30 or more petals?

22. Approximately how many fl owers have fewer than 34 petals?

Suppose a factory produces 50,000 packages of mints each week. Assume that the number of mints in a package is normally distributed with a mean of 120 mints and a standard deviation of 3 mints.

23. Approximately how many packages each week have between 120 and 129 mints?

24. Approximately how many packages each week have between 111 and 117 mints?

25. Approximately how many packages each week have more than 123 mints?

26. Approximately how many packages each week have fewer than 126 mints?

Suppose a manufacturer makes nails that are normally distributed with a length of 150 millimeters and a standard deviation of 1 millimeter.

27. Challenge In a certain batch of nails, 3900 nails are between 149 millimeters and 151 millimeters long. What is the approximate number of nails in the entire batch? Round your answer to the nearest whole number.

28. Challenge In a certain batch of nails, 25 nails are longer than 153 millimeters. What is the approximate number of nails in the entire batch? Round your answer to the nearest whole number.




Standardizing DataA standard score tells us how far a data value is from the mean.

Converting Raw Scores into z-Scores

Antibiotics are strong medicines that are prescribed by a doctor and often used to treat infections. Once a patient begins taking an antibiotic, the length of time that it takes to cure the infection can vary.

Suppose that the time it takes a certain antibiotic to cure an infection is known to be normally distributed with a mean of 5.2 days and a standard deviation of 1.1 days.

To get an idea of how effective the new antibiotic is with certain patients, actual results can be converted into z-scores. A z-score tells us the number of standard deviations that a data value is from the mean of the data set.

When a data value (called a raw score) is greater than the mean, its corresponding z-score will be positive. When a data value is less than the mean, its corresponding z-score will be negative.

A z-score is the number of standard deviations that a data value is from the mean. Every data value in a data set has a corresponding z-score and is obtained by the formula

z = x − μ

_____ σ .

In this formula, z is the z-score, x is the raw data value, μ is the mean of the data set, and σ is the standard deviation.

CONVERTING RAW SCORES INTO z-SCORES

Suppose the antibiotic cures an infection in a patient in 3.8 days. What is the corresponding z-score?

Here, x = 3.8, μ = 5.2, and σ = 1.1. These values can be substituted into the formula to find the corresponding z-score.

z = 3.8 − 5.2 ________ 1.1

≈ −1.3

A z-score of −1.3 means that the antibiotic cures the infection in a time that is 1.3 standard deviations below the mean.

REMEMBERz-scores are sometimes called standard scores.





Converting z-Scores into Raw Scores

Suppose a patient takes the antibiotic and is cured of the infection according to a z-score of 2.3. From what we know about z-scores, this means that the patient was cured in an amount of time that is about 2.3 standard deviations greater than the mean cure time.

But suppose we want to know how many days it took the antibiotic to cure the infection in this patient. In other words, what is the raw score if the z-score is 2.3?

The z-score formula can be rearranged to create a formula for converting z-scores back into raw scores.

A z-score can be converted back into its original value, called a raw score, using the formula

x = z ∙ σ + μ.

In this formula, x is the raw score, z is the z-score, μ is the mean of the data set, and σ is the standard deviation.

CONVERTING z-SCORES INTO RAW SCORES

Now, we can convert the z-score of 2.3 back into a raw score. Using z = 2.3, μ = 5.2, and σ = 1.1, we can find the value of the raw score, x.

x = 2.3 · (1.1) + 5.2

≈ 7.7

For this patient, the antibiotic cured the infection in about 8 days—more than 2 days longer than the average cure time for this antibiotic.

REMEMBERA raw score refers to the original data value.

STANDARDIZING DATA 161





Problem Set

Suppose a normal distribution has a mean of 20 and a standard deviation of 4.

1. A value of 24 is how many standard deviations away from the mean?

2. A value of 26 is how many standard deviations away from the mean?

3. What is the z-score of 22?




7. What is the z-score of a value that is 1.43 standard deviations greater than the mean?

8. What is the z-score of a value that is 0.52 standard deviations less than the mean?

Suppose a manufacturer makes disposable peppercorn grinders. The number of peppercorns in the grinders are normally distributed with a mean of 322 peppercorns and a standard deviation of 5.3 peppercorns.

13. Suppose a peppercorn grinder contains 315 peppercorns. What is the z-score for this grinder?

14. Suppose a peppercorn grinder contains 323 peppercorns. What is the z-score for this grinder?

15. Suppose the manufacturer will only sell peppercorn grinders with a z-score between −2 and 2. What are the least and most peppercorns a grinder can contain?

16. Suppose the manufacturer will only sell peppercorn grinders with a z-score between −0.9 and 0.9. What are the least and most peppercorns a grinder can contain?

Suppose the ages of cars driven by employees at a company are normally distributed with a mean of 8 years and a standard deviation of 3.2 years.

9. What is the z-score of a car that is 12 years old?

10. What is the z-score of a car that is 6 years old?

11. What is the z-score of a car that is 9.1 years old?

12. DeSean, a new employee, drives a brand new car. What is the z-score of the car?




STANDARDIZING DATA 163

Solve by completing the table. Round your answer to the nearest hundredth.

17. The healing time for a broken ankle is normally distributed with a mean of 42 days and a standard deviation of 9.25 days.

18. With new technology, the healing time for a broken ankle now takes a mean of 36 days with a standard deviation of 8.67 days.

Suppose the velocities of golf swings for amateur golfers are normally distributed with a mean of 95 miles per hour and a standard deviation of 3.9 miles per hour.

19. What is the difference in velocities between a golfer whose z-score is 0 and another golfer whose z-score is 1?

20. What is the difference in velocities between a golfer whose z-score is 0 and another golfer whose z-score is −1?

21. What is the difference in velocities between a golfer whose z-score is −0.5 and another golfer whose z-score is 0.5?

22. What is the difference in velocities between a golfer whose z-score is −1 and another golfer whose z-score is −2?

23. What is the difference in velocities between a golfer with a z-score of a and another golfer whose z-score is a + 1?

Suppose a manufacturer makes nails that are normally distributed with a mean length of x millimeters and a standard deviation of y millimeters.

24. Challenge In a certain batch of nails, an 83-millimeter nail has a z-score of 1 and an 84.5-millimeter nail has a z-score of 2. Determine x and y.

Name Healing Time z-score

Ayla 35 days

Miranda 1.4

Stacy –2.2

Name Healing Time z-score

Alfredo 35 days

Josh 1.4

Sergio –2.2




Mixture A Mixture B

Mean: μ = 55 mg

Standard Deviation: σ = 4.1 mg

Mean: μ = 52 mg

Standard Deviation: σ = 3.3 mg

Mixture A Mixture B

z = 50 − 55 _______ 4.1

≈ −1.22

z = 50 − 52 _______ 3.3

≈ −0.61

Comparing ScoresStandard scores can be used to compare normal distributions.

Comparing z-Scores from Two Distributions

Black tea has been studied for its medicinal properties and for the amount of caffeine it contains (which can vary depending on the type of tea used and the amount of time brewed).

The amount of caffeine in single servings of tea is normally distributed.For a particular comparison, single servings of two different Ceylon tea mixtures are compared.

Suppose you are given a serving of Ceylon tea which contains 50 milligrams of caffeine. Which mixture is the tea more likely to come from?

To determine whether the serving with 50 milligrams of caffeine is more likely to come from Mixture A or Mixture B, you can determine the z-score for each mixture.

For a normal distribution, any value that is closer to the mean will be a more likely outcome than any value farther away from the mean. Since −0.61 is closer to 0 than −1.22, Mixture B is more likely to produce a serving of Ceylon tea with 50 milligrams of caffeine.

Paulo is doing his own research about the caffeine content of these two Ceylon tea mixtures. The brew he makes with Mixture A has 57 milligrams of caffeine in a single serving. The brew he makes with Mixture B has 49 milligrams of caffeine in a single serving. Which result is more unusual?

BY THE WAYCeylon tea is a type of black tea grown in Sri Lanka.

THINK ABOUT ITThe z-score of the mean is 0.00 because it is zero standard deviations away from the mean.

REMEMBER

z = x − μ

_____ σ





Mixture A Mixture B

z = 57 − 55 _______ 4.1

≈ 0.49

z = 49 − 52 _______ 3.3

≈ −0.91

Mixture A Mixture B

x = −3 · (4.1) + 55

= 42.7

x = −3 · (3.3) + 52

= 42.1

To determine whether the serving with 57 milligrams of caffeine from Mixture A is more unusual than the serving with 49 milligrams of caffeine from Mixture B, you can determine the z-score for each mixture.

The z-scores for each mixture have different signs (one positive and one negative). However, we are only interested in knowing which z-score is farthest from 0.

Since Mixture A’s z-score of 0.49 is closer to 0 than Mixture B’s z-score of −0.91, the serving with 49 milligrams of caffeine from Mixture B is more unusual than the serving with 57 milligrams of caffeine from Mixture A.

Interpreting a z-Score in Two Distributions

Two normally distributed data sets can also be compared for the same z-score. For instance, in looking at the two mixtures of Ceylon tea, which mixture has more caffeine for a z-score of −3.00?

We know that the z-score is −3.00, and we know the mean and standard deviation for each mixture. We can use the appropriate formula to determine the actual amount of caffiene in a serving from each mixture.

At a distance of three standard deviations below the mean, the two mixtures have about the same amounts of caffeine. However, Mixture A has more caffeine compared to Mixture B.

REMEMBERx = z · σ + μ

COMPARING SCORES 165





Problem Set

The bar graph shows the z-score results of four students on two different math tests. The students took Test 1, and then a month later took Test 2.

1. What are Carla’s z-scores for both tests?

2. How many standard deviations away from the mean did Dave score on Test 2? Was the score above the mean or below the mean? Explain.

3. Which student scored the farthest away from the mean? On which test did this occur? Explain.

4. Which student had the lowest score on Test 2? Explain.

5. Which student had the best overall performance on both tests? Explain.

6. Which student showed the most improvement from Test 1 to Test 2? Explain.

Carla Dave Euan Felicia

1.75

1.5

1.25

1

0.75

0.5

0.25

0

−0.25

−0.5

−0.75

−1

−1.25

z-S

core

Students

Math Test z-Scores

Test 1

Test 2

Suppose the growth of Bermuda grass from germination to lawn care length is normally distributed with a mean of 20 days with a standard deviation of 3.3 days. The growth of buffalo grass from germination to lawn care length is normally distributed with a mean of 21 days with a standard deviation of 2.3 days.

7. A gardener grew a lawn using Bermuda grass in 23 days. Determine the z-score.

8. A gardener grew a lawn using buffalo grass in 15 days. Determine the z-score.

9. On which lawn would a z-score of 3 give a longer growing period? Explain.

10. On which lawn would a z-score of −2.5 give a longer growing period? Explain.




COMPARING SCORES 167

Suppose scores on a biology test are normally distributed with a mean of 85 and a standard deviation of 4. Scores on a history test are normally distributed with a mean of 85 and a standard deviation of 2.

11. Ayla scored 90 on both tests. Determine her z-score for each test. On which test did she have the higher z-score?

12. Maya scored 80 on both tests. Determine her z-score for each test. On which test did she have the higher z-score?

13. Eric scored 89 on the biology test and 87 on the history test. On which test did he have the better performance? Explain.

14. Paulo scored 90 on the biology test and 89 on the history test. On which test did he have the better performance? Explain.

15. On which test would a z-score of 3 give a higher raw score? Explain.

16. On which test would a z-score of −2.5 give a higher raw score? Explain.

17. On which test would a z-score of 0.5 give a higher raw score? Explain.

18. On which test would you expect a higher range of scores? Explain.

19. On which test would a score of 87 be a better score? Explain.

20. On which test would a score of 80 be a better score? Explain.

Suppose Company A produces packages of throat lozenges that are normally distributed with a mean of 38.2 individual lozenges and a standard deviation of 1.7 lozenges. Company B produces packages of throat lozenges that are normally distributed with a mean of 36.9 individual lozenges and a standard deviation of 2.2 lozenges.

21. Which company would be more likely to produce a package of 37 throat lozenges? Explain using z-scores.

22. Which company would be more likely to produce a package of 43 throat lozenges? Explain using z-scores.

23. Would it be more likely for Company A to produce a package of 35 throat lozenges or Company B to produce a package of 40 throat lozenges? Explain using z-scores.

24. Challenge Suppose each company produces a package of x lozenges so that the z-scores are identical. Find the value of x.




Standard Normal Curve

z-scores–2 0 2–3 –1 1 3

?

z-scores0 0.54

The standard normal curve is used to analyze data.

Standard Normal Curve Properties

When all of the data values of a normal distribution are turned into z-scores, the resulting distribution is called the standard normal curve.

Each value on the horizontal axis of the standard normal curve is a z-score, and the mean is always 0.

For instance, a data value with a z-score of 0.54 would correspond to 0.54 on the horizontal axis of the standard normal curve.

The Standard Normal Curve

REMEMBERIf a z-score is positive, then it is above the mean. If a z-score is negative, then it is below the mean.

The standard normal curve is a probability distribution with the following properties:• The mean is 0.• The standard deviation is 1.• The area under the curve is 1.

PROPERTIES OF THE STANDARD NORMAL CURVE

Using the z-Distribution Table

Because the standard normal curve is a probability distribution, it can be used to find probabilities. These probabilities correspond to areas under the standard normal curve.

Suppose a data value has a z-score of 0.54. What percent of the data values are less than this value?





z0.00.10.20.30.40.50.6

.000.50000.53980.57930.61790.65540.69150.7257

.010.50400.54380.58320.62170.65910.69500.7291

.020.50800.54780.58710.62550.66280.69850.7324

.030.51200.55170.59100.62930.66640.70190.7357

.040.51600.55570.59480.63310.67000.70540.7389

.050.51990.55960.59870.63680.67360.70880.7422

z-scores0 0.68 2.05

0.2281

z-scores0 0.54

Area = 0.7054

To answer this question, we can use the Table of z-Scores for Normal Distribution (pages A-1 and A-2). This table gives the areas under the standard normal curve that are less than a given z-score.

Using the Standard Normal Curve to Solve Problems

Suppose scores on a standardized test for future doctors are normally distributed with a mean of 10.5 and a standard deviation of 2.2. If 150 future doctors take the test, how many are likely to score between 12 and 15?

The score 12 has a z-score of about 0.68, and the score 15 has a z-score of about 2.05. To help us answer the question, we need to know the area that corresponds to P (0.68 < z < 2.05).

The area under the standard normal curve less than z = 0.68 is 0.7517, and the area less than z = 2.05 is 0.9798.

So P (0.68 < z < 2.05) = 0.9798 − 0.7517 = 0.2281.

Using the z-distribution table, look for the area that corresponds to the row and column of a z-score of 0.54. We can see that P (z < 0.54) = 0.7054, which is about 71%.

This means that about 71% of the data values are less than a value with a z-score of 0.54.

Multiply the area by the number of doctors taking the exam. 0.2281 · 150 ≈ 34

So about 34 of the 150 future doctors taking the exam are likely to score between 12 and 15.

REMEMBERz-scores are obtained using the following formula:

z = x − μ

_____ σ

BY THE WAYTechnology, such as calculators and computers, can also be used to fi nd areas under the standard normal curve.

THE STANDARD NORMAL CURVE 169





Problem Set

The expression represents the area under the standard normal curve below a given value of z. Use the z-distribution table on pp. A-1 to A-2 or technology to find the indicated area, and include a sketch of the standard normal curve with the appropriate area shaded.

1. P (z < 0.00)

2. P (z < 1.00)

3. P (z < 1.57)

4. P (z < 2.87)

5. P (z < −1.00)

6. P (z < −1.53)

7. P (z < −2.02)

8. P (z < −2.77)

The expression represents the area under the standard normal curve above a given value of z. Use the z-distribution table on pp. A-1 to A-2 or technology to find the indicated area, and include a sketch of the standard normal curve with the appropriate area shaded.

9. P (z > 0.00)

10. P (z > 1.00)

11. P (z > 1.57)

12. P (z > 2.87)

13. P (z > −1.00)

14. P (z > −1.53)

15. P (z > −2.05)

16. P (z > −2.87)

The expression represents the area under the standard normal curve above a given value of z. Use the z-distribution table on pp. A-1 to A-2 or technology to find the indicated area, and include a sketch of the standard normal curve with the appropriate area shaded.

17. P (0.00 < z < 1.00)

18. P (0.00 < z < 2.00)

19. P (1.00 < z < 1.00)

20. P (1.06 < z < 2.63)

21. P (−1.33 < z < 1.59)

22. P (−2.00 < z < 2.00)

23. P (−1.61 < z < −0.09)

24. P (−2.53 < z < −0.55)




THE STANDARD NORMAL CURVE 171

The scores on a standardized test are normally distributed with a mean of 500 and a standard deviation of 100. Use the z-distribution table on pp. A-1 to A-2 or technology to solve.

25. Sofi a scored 632 on the test. What percent of students scored below Sofi a?

26. Jake scored 500 on the test. What percent of students scored below Jake?

27. Benita scored 432 on the test. What percent of students scored below Benita?

28. Kendall scored 380 on the test. What percent of students scored below Kendall?

29. Suppose 6000 students took the test. Approximately how many students scored below 632?

30. Suppose 6000 students took the test. Approximately how many students scored above 490?

31. Suppose 6000 students took the test. Approximately how many students scored between 450 and 550?

32. Suppose 6000 students took the test. Approximately how many students scored between 500 and 600?

Lengths of newborn girls are normally distributed with a mean of 49.2 centimeters and a standard deviation of 1.8 centimeters. Use the z-distribution table on pp. A-1 to A-2 or technology to solve.

33. What percent of newborn girls will be 50 centimeters or longer?



36. What percent of newborn girls will be 45 centimeters or shorter?

37. Consider a group of 2000 newborn girls. How many girls will be 50.5 centimeters or longer?

38. Consider a group of 2000 newborn girls. How many girls will be 51 centimeters or shorter?

39. Consider a group of 2000 newborn girls. How many girls will be between 49 centimeters and 51 centimeters?

40. Consider a group of 2000 newborn girls. How many girls will be between 44 centimeters and 45 centimeters?

41. Challenge Consider a group of 2000 newborn girls. If 500 of the newborn girls are between 47 centimeters and x centimeters long, what is the value of x?




Standard Normal Curve

z-scoresz = ?

20%

z–1.4–1.3–1.2–1.1–1.0–0.9–0.8–0.7

.000.08080.09680.11510.13570.15870.18410.21190.2420

.010.07930.09510.11310.13350.15620.18140.20900.2389

.020.07780.09340.11120.13140.15390.17880.20610.2358

.030.07640.09180.10930.12920.15150.17620.20330.2327

.040.07490.09010.10750.12710.14920.17360.20050.2296

.050.07350.08850.10560.12510.14690.17110.19770.2266

Finding Standard ScoresAreas under the standard normal curve can be used to find standard scores and percentiles.

Using Area To Find a z-Score

While the straightforward way of reading the Table of z-Scores Normal Distribution is to look up the z-score and then find the corresponding area under the standard normal curve, we can also read the table in the other direction.

Suppose that about 20% of the data in a data set is below a certain value. What z-score corresponds to this value?

Since 20% = 0.2, we want to find an area in the body of the z-distribution table that is closest to 0.2.

The area 0.2005 is closest to 20%. Use the row and column of the area to locate the corresponding z-score. This corresponds to a z-score of −0.84.

REMEMBERz-scores are sometimes called standard scores.





z-scoresz = ?

15%

85%

z0.60.70.80.91.01.1

.000.72570.75800.78810.81590.84130.8643

.010.72910.76110.79100.81860.84380.8665

.020.73240.76420.79390.82120.84610.8686

.030.73570.76730.79670.82380.84850.8708

.040.73890.77040.79950.82640.85080.8729

.050.74220.77340.80230.82890.85310.8749

Since the z-distribution table only gives the areas that are below a particular z-score, we need to find the percent of data below the value: 100% − 15% = 85%. Since 85% = 0.85, we want to find an area in the body of the z-distribution table that is closest to 0.85.

Determining Area Above a z-Score

Suppose that 15% of the data in a data set is above a certain value. What z-score corresponds to this value?

The area 0.8508 is closest to 85%. This corresponds to a z-score of 1.04.

Understanding Percentiles

A percentile rank is the percentage of data that falls below a particular value. For example, if a student scores higher than 60% of the other students who took an exam, then the student’s percentile rank is at the 60th percentile.

Scores on the math section of the Scholastic Aptitude Test (SAT) are normally distributed with a mean of about 500 and a standard deviation of about 100. According to this information, what score would be at the 80th percentile?

Using the z-distribution table, we see that the area 0.8023 is closest to 80%. This corresponds to a z-score of 0.85. We can use the appropriate formula to find the corresponding score.

x = z · σ + µ

= 0.85 · 100 + 500

= 585

So a score of 585 on the math section of the SAT is at the 80th percentile.

RemembeRThe nth percentile is the value (or score) below which n percent of the data may be found.

FindinG stAndArd scores 173



Problem Set

The number represents the area under the standard normal curve below a particular z-score. Find the z-score, and include a sketch of the standard normal curve with the appropriate area shaded.

1. 0.2358

2. 0.9967

3. 0.5000

4. 0.1170

5. 0.9115

6. 0.8869

7. 0.1003

8. 0.2389

Suppose a set of data is normally distributed. Use the z-distribution table on pp. A-1 to A-2 or technology to solve.

9. Below which z-score do approximately half of the data lie?

10. Above which z-score do approximately half of the data lie?

11. Below which z-score do approximately 23.58% of the data lie?

12. Above which z-score do approximately 23.58% of the data lie?

13. Below which z-score do approximately 88.3% of the data lie?

14. Above which z-score do approximately 88.3% of the data lie?

15. Above which z-score do approximately 3 __ 4 of the data lie?

16. Below which z-score do approximately 3 __ 4 of the data lie?

Suppose scores on a standardized test are normally distributed with a mean of 500 and a standard deviation of 100. Use the z-distribution table on pp. A-1 to A-2 or technology to solve.

17. Approximately 50% of the scores are above which z-score? Which test score corresponds to the z-score?

18. Approximately 50% of the scores are below which z-score? Which test score corresponds to the z-score?








FINDING STANDARD SCORES 175

Suppose ages of cars driven by company employees are normally distributed with a mean of 8 years and a standard deviation of 3.2 years. Use the z-distribution table on pp. A-1 to A-2 or technology to solve.

23. Approximately 15% of cars driven by company employees are older than what age?

24. Approximately 75% of cars driven by company employees are older than what age?

25. In a sample of 8000 company employees who drive cars, 6000 of them drive cars younger than x years old. Determine x.

26. In a sample of 10,000 company employees who drive cars, 1350 of them drive cars younger than y years old. Determine y.

Assume that lengths of newborn girls are normally distributed with a mean of 49.2 centimeters and a standard deviation of 1.8 centimeters.

27. Find the length of a newborn girl whose length is at the 50th percentile.



30. What is the percentile rank of a baby girl whose length is 51 centimeters?






X 0 1 2 3

P 0.125 0.375 0.375 0.125

40

35

30

25

20

15

10

5

0 1 2Number of boys

Number of Boys in Family(Predicted)

30

Pro

babi

lity

(%)

Chapter 4 Wrap-UpProbability Distributions

Though there are slightly more boys than girls under the age of 15 in the United States, it is often estimated that there is a 50% chance that a newborn baby will be a boy (or girl as the case may be).

In general, there should be an equal probability of a newborn being either a boy or a girl. To test this, a binomial distribution can be created for a family with 3 children.

This is a binomial distribution with P(success) = 0.5 and 3 trials. The probability distribution is given in the table and the histogram.

A few conclusions can be drawn from this distribution. In a family of 3 children,

• There is a 37.5% chance that the family will have 1 boy.

• There is a 75% chance that the family will have boys and girls.

• There is a 25% chance that the family will have all boys or all girls.


HS_PS_S1_04_WU_RG.indd 176HS_PS_S1_04_WU_RG.indd 176 9/17/11 4:59 AM9/17/11 4:59 AM



Normal Distributions

In general, the length of a full-term newborn baby is distributed normally with a mean of 20.1 inches with a standard deviation of 0.4 inches. According to the properties of a normal distribution, this means that approximately 68% of newborn babies are between 19.7 inches and 20.5 inches; 95% of newborn babies are between 19.3 inches and 20.9 inches; and 99.7% of newborn babies are between 18.9 inches and 21.3 inches.

What is the 80th percentile of a newborn baby’s length? According to the standard normal distribution, the value in the z-distribution table closest to 0.8 is when the z-score is z = 0.84.

To convert the z-score into the actual length, multiply 0.84 by the standard deviation (0.4), then add the mean (20.1): 0.84 · (0.4) + 20.1 = 20.44.

This means that 80% of newborn babies will be no more than 20.44 inches long.

In Summary

Probability distributions give probabilities over a whole set of outcomes. A binomial distribution explains repeated trials of the same event when the probability of success is the same for each trial. A normal distribution has properties that can be used to analyze data sets.

CHAPTER 4 WRAP-UP 177

HS_PS_S1_04_WU_RG.indd 177HS_PS_S1_04_WU_RG.indd 177 9/17/11 4:59 AM9/17/11 4:59 AM



Hourly Wage $7 $8 $9 $10 $11 $12 $17

Number of High School Students 6 4 3 4 1 1 1

X 7 8 9 10 11 12 17

P

Random Variables and Distributions

The frequency table shows the summer employment hourly wages (to the nearest dollar) for a group of high school students.

Practice Problems

1. Explain why the table is not a probability distribution table.

2. If X is the hourly wage, complete the probability distribution table. Round probabilities to the nearest hundredth.

3. If a high school student is selected at random, determine P(X < 10).

4. Based on these data, what would be the expected value of a high school student’s summer hourly wage?

The Personal Computers data set on p. A-9 shows the number of personal computers owned per 100 inhabitants of South American countries from 1998 to 2005.

5. Create a probability distribution table for 2004. Round probabilities to the nearest thousandth.

6. Suppose a personal computer is in use in South America in 2004. What is the probability, to the nearest thousandth, that the computer is being used in Chile?

7. Suppose there are 280,000 personal computers of a certain brand in use in South America in 2004. How many would you expect to be used in Peru?

Determine if X has a binomial distribution. If it does, create a binomial distribution histogram. Otherwise, explain why X does not have a binomial distribution.

8. When a coin is tossed twice, X is the number of heads.

9. A tennis player’s probability of winning a match is 0.6. X is the number of wins for the next 3 matches.

10. Over the next 5 days, X is the number of days it will snow.


HS_PS_S1_04_WU_PS.indd 178HS_PS_S1_04_WU_PS.indd 178 9/17/11 5:03 AM9/17/11 5:03 AM



Chapter 4 Wrap-Up

318 320 322 324Temperature (°F)

X326 328 330 332

P

0 1

0.050.1

0.150.2

0.250.3

2 3 4 5 6 7X

Use the z-distribution table on pp. A-1 to A-2 or technology to solve. The number of seeds in a certain type of strawberry are distributed normally with a mean of 200 seeds and a standard deviation of 9 seeds.

22. What percent of strawberries have fewer than 200 seeds?

23. What percent of strawberries have more than 191 seeds?

24. If a strawberry has a z-score of 1.11, about how many seeds does it contain?

25. If X is the number of seeds on a strawberry, what is P (185 < X < 205)?

26. How many seeds would a strawberry contain if it were at the 20th percentile?

27. How many seeds would a strawberry contain if it were at the 90th percentile?

Use the z-distribution table on pp. A-1 to A-2 or technology to solve.Scores on a college entrance test of verbal skills are normally distributed with a mean of 78 and a standard deviation of 5. At the same college, scores on the entrance test of mathematical skills are normally distributed with a mean of 84 and a standard deviation of 2.

28. Henry had a z-score of −0.1 on the verbal skills test and a z-score of 1.8 on the mathematical skills test. What raw score did he get on each test?

29. Millie scored an 83 on the verbal skills test and an 86 on the mathematical skills test. Explain why Millie did equally well on both tests.

30. On which test is a score of 80 more likely? Explain.

The uniform probability distribution shows temperature X of an oven set at 325°F at random times.


12. What is the height of the distribution?


14. Suppose the oven temperature is taken at 6 p.m. What is probability that the temperature will be within 1.5°F of 325°F?

15. If P (X > a) = 0.7, what is the value of a?

Use the probability distribution to find the probability indicated.



18. Determine P (X < 2 or X > 4).

19. Determine P (3 < X ≤ 6).



CHAPTER 4 WRAP-UP 179

HS_PS_S1_04_WU_PS.indd 179HS_PS_S1_04_WU_PS.indd 179 9/17/11 5:03 AM9/17/11 5:03 AM



chapter 4 random variables and...

Documents