184 chapter 4 probability and counting rules -...

184 Chapter 4 Probability and Counting Rules

4–4

Figure 4–1

Sample Space for

Rolling Two Dice

(Example 4–1)(1, 1)

(2, 1)

(3, 1)

(4, 1)

(5, 1)

(6, 1)

1

1

2

3

4

5

6

Die 1

(1, 2)

(2, 2)

(3, 2)

(4, 2)

(5, 2)

(6, 2)

2

(1, 3)

(2, 3)

(3, 3)

(4, 3)

(5, 3)

(6, 3)

3

(1, 4)

(2, 4)

(3, 4)

(4, 4)

(5, 4)

(6, 4)

4

Die 2

(1, 5)

(2, 5)

(3, 5)

(4, 5)

(5, 5)

(6, 5)

5

(1, 6)

(2, 6)

(3, 6)

(4, 6)

(5, 6)

(6, 6)

6

Figure 4–2

Sample Space for

Drawing a Card

(Example 4–2)

A 2 3 4 5 6 7 8 9 10 J Q K

A 2 3 4 5 6 7 8 9 10 J Q K

A 2 3 4 5 6 7 8 9 10 J Q K

A 2 3 4 5 6 7 8 9 10 J Q K

Example 4–2 Drawing Cards

Find the sample space for drawing one card from an ordinary deck of cards.

Solution

Since there are 4 suits (hearts, clubs, diamonds, and spades) and 13 cards for each suit

(ace through king), there are 52 outcomes in the sample space. See Figure 4–2.

Example 4–3 Gender of Children

Find the sample space for the gender of the children if a family has three children. Use

B for boy and G for girl.

Solution

There are two genders, male and female, and each child could be either gender. Hence,

there are eight possibilities, as shown here.

BBB BBG BGB GBB GGG GGB GBG BGG

In Examples 4–1 through 4–3, the sample spaces were found by observation and rea-

soning; however, another way to find all possible outcomes of a probability experiment

is to use a tree diagram.

Solution

Since each die can land in six different ways, and two dice are rolled, the sample space

can be presented by a rectangular array, as shown in Figure 4–1. The sample space is the

list of pairs of numbers in the chart.

A–1 Factorials

A–2 Summation Notation

A–3 The Line

A–1 Factorials

Definition and Properties of Factorials

The notation called factorial notation is used in probability.

Factorial notation uses the exclamation point and involves

multiplication. For example,

5! 5 5 ? 4 ? 3 ? 2 ? 1 5 120

4! 5 4 ? 3 ? 2 ? 1 5 24

3! 5 3 ? 2 ? 1 5 6

2! 5 2 ? 1 5 2

1! 5 1

In general, a factorial is evaluated as follows:

n! 5 n(n 2 1)(n 2 2) ? ? ? 3 ? 2 ? 1

Note that the factorial is the product of n factors, with the

number decreased by 1 for each factor.

One property of factorial notation is that it can be

stopped at any point by using the exclamation point. For

example,

5! 5 5 ? 4! since 4! 5 4 ? 3 ? 2 ? 1

5 5 ? 4 ? 3! since 3! 5 3 ? 2 ? 1

5 5 ? 4 ? 3 ? 2! since 2! 5 2 ? 1

5 5 ? 4 ? 3 ? 2 ? 1

Thus, n! 5 n(n 2 1)!

5 n(n 2 1)(n 2 2)!

5 n(n 2 1)(n 2 2)(n 2 3)! etc.

Another property of factorials is

0! 5 1

This fact is needed for formulas.

Operations with Factorials

Factorials cannot be added or subtracted directly. They

must be multiplied out. Then the products can be added or

subtracted.

Example A–1

Evaluate 3! 1 4!.

Solution

3! 1 4! 5 (3 ? 2 ? 1) 1 (4 ? 3 ? 2 ? 1)

5 6 1 24 5 30

Note: 3! 1 4! Þ 7!, since 7! 5 5040.

Example A–2

Evaluate 5! 2 3!.

Solution

5! 2 3! 5 (5 ? 4 ? 3 ? 2 ? 1) 2 (3 ? 2 ? 1)

5 120 2 6 5 114

Note: 5! 2 3! Þ 2!, since 2! 5 2.

Factorials cannot be multiplied directly. Again, you must

multiply them out and then multiply the products.

Example A–3

Evaluate 3! ? 2!.

Solution

3! ? 2! 5 (3 ? 2 ? 1) ? (2 ? 1) 5 6 ? 2 5 12

Note: 3! ? 2! Þ 6!, since 6! 5 720.

Finally, factorials cannot be divided directly unless they

are equal.

Example A–4

Evaluate 6! 4 3!.

Solution

Note:

But3!

3!5

3 • 2 • 1

3 • 2 • 15

6

65 1

6!

3!Þ 2! since 2! 5 2

6!

3!5

6 • 5 • 4 • 3 • 2 • 1

3 • 2 • 15

720

65 120

Appendix A

Algebra Review

A–1

A–2

In division, you can take some shortcuts, as shown:

Another shortcut that can be used with factorials is

cancellation, after factors have been expanded. For

example,

Now cancel both instances of 4!. Then cancel the 3 ? 2 in

the denominator with the 6 in the numerator.

Example A–5

Evaluate 10! 4 (6!)(4!).

Solution

Exercises

Evaluate each expression.

A–1. 9! 362,880 A–9. 20

A–2. 7! 5040 A–10. 7920

A–3. 5! 120 A–11. 126

A–4. 0! 1 A–12. 120

A–5. 1! 1 A–13. 70

A–6. 3! 6 A–14. 455

A–7. 1320 A–15. 1

A–8. 1,814,400 A–16. 105!

_3!+_2!+_1!+

10!

2!

10!

_10!+_0!+

12!

9!

15!

_12!+_3!+

8!

_4!+_4!+

10!

_7!+_3!+

9!

_4!+_5!+

11!

7!

5!

3!

10!

_6!+_4!+5

10 • 93

• 81

• 7 • 61

!

41

• 31

• 21

• 1 • 61

!5 10 • 3 • 7 5 210

7 • 61

• 5 • 41

!

31

• 21

• 1 • 41

!5 7 • 5 5 35

7!

_4!+_3!+5

7 • 6 • 5 • 4!

3 • 2 • 1 • 4!

5 8 • 7 5 56

8!

6!5

8 • 7 • 6!

6! and

6!

6!5 1

5 6 • 5 • 4 5 120

6!

3!5

6 • 5 • 4 • 3!

3! and

3!

3!5 1

A–17. 560 A–19. 2520

A–18. 1980 A–20. 90

A–2 Summation Notation

In mathematics, the symbol o (Greek capital letter sigma)

means to add or find the sum. For example, oX means

to add the numbers represented by the variable X. Thus,

when X represents 5, 8, 2, 4, and 6, then oX means

5 1 8 1 2 1 4 1 6 5 25.

Sometimes, a subscript notation is used, such as

This notation means to find the sum of five numbers

represented by X, as shown:

When the number of values is not known, the unknown

number can be represented by n, such as

There are several important types of summation used in

statistics. The notation oX 2 means to square each value

before summing. For example, if the values of the X’s are

2, 8, 6, 1, and 4, then

The notation (oX)2 means to find the sum of X’s and then

square the answer. For instance, if the values for X are 2, 8,

6, 1, and 4, then

Another important use of summation notation is in

finding the mean (shown in Section 3–1). The mean is

defined as

For example, to find the mean of 12, 8, 7, 3, and 10, use the

formula and substitute the values, as shown:

X 5oX

n5

12 1 8 1 7 1 3 1 10

55

40

55 8

X 5oX

n

X

5 _21+2 5 441

_oX+2 5 _2 1 8 1 6 1 1 1 4+2

5 4 1 64 1 36 1 1 1 16 5 121

oX 2 5 221 82

1 621 12

1 42

on

i51

Xi 5 X1 1 X2 1 X3 1 . . . 1 Xn

o5

i51

Xi 5 X1 1 X2 1 X3 1 X4 1 X5

o5

i51

Xi

6!

_2!+_2!+_2!+

11!

_7!+_2!+_2!+

10!

_3!+_2!+_5!+

8!

_3!+_3!+_2!+

754 Appendix A Algebra Review

The notation o(X 2 )2 means to perform the following

steps.

STEP 1 Find the mean.

STEP 2 Subtract the mean from each value.

STEP 3 Square the answers.

STEP 4 Find the sum.

Example A–6

Find the value of o(X 2 )2 for the values 12, 8, 7, 3, and

10 of X.

Solution

STEP 1 Find the mean.

STEP 2 Subtract the mean from each value.

12 2 8 5 4 7 2 8 5 21 10 2 8 5 2

8 2 8 5 0 3 2 8 5 25

STEP 3 Square the answers.

425 16 (21)2

5 1 225 4

025 0 (25)2

5 25

STEP 4 Find the sum.

16 1 0 1 1 1 25 1 4 5 46

Example A–7

Find o(X 2 )2 for the following values of X: 5, 7, 2, 1, 3, 6.

Solution

Find the mean.

Then the steps in Example A–6 can be shortened as

follows:

Exercises

For each set of values, find oX, oX 2, (oX)2, and o(X 2 )2.

A–21. 9, 17, 32, 16, 8, 2, 9, 7, 3, 18 121; 2181; 14,641; 716.9

A–22. 4, 12, 9, 13, 0, 6, 2, 10 56; 550; 3136; 158

A–23. 5, 12, 8, 3, 4 32; 258; 1024; 53.2

X

5 1 1 9 1 4 1 9 1 1 1 4 5 28

1 _21+2 1 22

5 121 32

1 _22+21 _23+2

1 _1 2 4+2 1 _3 2 4+2 1 _6 2 4+2

o_X 2 X+2 5 _5 2 4+2 1 _7 2 4+2 1 _2 2 4+2

X 55 1 7 1 2 1 1 1 3 1 6

65

24

65 4

X

X 512 1 8 1 7 1 3 1 10

55

40

55 8

X

X A–24. 6, 2, 18, 30, 31, 42, 16, 5 150; 4270; 22,500; 1457.5

A–25. 80, 76, 42, 53, 77 328; 22,678; 107,584; 1161.2

A–26. 123, 132, 216, 98, 146, 114829; 123,125; 687,241; 8584.8333

A–27. 53, 72, 81, 42, 63, 71, 73, 85, 98, 55693; 50,511; 480,249; 2486.1

A–28. 43, 32, 116, 98, 120 409; 40,333; 167,281; 6876.80

A–29. 12, 52, 36, 81, 63, 74 318; 20,150; 101,124; 3296

A–30. 29, 212, 18, 0, 22, 215 220; 778; 400; 711.3334

A–3 The Line

The following figure shows the rectangular coordinate

system, or Cartesian plane. This figure consists of two axes:

the horizontal axis, called the x axis, and the vertical axis,

called the y axis. Each axis has numerical scales. The point

of intersection of the axes is called the origin.

Points can be graphed by using coordinates. For example,

the notation for point P(3, 2) means that the x coordinate is

3 and the y coordinate is 2. Hence, P is located at the

intersection of x 5 3 and y 5 2, as shown.

Other points, such as Q(25, 2), R(4, 1), and S(23, 24),

can be plotted, as shown in the next figure.

When a point lies on the y axis, the x coordinate is 0, as

in (0, 6)(0, 23), etc. When a point lies on the x axis, the y

coordinate is 0, as in (6, 0)(28, 0), etc., as shown at the top

of the next page.

Appendix A Algebra Review 755

A–3

Appendix B–2

Bayes’ Theorem

A–9

Figure B–1

Tree Diagram for

Example 4–31

P(A1 and R ) = P(A1) ? P(R | A1)

P(R | A1) =

P(A1 and B ) = P(A1) ? P(B | A1)

P(A2 and R ) = P(A2) ? P(R | A2)

P(A2 and B ) = P(A2) ? P(B | A2)

12

23

23

P(A 1) =

12

P(A2 ) = 1

2

P(R | A2) =14

P(B | A1) = 1

3

P(B | A2) = 3

4

26

13

? ==

12

13

16

? =

12

14

18

? =

12

34

38

? =

Ball

Red

Blue

Red

Blue

Box 1

Box 2

Box

Given two dependent events Aand B, the previous formulas forconditional probability allow youto find P(A and B), or P(B A).Related to these formulas is arule developed by the EnglishPresbyterian minister ThomasBayes (1702–1761). The rule isknown as Bayes’ theorem.

It is possible, given theoutcome of the second event ina sequence of two events, todetermine the probability ofvarious possibilities for the firstevent. In Example 4–31, therewere two boxes, each containingred balls and blue balls. A boxwas selected and a ball wasdrawn. The example asked forthe probability that the ballselected was red. Now a

different question can be asked: If the ball is red, what isthe probability it came from box 1? In this case, theoutcome is known, a red ball was selected, and you areasked to find the probability that it is a result of a previousevent, that it came from box 1. Bayes’ theorem can enable

|

you to compute this probability and can be explained byusing tree diagrams.

The tree diagram for the solution of Example 4–31 isshown in Figure B–1, along with the appropriate notationand the corresponding probabilities. In this case, A1 is theevent of selecting box 1, A2 is the event of selecting box 2,R is the event of selecting a red ball, and B is the event ofselecting a blue ball.

To answer the question “If the ball selected is red, whatis the probability that it came from box 1?” two formulas

(1)

(2)

can be used. The notation that will be used is that ofExample 4–31, shown in Figure B–1. Finding theprobability that box 1 was selected given that the ballselected was red can be written symbolically as P(A1 R). By formula 1,

Note: P(R and A1) 5 P(A1 and R).

P_ A1|R+ 5P_R and A1+

P_R1+

|

P_ A and B+ 5 P_A+ • P_B|A+

P_B|A+ 5P_A and B+

P _ A+

Historical Notes

Thomas Bayes was

born around 1701

and lived in London.

He was an ordained

minister who dabbled

in mathematics and

statistics. All his

findings and writings

were published after

his death in 1761.

Objective B-1

Find the probability of

an event, using Bayes’

theorem.

By formula 2,

P(A1 and R) 5 P(A1) . P(R A1)

and

P(R) 5 P(A1 and R) 1 P(A2 and R)

as shown in Figure B–1; P(R) was found by adding theproducts of the probabilities of the branches in which a redball was selected. Now,

P(A1 and R) 5 P(A1) . P(R A1)

P(A2 and R) 5 P(A2) . P(R A2)

Substituting these values in the original formula for P(A1 R), you get

Refer to Figure B–1. The numerator of the fraction is theproduct of the top branch of the tree diagram, whichconsists of selecting a red ball and selecting box 1. And thedenominator is the sum of the products of the two branchesof the tree where the red ball was selected.

Using this formula and the probability values shown inFigure B–1, you can find the probability that box 1 wasselected given that the ball was red, as shown.

This formula is a simplified version of Bayes’ theorem.Before Bayes’ theorem is stated, another example is

shown.

51

34

11

245

1

31

•

248

115

8

11

512 •

23

12 •

23 1

12 •

14

5

13

13 1

18

5

13

824 1

324

5

131124

P_A1|R+ 5P_A1+ • P_R|A1+

P_A1+ • P_R|A1+ 1 P_A2+ • P_R|A2+

P_A1|R+ 5P_A1+ • P_R|A1+

P_A1+ • P_R|A1+ 1 P_A2+ • P_R|A2+

|

|

|

|

762 Appendix B–2 Bayes’ Theorem

A–10

Figure B–2

Tree Diagram for

Example B–1P(A1) ? P(D | A1)

A2

A1

P(D | A1) =16

P(A 1) =

12

P(A2 ) = 1

2

P(D | A2) =26

P(ND | A1) = 5

6

P(ND | A2) = 4

6

216

112? =

1=

P(A2) ? P(D | A2) 226

212? =

1=

61

=

Phone

D

ND

D

ND

Box

Example B–1

A shipment of two boxes, each containing six telephones,is received by a store. Box 1 contains one defective phone,and box 2 contains two defective phones. After the boxesare unpacked, a phone is selected and found to be defective.Find the probability that it came from box 2.

Solution

STEP 1 Select the proper notation. Let A1 represent box 1and A2 represent box 2. Let D represent adefective phone and ND represent a phone that isnot defective.

STEP 2 Draw a tree diagram and find the correspondingprobabilities for each branch. The probabilityof selecting box 1 is , and the probability ofselecting box 2 is . Since there is one defectivephone in box 1, the probability of selecting it is .The probability of selecting a nondefective phonefrom box 1 is .

Since there are two defective phones in box 2,the probability of selecting a defective phone frombox 2 is , or ; and the probability of selecting anondefective phone is , or . The tree diagram isshown in Figure B–2.

STEP 3 Write the corresponding formula. Since theexample is asking for the probability that, given a defective phone, it came from box 2, thecorresponding formula is as shown.

51

64

3

125

1

61

•

122

35

2

3

512 •

26

12 •

16 1

12 •

26

5

16

112 1

212

5

163

12

P_ A2|D + 5P_ A2+ • P_D|A2+

P_ A1+ • P_D|A1+ 1 P_ A2+ • P_D|A2+

23

46

13

26

56

16

12

12

random and selects a bill from the box at random. If a $100bill is selected, find the probability that it came from box 4.

Solution

STEP 1 Select the proper notation. Let B1, B2, B3, and B4

represent the boxes and 100 and 1 represent thevalues of the bills in the boxes.

STEP 2 Draw a tree diagram and find the correspondingprobabilities. The probability of selecting eachbox is , or 0.25. The probabilities of selectingthe $100 bill from each box, respectively, are

5 0.1, 5 0.2, 5 0.3, and 5 0.5. Thetree diagram is shown in Figure B–3.

STEP 3 Using Bayes’ theorem, write the correspondingformula. Since the example asks for the probabilitythat box 4 was selected, given that $100 wasobtained, the corresponding formula is as follows:

50.125

0.2755 0.455

50.125

0.025 1 0.05 1 0.075 1 0.125

1 P_B3+ • P_100|B3+ 1 P_B4+ • P_100|B4+]

P_B4|100+ 5P_B4+ • P_100|B4+

[P_B1+ • P_100|B1+ 1 P_B2+ • P_100|B2+

510

310

210

110

14

Appendix B–2 Bayes’ Theorem 763

A–11

Bayes’ theorem can be generalized to events with threeor more outcomes and formally stated as in the next box.

Bayes’ theorem For two events A and B, where event B

follows event A, event A can occur in A1, A2, . . . , An mutually

exclusive ways, and event B can occur in B1, B2, . . . , Bm

mutually exclusive ways,

for any specific events A1 and B1.

The numerator is the product of the probabilities on thebranch of the tree that consists of outcomes A1 and B1. Thedenominator is the sum of the products of the probabilitiesof the branches containing B1 and A1, B1 and A2, . . . , B1

and An.

Example B–2

On a game show, a contestant can select one of four boxes.Box 1 contains one $100 bill and nine $1 bills. Box 2contains two $100 bills and eight $1 bills. Box 3 containsthree $100 bills and seven $1 bills. Box 4 contains five$100 bills and five $1 bills. The contestant selects a box at

1 . . . 1 P _An+ • P _B1|An+]

P _A1|B1+ 5P _A1+ • P _B1|A1+

[P _A1+ • P _B1|A1+ 1 P _A2+ • P _B1|A2+

P(100 | B1) = 0.1

P(1 | B1) = 0.9

P(B1) ? P(100 | B1) = 0.025$100

$1

Box 1

Bill

Box

P(100 | B2) = 0.2

P(1 | B2) = 0.8

P(B2) ? P(100 | B2) = 0.05$100

$1

Box 2

P(100 | B3) = 0.3

P(1 | B3) = 0.7

P(B3) ? P(100 | B3) = 0.075

P(B

4 ) = 0.25

P(B

1) =

0.2

5

P(B3 ) = 0.25

P(B 2) =

0.25

$100

$1

Box 3

P(100 | B4) = 0.5

P(1 | B4) = 0.5

P(B4) ? P(100 | B4) = 0.125$100

$1

Box 4

Figure B–3

Tree Diagram for

Example B–2

In Example B–2, the original probability of selectingbox 4 was 0.25. However, once additional information wasobtained—and the condition was considered that a $100 billwas selected—the revised probability of selecting box 4became 0.455.

Bayes’ theorem can be used to revise probabilities ofevents once additional information becomes known. Bayes’theorem is used as the basis for a branch of statistics calledBayesian decision making, which includes the use ofsubjective probabilities in making statistical inferences.

Exercises

B–1. An appliance store purchases electric ranges fromtwo companies. From company A, 500 ranges arepurchased and 2% are defective. From company B,850 ranges are purchased and 2% are defective.Given that a range is defective, find the probabilitythat it came from company B. 0.65

B–2. Two manufacturers supply blankets to emergencyrelief organizations. Manufacturer A supplies3000 blankets, and 4% are irregular in workmanship.Manufacturer B supplies 2400 blankets, and 7%are found to be irregular. Given that a blanket isirregular, find the probability that it came frommanufacturer B. 0.579

B–3. A test for a certain disease is found to be 95%accurate, meaning that it will correctly diagnosethe disease in 95 out of 100 people who have theailment. For a certain segment of the population,the incidence of the disease is 9%. If a person testspositive, find the probability that the person actuallyhas the disease. The test is also 95% accurate for anegative result. 0.653

B–4. Using the test in Exercise B–3, if a person testsnegative for the disease, find the probability that theperson actually has the disease. Remember, 9% ofthe population has the disease. 0.005

B–5. A corporation has three methods of trainingemployees. Because of time, space, and location, itsends 20% of its employees to location A, 35% tolocation B, and 45% to location C. Location A hasan 80% success rate. That is, 80% of the employeeswho complete the course will pass the licensing

exam. Location B has a 75% success rate, andlocation C has a 60% success rate. If a person haspassed the exam, find the probability that the personwent to location B. 0.379

B–6. In Exercise B–5, if a person failed the exam, find theprobability that the person went to location C. 0.585

B–7. A store purchases baseball hats from three differentmanufacturers. In manufacturer A’s box, there are12 blue hats, 6 red hats, and 6 green hats. Inmanufacturer B’s box, there are 10 blue hats, 10 redhats, and 4 green hats. In manufacturer C’s box,there are 8 blue hats, 8 red hats, and 8 green hats. Abox is selected at random, and a hat is selected atrandom from that box. If the hat is red, find theprobability that it came from manufacturer A’s box.

B–8. In Exercise B–7, if the hat selected is green, find theprobability that it came from manufacturer B’s box.

B–9. A driver has three ways to get from one city toanother. There is an 80% probability of encounteringa traffic jam on route 1, a 60% probability onroute 2, and a 30% probability on route 3. Becauseof other factors, such as distance and speed limits,the driver uses route 1 fifty percent of the time androutes 2 and 3 each 25% of the time. If the drivercalls the dispatcher to inform him that she is in atraffic jam, find the probability that she has selectedroute 1. 0.64

B–10. In Exercise B–9, if the driver did not encounter atraffic jam, find the probability that she selectedroute 3. 0.467

B–11. A store owner purchases telephones from twocompanies. From company A, 350 telephones arepurchased and 2% are defective. From company B,1050 telephones are purchased and 4% are defective.Given that a phone is defective, find the probabilitythat it came from company B. 0.857

B–12. Two manufacturers supply food to a large cafeteria.Manufacturer A supplies 2400 cans of soup, and 3%are found to be dented. Manufacturer B supplies3600 cans, and 1% are found to be dented. Giventhat a can of soup is dented, find the probability thatit came from manufacturer B. 0.33

2

9

1

4

764 Appendix B–2 Bayes’ Theorem

A–12

Appendix B–3

Alternate Approach to the Standard Normal Distribution

A–13

The following procedure may be used to replace thecumulative area to the left procedure shown in Section 6–1.This method determines areas from the mean where z 5 0.

Finding Areas Under the Standard Normal

Distribution

For the solution of problems using the standard normaldistribution, a four-step procedure may be used with the useof the Procedure Table shown.

STEP 1 Sketch the normal curve and label.

STEP 2 Shade the area desired.

STEP 3 Find the figure that matches the shaded area fromthe following procedure table.

STEP 4 Follow the directions given in the appropriateblock of the procedure table to get the desired area.

Note: Table B–1 gives the area between 0 and any z scoreto the right of 0, and all areas are positive.

There are seven basic types of problems and all sevenare summarized in the Procedure Table, with appropriateexamples.

766 Appendix B–3 Alternate Approach to the Standard Normal Distribution

A–14

Pro

ced

ure

Tab

leE

xam

ple

s

01

z

01

z0

2z

0z

2z

10

2z

22

z1

01

z2

z

B–3–1:

Fin

d t

he

area

bet

wee

n z

50

an

d z

51

.23

.

Look u

p a

rea

from

z5

0 t

o z

51

.23

on

Tab

leB

–1,

as s

how

n b

elow

.

2.

In a

ny t

ail:

a.

Lo

ok

up

th

e z

score

to g

et t

he

area

.

b.

Subtr

act

the

area

fro

m 0

.5000.

B–3–2:

Fin

d t

he

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on

Tab

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sh

ow

n b

elo

w.

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0 t

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51

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50

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18

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01.

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.

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z.0

7

. . .

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0.0

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0.49

11

02.

371.

23

02.

372

1.23

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n z

50

an

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51

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m z

50

to

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m z

50

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he

left

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z5

22

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50

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00

20

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11 5

0.0

08

9.

02

z

Appendix B–3 Alternate Approach to the Standard Normal Distribution 767

A–15

5.

To t

he

left

of

any z

score

, w

her

e z

is g

reat

er t

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s 0

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6.

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50

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the

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2

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3.

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a to

the

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00

0 2

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50

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.

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the

right

of

z5

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75

0.1

093

10.0

089

50.1

182.

01

z

02

z

01

z2

z

02.

37

02

2.37

02.

372

1.23

768 Appendix B–3 Alternate Approach to the Standard Normal Distribution

A–16

0 z

Area given in table

Table B–1 The Standard Normal Distribution

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359

0.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753

0.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141

0.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517

0.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879

0.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224

0.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549

0.7 .2580 .2611 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852

0.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133

0.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389

1.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .3621

1.1 .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .3830

1.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .4015

1.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .4177

1.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319

1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441

1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545

1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633

1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706

1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767

2.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .4817

2.1 .4821 .4826 .4830 .4834 .4838 .4842 .4846 .4850 .4854 .4857

2.2 .4861 .4864 .4868 .4871 .4875 .4878 .4881 .4884 .4887 .4890

2.3 .4893 .4896 .4898 .4901 .4904 .4906 .4909 .4911 .4913 .4916

2.4 .4918 .4920 .4922 .4925 .4927 .4929 .4931 .4932 .4934 .4936

2.5 .4938 .4940 .4941 .4943 .4945 .4946 .4948 .4949 .4951 .4952

2.6 .4953 .4955 .4956 .4957 .4959 .4960 .4961 .4962 .4963 .4964

2.7 .4965 .4966 .4967 .4968 .4969 .4970 .4971 .4972 .4973 .4974

2.8 .4974 .4975 .4976 .4977 .4977 .4978 .4979 .4979 .4980 .4981

2.9 .4981 .4982 .4982 .4983 .4984 .4984 .4985 .4985 .4986 .4986

3.0 .4987 .4987 .4987 .4988 .4988 .4989 .4989 .4989 .4990 .4990

3.1 .4990 .4991 .4991 .4991 .4992 .4992 .4992 .4992 .4993 .4993

3.2 .4993 .4993 .4994 .4994 .4994 .4994 .4994 .4995 .4995 .4995

3.3 .4995 .4995 .4995 .4996 .4996 .4996 .4996 .4996 .4996 .4997

3.4 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997 .4998

For z values greater than 3.49, use 0.4999.

A–17

Appendix C

Tables

Table A Factorials

Table B The Binomial Distribution

Table C The Poisson Distribution

Table D Random Numbers

Table E The Standard Normal Distribution

Table F The t Distribution

Table G The Chi-Square Distribution

Table H The F Distribution

Table I Critical Values for the PPMC

Table J Critical Values for the Sign Test

Table K Critical Values for the Wilcoxon Signed-Rank Test

Table L Critical Values for the Rank Correlation Coefficient

Table M Critical Values for the Number of Runs

Table N Critical Values for the Tukey Test

Table A Factorials

n n!

0 1

1 1

2 2

3 6

4 24

5 120

6 720

7 5,040

8 40,320

9 362,880

10 3,628,800

11 39,916,800

12 479,001,600

13 6,227,020,800

14 87,178,291,200

15 1,307,674,368,000

16 20,922,789,888,000

17 355,687,428,096,000

18 6,402,373,705,728,000

19 121,645,100,408,832,000

20 2,432,902,008,176,640,000

770 Appendix C Tables

A–18

Table B The Binomial Distribution

p

n x 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95

2 0 0.902 0.810 0.640 0.490 0.360 0.250 0.160 0.090 0.040 0.010 0.002

1 0.095 0.180 0.320 0.420 0.480 0.500 0.480 0.420 0.320 0.180 0.095

2 0.002 0.010 0.040 0.090 0.160 0.250 0.360 0.490 0.640 0.810 0.902

3 0 0.857 0.729 0.512 0.343 0.216 0.125 0.064 0.027 0.008 0.001

1 0.135 0.243 0.384 0.441 0.432 0.375 0.288 0.189 0.096 0.027 0.007

2 0.007 0.027 0.096 0.189 0.288 0.375 0.432 0.441 0.384 0.243 0.135

3 0.001 0.008 0.027 0.064 0.125 0.216 0.343 0.512 0.729 0.857

4 0 0.815 0.656 0.410 0.240 0.130 0.062 0.026 0.008 0.002

1 0.171 0.292 0.410 0.412 0.346 0.250 0.154 0.076 0.026 0.004

2 0.014 0.049 0.154 0.265 0.346 0.375 0.346 0.265 0.154 0.049 0.014

3 0.004 0.026 0.076 0.154 0.250 0.346 0.412 0.410 0.292 0.171

4 0.002 0.008 0.026 0.062 0.130 0.240 0.410 0.656 0.815

5 0 0.774 0.590 0.328 0.168 0.078 0.031 0.010 0.002

1 0.204 0.328 0.410 0.360 0.259 0.156 0.077 0.028 0.006

2 0.021 0.073 0.205 0.309 0.346 0.312 0.230 0.132 0.051 0.008 0.001

3 0.001 0.008 0.051 0.132 0.230 0.312 0.346 0.309 0.205 0.073 0.021

4 0.006 0.028 0.077 0.156 0.259 0.360 0.410 0.328 0.204

5 0.002 0.010 0.031 0.078 0.168 0.328 0.590 0.774

6 0 0.735 0.531 0.262 0.118 0.047 0.016 0.004 0.001

1 0.232 0.354 0.393 0.303 0.187 0.094 0.037 0.010 0.002

2 0.031 0.098 0.246 0.324 0.311 0.234 0.138 0.060 0.015 0.001

3 0.002 0.015 0.082 0.185 0.276 0.312 0.276 0.185 0.082 0.015 0.002

4 0.001 0.015 0.060 0.138 0.234 0.311 0.324 0.246 0.098 0.031

5 0.002 0.010 0.037 0.094 0.187 0.303 0.393 0.354 0.232

6 0.001 0.004 0.016 0.047 0.118 0.262 0.531 0.735

7 0 0.698 0.478 0.210 0.082 0.028 0.008 0.002

1 0.257 0.372 0.367 0.247 0.131 0.055 0.017 0.004

2 0.041 0.124 0.275 0.318 0.261 0.164 0.077 0.025 0.004

3 0.004 0.023 0.115 0.227 0.290 0.273 0.194 0.097 0.029 0.003

4 0.003 0.029 0.097 0.194 0.273 0.290 0.227 0.115 0.023 0.004

5 0.004 0.025 0.077 0.164 0.261 0.318 0.275 0.124 0.041

6 0.004 0.017 0.055 0.131 0.247 0.367 0.372 0.257

7 0.002 0.008 0.028 0.082 0.210 0.478 0.698

8 0 0.663 0.430 0.168 0.058 0.017 0.004 0.001

1 0.279 0.383 0.336 0.198 0.090 0.031 0.008 0.001

2 0.051 0.149 0.294 0.296 0.209 0.109 0.041 0.010 0.001

3 0.005 0.033 0.147 0.254 0.279 0.219 0.124 0.047 0.009

4 0.005 0.046 0.136 0.232 0.273 0.232 0.136 0.046 0.005

5 0.009 0.047 0.124 0.219 0.279 0.254 0.147 0.033 0.005

6 0.001 0.010 0.041 0.109 0.209 0.296 0.294 0.149 0.051

7 0.001 0.008 0.031 0.090 0.198 0.336 0.383 0.279

8 0.001 0.004 0.017 0.058 0.168 0.430 0.663

Appendix C Tables 771

A–19

Table B (continued)

p

n x 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95

9 0 0.630 0.387 0.134 0.040 0.010 0.002

1 0.299 0.387 0.302 0.156 0.060 0.018 0.004

2 0.063 0.172 0.302 0.267 0.161 0.070 0.021 0.004

3 0.008 0.045 0.176 0.267 0.251 0.164 0.074 0.021 0.003

4 0.001 0.007 0.066 0.172 0.251 0.246 0.167 0.074 0.017 0.001

5 0.001 0.017 0.074 0.167 0.246 0.251 0.172 0.066 0.007 0.001

6 0.003 0.021 0.074 0.164 0.251 0.267 0.176 0.045 0.008

7 0.004 0.021 0.070 0.161 0.267 0.302 0.172 0.063

8 0.004 0.018 0.060 0.156 0.302 0.387 0.299

9 0.002 0.010 0.040 0.134 0.387 0.630

10 0 0.599 0.349 0.107 0.028 0.006 0.001

1 0.315 0.387 0.268 0.121 0.040 0.010 0.002

2 0.075 0.194 0.302 0.233 0.121 0.044 0.011 0.001

3 0.010 0.057 0.201 0.267 0.215 0.117 0.042 0.009 0.001

4 0.001 0.011 0.088 0.200 0.251 0.205 0.111 0.037 0.006

5 0.001 0.026 0.103 0.201 0.246 0.201 0.103 0.026 0.001

6 0.006 0.037 0.111 0.205 0.251 0.200 0.088 0.011 0.001

7 0.001 0.009 0.042 0.117 0.215 0.267 0.201 0.057 0.010

8 0.001 0.011 0.044 0.121 0.233 0.302 0.194 0.075

9 0.002 0.010 0.040 0.121 0.268 0.387 0.315

10 0.001 0.006 0.028 0.107 0.349 0.599

11 0 0.569 0.314 0.086 0.020 0.004

1 0.329 0.384 0.236 0.093 0.027 0.005 0.001

2 0.087 0.213 0.295 0.200 0.089 0.027 0.005 0.001

3 0.014 0.071 0.221 0.257 0.177 0.081 0.023 0.004

4 0.001 0.016 0.111 0.220 0.236 0.161 0.070 0.017 0.002

5 0.002 0.039 0.132 0.221 0.226 0.147 0.057 0.010

6 0.010 0.057 0.147 0.226 0.221 0.132 0.039 0.002

7 0.002 0.017 0.070 0.161 0.236 0.220 0.111 0.016 0.001

8 0.004 0.023 0.081 0.177 0.257 0.221 0.071 0.014

9 0.001 0.005 0.027 0.089 0.200 0.295 0.213 0.087

10 0.001 0.005 0.027 0.093 0.236 0.384 0.329

11 0.004 0.020 0.086 0.314 0.569

12 0 0.540 0.282 0.069 0.014 0.002

1 0.341 0.377 0.206 0.071 0.017 0.003

2 0.099 0.230 0.283 0.168 0.064 0.016 0.002

3 0.017 0.085 0.236 0.240 0.142 0.054 0.012 0.001

4 0.002 0.021 0.133 0.231 0.213 0.121 0.042 0.008 0.001

5 0.004 0.053 0.158 0.227 0.193 0.101 0.029 0.003

6 0.016 0.079 0.177 0.226 0.177 0.079 0.016

7 0.003 0.029 0.101 0.193 0.227 0.158 0.053 0.004

8 0.001 0.008 0.042 0.121 0.213 0.231 0.133 0.021 0.002

9 0.001 0.012 0.054 0.142 0.240 0.236 0.085 0.017

10 0.002 0.016 0.064 0.168 0.283 0.230 0.099

11 0.003 0.017 0.071 0.206 0.377 0.341

12 0.002 0.014 0.069 0.282 0.540


A–20

Table B (continued)

p

n x 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95

13 0 0.513 0.254 0.055 0.010 0.001

1 0.351 0.367 0.179 0.054 0.011 0.002

2 0.111 0.245 0.268 0.139 0.045 0.010 0.001

3 0.021 0.100 0.246 0.218 0.111 0.035 0.006 0.001

4 0.003 0.028 0.154 0.234 0.184 0.087 0.024 0.003

5 0.006 0.069 0.180 0.221 0.157 0.066 0.014 0.001

6 0.001 0.023 0.103 0.197 0.209 0.131 0.044 0.006

7 0.006 0.044 0.131 0.209 0.197 0.103 0.023 0.001

8 0.001 0.014 0.066 0.157 0.221 0.180 0.069 0.006

9 0.003 0.024 0.087 0.184 0.234 0.154 0.028 0.003

10 0.001 0.006 0.035 0.111 0.218 0.246 0.100 0.021

11 0.001 0.010 0.045 0.139 0.268 0.245 0.111

12 0.002 0.011 0.054 0.179 0.367 0.351

13 0.001 0.010 0.055 0.254 0.513

14 0 0.488 0.229 0.044 0.007 0.001

1 0.359 0.356 0.154 0.041 0.007 0.001

2 0.123 0.257 0.250 0.113 0.032 0.006 0.001

3 0.026 0.114 0.250 0.194 0.085 0.022 0.003

4 0.004 0.035 0.172 0.229 0.155 0.061 0.014 0.001

5 0.008 0.086 0.196 0.207 0.122 0.041 0.007

6 0.001 0.032 0.126 0.207 0.183 0.092 0.023 0.002

7 0.009 0.062 0.157 0.209 0.157 0.062 0.009

8 0.002 0.023 0.092 0.183 0.207 0.126 0.032 0.001

9 0.007 0.041 0.122 0.207 0.196 0.086 0.008

10 0.001 0.014 0.061 0.155 0.229 0.172 0.035 0.004

11 0.003 0.022 0.085 0.194 0.250 0.114 0.026

12 0.001 0.006 0.032 0.113 0.250 0.257 0.123

13 0.001 0.007 0.041 0.154 0.356 0.359

14 0.001 0.007 0.044 0.229 0.488

15 0 0.463 0.206 0.035 0.005

1 0.366 0.343 0.132 0.031 0.005

2 0.135 0.267 0.231 0.092 0.022 0.003

3 0.031 0.129 0.250 0.170 0.063 0.014 0.002

4 0.005 0.043 0.188 0.219 0.127 0.042 0.007 0.001

5 0.001 0.010 0.103 0.206 0.186 0.092 0.024 0.003

6 0.002 0.043 0.147 0.207 0.153 0.061 0.012 0.001

7 0.014 0.081 0.177 0.196 0.118 0.035 0.003

8 0.003 0.035 0.118 0.196 0.177 0.081 0.014

9 0.001 0.012 0.061 0.153 0.207 0.147 0.043 0.002

10 0.003 0.024 0.092 0.186 0.206 0.103 0.010 0.001

11 0.001 0.007 0.042 0.127 0.219 0.188 0.043 0.005

12 0.002 0.014 0.063 0.170 0.250 0.129 0.031

13 0.003 0.022 0.092 0.231 0.267 0.135

14 0.005 0.031 0.132 0.343 0.366

15 0.005 0.035 0.206 0.463


A–21

Table B (continued)

p

n x 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95

16 0 0.440 0.185 0.028 0.003

1 0.371 0.329 0.113 0.023 0.003

2 0.146 0.275 0.211 0.073 0.015 0.002

3 0.036 0.142 0.246 0.146 0.047 0.009 0.001

4 0.006 0.051 0.200 0.204 0.101 0.028 0.004

5 0.001 0.014 0.120 0.210 0.162 0.067 0.014 0.001

6 0.003 0.055 0.165 0.198 0.122 0.039 0.006

7 0.020 0.101 0.189 0.175 0.084 0.019 0.001

8 0.006 0.049 0.142 0.196 0.142 0.049 0.006

9 0.001 0.019 0.084 0.175 0.189 0.101 0.020

10 0.006 0.039 0.122 0.198 0.165 0.055 0.003

11 0.001 0.014 0.067 0.162 0.210 0.120 0.014 0.001

12 0.004 0.028 0.101 0.204 0.200 0.051 0.006

13 0.001 0.009 0.047 0.146 0.246 0.142 0.036

14 0.002 0.015 0.073 0.211 0.275 0.146

15 0.003 0.023 0.113 0.329 0.371

16 0.003 0.028 0.185 0.440

17 0 0.418 0.167 0.023 0.002

1 0.374 0.315 0.096 0.017 0.002

2 0.158 0.280 0.191 0.058 0.010 0.001

3 0.041 0.156 0.239 0.125 0.034 0.005

4 0.008 0.060 0.209 0.187 0.080 0.018 0.002

5 0.001 0.017 0.136 0.208 0.138 0.047 0.008 0.001

6 0.004 0.068 0.178 0.184 0.094 0.024 0.003

7 0.001 0.027 0.120 0.193 0.148 0.057 0.009

8 0.008 0.064 0.161 0.185 0.107 0.028 0.002

9 0.002 0.028 0.107 0.185 0.161 0.064 0.008

10 0.009 0.057 0.148 0.193 0.120 0.027 0.001

11 0.003 0.024 0.094 0.184 0.178 0.068 0.004

12 0.001 0.008 0.047 0.138 0.208 0.136 0.017 0.001

13 0.002 0.018 0.080 0.187 0.209 0.060 0.008

14 0.005 0.034 0.125 0.239 0.156 0.041

15 0.001 0.010 0.058 0.191 0.280 0.158

16 0.002 0.017 0.096 0.315 0.374

17 0.002 0.023 0.167 0.418


A–22

Table B (continued)

p

n x 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95

18 0 0.397 0.150 0.018 0.002

1 0.376 0.300 0.081 0.013 0.001

2 0.168 0.284 0.172 0.046 0.007 0.001

3 0.047 0.168 0.230 0.105 0.025 0.003

4 0.009 0.070 0.215 0.168 0.061 0.012 0.001

5 0.001 0.022 0.151 0.202 0.115 0.033 0.004

6 0.005 0.082 0.187 0.166 0.071 0.015 0.001

7 0.001 0.035 0.138 0.189 0.121 0.037 0.005

8 0.012 0.081 0.173 0.167 0.077 0.015 0.001

9 0.003 0.039 0.128 0.185 0.128 0.039 0.003

10 0.001 0.015 0.077 0.167 0.173 0.081 0.012

11 0.005 0.037 0.121 0.189 0.138 0.035 0.001

12 0.001 0.015 0.071 0.166 0.187 0.082 0.005

13 0.004 0.033 0.115 0.202 0.151 0.022 0.001

14 0.001 0.012 0.061 0.168 0.215 0.070 0.009

15 0.003 0.025 0.105 0.230 0.168 0.047

16 0.001 0.007 0.046 0.172 0.284 0.168

17 0.001 0.013 0.081 0.300 0.376

18 0.002 0.018 0.150 0.397

19 0 0.377 0.135 0.014 0.001

1 0.377 0.285 0.068 0.009 0.001

2 0.179 0.285 0.154 0.036 0.005

3 0.053 0.180 0.218 0.087 0.017 0.002

4 0.011 0.080 0.218 0.149 0.047 0.007 0.001

5 0.002 0.027 0.164 0.192 0.093 0.022 0.002

6 0.007 0.095 0.192 0.145 0.052 0.008 0.001

7 0.001 0.044 0.153 0.180 0.096 0.024 0.002

8 0.017 0.098 0.180 0.144 0.053 0.008

9 0.005 0.051 0.146 0.176 0.098 0.022 0.001

10 0.001 0.022 0.098 0.176 0.146 0.051 0.005

11 0.008 0.053 0.144 0.180 0.098 0.071

12 0.002 0.024 0.096 0.180 0.153 0.044 0.001

13 0.001 0.008 0.052 0.145 0.192 0.095 0.007

14 0.002 0.022 0.093 0.192 0.164 0.027 0.002

15 0.001 0.007 0.047 0.149 0.218 0.080 0.011

16 0.002 0.017 0.087 0.218 0.180 0.053

17 0.005 0.036 0.154 0.285 0.179

18 0.001 0.009 0.068 0.285 0.377

19 0.001 0.014 0.135 0.377


A–23

Table B (concluded)

p

n x 0.05 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95

20 0 0.358 0.122 0.012 0.001

1 0.377 0.270 0.058 0.007

2 0.189 0.285 0.137 0.028 0.003

3 0.060 0.190 0.205 0.072 0.012 0.001

4 0.013 0.090 0.218 0.130 0.035 0.005

5 0.002 0.032 0.175 0.179 0.075 0.015 0.001

6 0.009 0.109 0.192 0.124 0.037 0.005

7 0.002 0.055 0.164 0.166 0.074 0.015 0.001

8 0.022 0.114 0.180 0.120 0.035 0.004

9 0.007 0.065 0.160 0.160 0.071 0.012

10 0.002 0.031 0.117 0.176 0.117 0.031 0.002

11 0.012 0.071 0.160 0.160 0.065 0.007

12 0.004 0.035 0.120 0.180 0.114 0.022

13 0.001 0.015 0.074 0.166 0.164 0.055 0.002

14 0.005 0.037 0.124 0.192 0.109 0.009

15 0.001 0.015 0.075 0.179 0.175 0.032 0.002

16 0.005 0.035 0.130 0.218 0.090 0.013

17 0.001 0.012 0.072 0.205 0.190 0.060

18 0.003 0.028 0.137 0.285 0.189

19 0.007 0.058 0.270 0.377

20 0.001 0.012 0.122 0.358

Note: All values of 0.0005 or less are omitted.

Source: J. Freund and G. Simon, Modern Elementary Statistics, Table “The Binomial Distribution,” © 1992 Prentice-Hall, Inc. Reproduced by permission of Pearson

Education, Inc.


A–32

Table E The Standard Normal Distribution

Cumulative Standard Normal Distribution

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

23.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002

23.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .0003

23.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005

23.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007

23.0 .0013 .0013 .0013 .0012 .0012 .0011 .0011 .0011 .0010 .0010

22.9 .0019 .0018 .0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014

22.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019

22.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026

22.6 .0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036

22.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048

22.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064

22.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084

22.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110

22.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143

22.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188 .0183

21.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233

21.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294

21.7 .0446 .0436 .0427 .0418 .0409 .0401 .0392 .0384 .0375 .0367

21.6 .0548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 .0465 .0455

21.5 .0668 .0655 .0643 .0630 .0618 .0606 .0594 .0582 .0571 .0559

21.4 .0808 .0793 .0778 .0764 .0749 .0735 .0721 .0708 .0694 .0681

21.3 .0968 .0951 .0934 .0918 .0901 .0885 .0869 .0853 .0838 .0823

21.2 .1151 .1131 .1112 .1093 .1075 .1056 .1038 .1020 .1003 .0985

21.1 .1357 .1335 .1314 .1292 .1271 .1251 .1230 .1210 .1190 .1170

21.0 .1587 .1562 .1539 .1515 .1492 .1469 .1446 .1423 .1401 .1379

20.9 .1841 .1814 .1788 .1762 .1736 .1711 .1685 .1660 .1635 .1611

20.8 .2119 .2090 .2061 .2033 .2005 .1977 .1949 .1922 .1894 .1867

20.7 .2420 .2389 .2358 .2327 .2296 .2266 .2236 .2206 .2177 .2148

20.6 .2743 .2709 .2676 .2643 .2611 .2578 .2546 .2514 .2483 .2451

20.5 .3085 .3050 .3015 .2981 .2946 .2912 .2877 .2843 .2810 .2776

20.4 .3446 .3409 .3372 .3336 .3300 .3264 .3228 .3192 .3156 .3121

20.3 .3821 .3783 .3745 .3707 .3669 .3632 .3594 .3557 .3520 .3483

20.2 .4207 .4168 .4129 .4090 .4052 .4013 .3974 .3936 .3897 .3859

20.1 .4602 .4562 .4522 .4483 .4443 .4404 .4364 .4325 .4286 .4247

20.0 .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4721 .4681 .4641

For z values less than 23.49, use 0.0001.

0z

Area


A–33

Table E (continued )

Cumulative Standard Normal Distribution

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141

0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517

0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879

0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852

0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177

1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633

1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706

1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767

2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817

2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857

2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890

2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916

2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936

2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952

2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964

2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974

2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981

2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986

3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990

3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993

3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995

3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997

3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998

For z values greater than 3.49, use 0.9999.

0 z

Area

184 chapter 4 probability and counting rules -...

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