184 chapter 4 probability and counting rules -...
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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
area
to t
he
left
of
z5
22
.37
.
Look u
p a
rea
from
z5
0 t
o z
52
.37
on
Tab
leB
–1
, as
sh
ow
n b
elo
w.
3.
Bet
wee
n t
wo z
score
s on t
he
sam
e si
de
of
the
mea
n:
a.
Lo
ok
up
bo
th z
score
s to
get
the
area
s.
b.
Subtr
act
the
smal
ler
area
fro
m t
he
larg
er a
rea.
B–3–3:
Fin
d t
he
area
bet
wee
n z
51
.23
an
d z
52
.37
.
Look u
p a
reas
for
z5
0 t
o z
51
.23
an
d
z5
0 t
o z
52
.37
, as
sh
ow
n i
n B
–3
–1
and
B–3–2,
resp
ecti
vel
y.
The
area
bet
wee
n z
51
.23
an
d z
5
2.3
7 5
0.4
911
20
.39
07
50
.10
04
.
4.
Bet
wee
n t
wo z
score
s on o
pposi
te s
ides
of
the
mea
n:
a.
Lo
ok
up
bo
th z
score
s to
get
the
area
s.
b.
Add t
he
area
s.
B–3–4:
Fin
d t
he
area
bet
wee
n z
52
1.2
3 a
nd
z5
2.3
7.
Look u
p a
reas
for
z5
0 t
o z
51
.23
an
d
z5
0 t
o z
52
.37
, as
sh
ow
n i
n B
–3
–1
and
B–3–2,
resp
ecti
vel
y.
The
area
bet
wee
n z
52
1.2
3 a
nd
z5
2.3
7 5
0.3
90
7 1
0.4
911
50
.88
18
.
1.
Bet
wee
n 0
and a
ny z
score
:
Lo
ok
up
th
e z
score
in t
he
table
to g
et t
he
area
.
01.
23
z.0
3. .
.
. . .
0.0
1.2
0.39
07
02
2.37
z.0
7
. . .
. . .
0.0
2.3
0.49
11
02.
371.
23
02.
372
1.23
The
area
bet
wee
n z
50
an
d z
51
.23
is
0.3
90
7.
The
area
fro
m z
50
to
z5
2.3
7 i
s th
e sa
me
as t
he
area
fro
m z
50
to
z5
22
.37
.
Ther
efore
, th
e ar
ea t
o t
he
left
of
z5
22
.37
50
.50
00
20
.49
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
han
the
mea
n:
a.
Lo
ok
up
th
e z
score
to g
et t
he
area
.
b.
Add 0
.5000 t
o t
he
area
.
B–3–5:
Fin
d t
he
area
to t
he
left
of
z5
2.3
7.
Look u
p a
rea
for
z5
2.3
7,
as s
ho
wn
in
B–
3–
2.
The
area
to t
he
left
of
z5
0 i
s 0
.50
00
.
The
area
to t
he
left
of
z5
2.3
7 5
0.4
911
1
0.5
00
0 5
0.9
911
.
6.
To t
he
right
of
any z
score
, w
her
e z
is l
ess
than
the
mea
n:
a.
Look u
p t
he
area
in t
he
table
to g
et t
he
area
.
b.
Add 0
.5000 t
o t
he
area
.
B–3–6:
Fin
d t
he
area
to t
he
right
of
z5
22
.37
.
Look u
p a
rea
for
z5
2.3
7,
as s
ho
wn
in
B–
3–
2.
The
area
to t
he
right
of
z5
0 i
s 0
.50
00
.
The
area
to t
he
right
of
z5
22.3
7 5
0.4
911
1
0.5
000 5
0.9
911
.
7.
In a
ny t
wo t
ails
:
a.
Lo
ok
up
th
e z
score
s in
the
table
to g
et t
he
area
s.
b.
Subtr
act
both
are
as f
rom
0.5
000.
c.A
dd t
he
answ
ers.
B–3–7:
Fin
d t
he
area
to t
he
left
of
z5
21.2
3 a
nd t
o t
he
right
of
z5
2.3
7.
Look
up
area
sfo
rz
50
toz
51.2
3an
dz
50
toz
52.3
7,
assh
ow
nin
B–3–1
and
B–3–2,
resp
ecti
vel
y.
Are
a to
the
left
of
z5
21
.23
50
.50
00
2
0.3
90
7 5
0.1
09
3.
Are
a to
the
right
of
z5
2.3
7 5
0.5
00
0 2
0.4
91 1
50
.00
89
.
The
area
toth
ele
ftof
z5
21.2
3an
dto
the
right
of
z5
2.3
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
772 Appendix C Tables
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
Appendix C Tables 773
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
774 Appendix C Tables
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
Appendix C Tables 775
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.
784 Appendix C Tables
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
Appendix C Tables 785
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