1. Overview probability distributions
This chapter will deal with the construction of
probability distributions
Probability Distributions will describe what will probably happen instead of what actually did happen.
Emphasize the combination of the methods in Chapter 2 (descriptive statistics) with the methods in Chapter 3 (probability).
Chapter 2 one would conduct an actual experiment and find and observe the mean, standard deviation, variance, etc. and construct a frequency table or histogram
Chapter 3 finds the probability of each possible outcome
Chapter 4 presents the possible outcomes along with the relative frequencies we expect

2. Random Variables

3. Definitions Random Variable Probability Distribution
a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure
Probability Distribution
a graph, table, or formula that gives the probability for each value of the random variable
page 181 of text

4. Definitions Discrete random variable
has either a finite number of values or countable number of values, where ‘countable’ refers to the fact that there might be infinitely many values, but they result from a counting process.
Continuous random variable
has infinitely many values, and those values can be associated with measurements on a continuous scale with no gaps or interruptions.
This chapter deals exclusively with discrete random variables - experiments where the data observed is a ‘countable’ value. Give examples.
Following chapters will deal with continuous random variables.

5. Requirements for Probability Distribution
P(x) = 1
where x assumes all possible values
0  P(x)  1
for every value of x

6. Mean, Variance and Standard Deviation of a Probability Distribution
Formula 6-1
µ =  [x • P(x)]
Formula 6-2
2 =  [(x - µ)2 • P(x)]

7. The probability that a car insurance company
pays $200,000 is .01, $100,000 is .02 and
$50,000 is .04.
.93
Draw the probability density function.
Determine the minimum value the insurance
company should charge to break even.
$50000
.04
.02
$100000
.01
$200000
µA= ($200000)(.01)+($100000)(.02)+($50000)(.04)+(0)(.93) = $6000

8. Page 243 #5 Draw the probability density function.
.70
Draw the probability density function.
Determine the minimum value the insurance
company should charge to break even. 10
.12
.09 25
.07 60 90
.02
µA= ($10000)(.12)+($25000)(.09)+($60000)(.07)+ (.022) ($90000)+ (0)(.93) = $9450

9. The average value of outcomes
Definition
Expected Value
The average value of outcomes
E =  [x • P(x)]
Also called expectation or mathematical expectation
Plays a very important role in decision theory
page 189 of text

10. E =  [x • P(x)] Event Win Lose
What is the expected value for a $1.00 ticket that pays off a $ prize if 1000 tickets are sold?
E =  [x • P(x)]
Event
Win
Lose
Example on page 189 of text

11. E =  [x • P(x)]
Event
Win
Lose x
$499
- $1

12. E =  [x • P(x)]
Event
Win
Lose x
$499
- $1
P(x)
0.001
0.999

13. E =  [x • P(x)] Event x P(x) x • P(x) Win Lose $499 - $1 0.001 0.999
0.499

14. E = -$.50 E =  [x • P(x)] Event x P(x) x • P(x) Win Lose $499 - $1
0.001
0.999
x • P(x)
0.499
E = -$.50

15. Definitions Binomial Probability Distribution
1. The experiment must have a fixed number of trials.
The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in the other trials.)
3. Each trial must have all outcomes classified into two categories.
4. The probabilities must remain constant for each trial.
Binomial probability distributions are important because they allow us to deal with circumstances in which the outcomes belong to TWO categories, such as pass/fall, acceptable/defective, etc.
page 194 of text

16. Notation for Binomial Probability Distributions
n = fixed number of trials
r = specific number of successes in n trials
p = probability of success in one of n trials
q = probability of failure in one of n trials
(q = 1 - p )
P(r) = probability of getting exactly r success among n trials
Be sure that r and p both refer to the same category being called a success.
The word success is arbitrary and does not necessarily represent something GOOD. If you are trying to find the probability of deaths from hang-gliding, the ‘success’ probability is that of dying from hang-gliding.
Remind students to carefully look at the probability (percentage or rate) provided and whether it matches the probability desired in the question. It might be necessary to determine the complement of the given probability to establish the ‘p’ value of the problem.

17. Binomial Probability Formula

18. Binomial Probability Formula
P(x) = • pr • qn-r
n !
(n - r )! r!
page 196 of text

19. Binomial Probability Formula
Method 1
Binomial Probability Formula
P(r) = • pr • qn-r
n !
(n - r )! r!
P(r) = nCr • pr • qn-r
A minimal scientific calculator with factorial key will be required to complete the problems from this section. Most newer scientific calculators and all graphics calculators will have the ‘combinations’ key (nCr) which makes the computations much easier.
Have students practice this formula with their calculator in class. Different calculators will have different key strokes to accomplish this formula.
Many students stop after only computing the ‘counting’ factor of the formula. Remind that the formula has three factors and that the answer (a probability) should be a number between 0 and 1, inclusive.
Remind students that for answers that are very small, their calculators might go into scientific notation. Look at the calculator notation very carefully.
for calculators with nCr key.

20. Example: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.
This is a binomial experiment where:
n = 5
r = 3
p = 0.90
q = 0.10
This shows the formula using the ‘combination’ notation.

21. Example: Find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in general, AT&T is correct 90% of the time.
This is a binomial experiment where:
n = 5
r = 3
p = 0.90
q = 0.10
Using the binomial probability formula to solve:
P(3) = 5C3 • 0.93 • =.0729
This shows the formula using the ‘combination’ notation.

22. Example: Using n = 5 and p = 0
Example: Using n = 5 and p = 0.90, find the following: a) The probability of exactly 3 successes b) The probability of at least 3 successes
a) P(3) = 0.073
b) P(at least 3) = P(3 or 4 or 5)
= P(3) or P(4) or P(5) =
= 0.991

23. For Binomial Distributions:
Formula µ = n • p
Formula 2 = n • p • q
or = n • p • q
The obvious advantage to the formulas for the mean, standard deviation, and variance of a binomial distribution is that you do not need the values in the distribution table. You need only the n, p, and q values.
A common error students tend to make is to forget to take the square root of (n)(p)(q) to find the standard deviation, especially if they used L1 and L2 lists in a graphical calculator to find the standard deviation with non-binomial distributions. The square root is built into the programming of the calculator and students do not have to remember it. These formulas do require the student to remember to take the square root of the three factors.

24. Using the binomial distribution formulas:
Example: Find the mean and standard deviation for the number of girls in groups of 14 births.
n = 14
p = 0.5
q = 0.5
Using the binomial distribution formulas:
Example on page 205 of text

25. Example: Find the mean and standard
Example: Find the mean and standard deviation for the number of girls in groups of 14 births.
n = 14
p = 0.5
q = 0.5
Using the binomial distribution formulas:
µ = (14)(0.5) = 7 girls
= 1.9 girls

26. Reminder Maximum usual values = µ + 2  Minimum usual values = µ - 2 
The Empirical Rule (section 2-5, pages 80-82) and range rule of thumb (section 2-5, page 79) establishes these guidelines.

27. Example: Determine whether 68 girls among 100 babies could easily occur by chance.
For this binomial distribution,
µ = 50 girls
= 5 girls
µ + 2  = (5) = 60
µ - 2  = (5) = 40
The usual number girls among 100 births would be from 40 to 60. So 68 girls in 100 births is an unusual result.
Example on page 206 of text

28. Poisson Random Variable
Poisson process is the random occurrence
of a series of events over time (or space) in
such a way that:
The number of successes can be any whole number.
The probability of success is equally spread out.
the mean number of successes
r the desired number of successes

29. Example: On an average day, 6 people visit the barbershop
What is the probability that exactly 4 will visit the barbershop?
What is the probability that 2 or less will visit the barbershop?
What is the probability that more than 3 will visit the barbershop?

30. Example: A 100-foot of pipe has 20 leaks. A 25-foot
section is chosen at random
What is the probability it has exactly 3 leaks?
What is the probability it has at most 2 leaks?
What is the probability it has more than 3 leaks?
 = 5

31. Overview Continuous random variable Normal distribution
page 226 of text
This is the first time the symmetric bell-shaped curve has been called a ‘normal’ distribution.

32. Overview Continuous random variable Normal distribution µ
Curve is bell shaped
and symmetric
Score

33. Overview y = Continuous random variable Normal distribution ( ) 
Curve is bell shaped
and symmetric
Score
Even though the equation appears daunting, with the knowledge of and , it is the same as any other equation relating x and y. Remind student that e and  are constant values.
The result is bell shape curve.
With a graphing calculator, the instructor can show the graph of the equation with  = 0 and  = 1.
( )
x - µ 2 1  2
y = e
 2 p

34. Definitions Uniform Distribution
a probability distribution in which the continuous random variable values are spread evenly over the range of      possibilities; the graph results in a      rectangular shape.
page 227 of text

35. Definitions Density Curve (or probability density function)
  the graph of a continuous probability   distribution
Remind students that the area of a probability density curve must equal 1 as the total of all probabilities of a probability distribution equals 1.

36. Definitions Density Curve (or probability density function)
  the graph of a continuous probability   distribution
1. The total area under the curve must equal 1.
2. Every point on the curve must have a vertical height that is 0 or greater.
Remind students that the area of a probability density curve must equal 1 as the total of all probabilities of a probability distribution equals 1.

37. Because the total area under the density curve is equal to 1, there is a correspondence between area and probability.
page 228 of text
This is an important fact for students to understand.

38. Times in First or Last Half Hours
Figure shows the area of example on page The distribution is uniform since all possible times are equally likely. Students will need to be reminded how the probability of 0.2 was obtained: the total area must be 1 and the width of the rectangle is 5. What number times 5 equals 1? Answer: 0.2

39. Graph the probability distribution. Find P(x>6) Find P(5<x<8)
A continuous random variable,T, has a uniform (rectangular) probability distribution over the interval [2,10].
Graph the probability distribution.
Find P(x>6)
Find P(5<x<8) 2 10
A = b*h, 1=b*8, h = .125
P(x>6) A = b*h = 4*.125 = .5
P(5<x<8) A = b*h = 3*.125 = .375

40. P(x > 4) A = ½ b*h = ½ *2*(1/9) = 1/9
A continuous random variable,T, has a uniform (triangular) probability distribution over the interval [0,6].
(0, 1/3)
(1, 5/18)
Graph the probability distribution.
Find P(x > 4)
Find P(1 < x < 4)
(2, 2/9) 6
(3, 1/6)
(4, 1/9)
(5, 1/18)
A = ½ b*h, 1= ½* 6*h, h = 1/3
P(x > 4) A = ½ b*h = ½ *2*(1/9) = 1/9
P(1 < x < 4) A = ½ b1*h1 - ½ b2*h2 =
½* 5*(5/18) – ½ *2*(1/9) = /36 – 1/9 = 21/36 =
7/12

41. Heights of Adult Men and Women
µ = 63.6
 = 2.5
Men:
µ = 69.0
 = 2.8
Normal distributions that have different means will be positioned differently on the number line.
Normal distributions that have different standard deviations will be ‘spread’ differently and thus will have different heights for the bell curve.
63.6
69.0
Figure 5-4
Height (inches)

42. Definition Standard Normal Deviation
a normal probability distribution that has a
mean of 0 and a standard deviation of 1
Again showing this graph with a graphing calculator will be very a very effective illustration for students.
Changing one or both of the parameters,  or , while leaving the standard normal curve in place, makes a good demonstration.

43. Table A-2
Formulas and Tables card
Appendix A2

44. Table A-2 Standard Normal Distribution
µ =  = 1
Emphasize that Table A-2 gives the area (probability) from the mean (=0) to the z-score only. Students will need to be reminded of this often in the initial stages of this topic. x z

45. Definition Standard Normal Deviation
a normal probability distribution that has a
mean of 0 and a standard deviation of 1
Area found in
Table A-2
Area = 1 2 3 -1 -2 -3
page 229 and 231 of text
The areas have been calculated by mathematicians using the calculus of the area between the curve and the x-axis.
Fortunately, for the students in this course, the calculus knowledge is not necessary with the inclusion of Table A-2 (z-distribution) in the book.
0.4429
z = 1.58
Score (z )
Figure 5-6
Figure 5-5

46. Table A-2 Standard Normal (z) Distribution
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
.0000
.0398
.0793
.1179
.1554
.1915
.2257
.2580
.2881
.3159
.3413
.3643
.3849
.4032
.4192
.4332
.4452
.4554
.4641
.4713
.4772
.4821
.4861
.4893
.4918
.4938
.4953
.4965
.4974
.4981
.4987
.0040
.0438
.0832
.1217
.1591
.1950
.2291
.2611
.2910
.3186
.3438
.3665
.3869
.4049
.4207
.4345
.4463
.4564
.4649
.4719
.4778
.4826
.4864
.4896
.4920
.4940
.4955
.4966
.4975
.4982
.4987
.0080
.0478
.0871
.1255
.1628
.1985
.2324
.2642
.2939
.3212
.3461
.3686
.3888
.4066
.4222
.4357
.4474
.4573
.4656
.4726
.4783
.4830
.4868
.4898
.4922
.4941
.4956
.4967
.4976
.4982
.4987
.0120
.0517
.0910
.1293
.1664
.2019
.2357
.2673
.2967
.3238
.3485
.3708
.3907
.4082
.4236
.4370
.4484
.4582
.4664
.4732
.4788
.4834
.4871
.4901
.4925
.4943
.4957
.4968
.4977
.4983
.4988
.0160
.0557
.0948
.1331
.1700
.2054
.2389
.2704
.2995
.3264
.3508
.3729
.3925
.4099
.4251
.4382
.4495
.4591
.4671
.4738
.4793
.4838
.4875
.4904
.4927
.4945
.4959
.4969
.4977
.4984
.4988
.0199
.0596
.0987
.1368
.1736
.2088
.2422
.2734
.3023
.3289
.3531
.3749
.3944
.4115
.4265
.4394
.4505
.4599
.4678
.4744
.4798
.4842
.4878
.4906
.4929
.4946
.4960
.4970
.4978
.4984
.4989
.0239
.0636
.1026
.1406
.1772
.2123
.2454
.2764
.3051
.3315
.3554
.3770
.3962
.4131
.4279
.4406
.4515
.4608
.4686
.4750
.4803
.4846
.4881
.4909
.4931
.4948
.4961
.4971
.4979
.4985
.4989
.0279
.0675
.1064
.1443
.1808
.2157
.2486
.2794
.3078
.3340
.3577
.3790
.3980
.4147
.4292
.4418
.4525
.4616
.4693
.4756
.4808
.4850
.4884
.4911
.4932
.4949
.4962
.4972
.4979
.4985
.4989
.0319
.0714
.1103
.1480
.1844
.2190
.2517
.2823
.3106
.3365
.3599
.3810
.3997
.4162
.4306
.4429
.4535
.4625
.4699
.4761
.4812
.4854
.4887
.4913
.4934
.4951
.4963
.4973
.4980
.4986
.4990
.0359
.0753
.1141
.1517
.1879
.2224
.2549
.2852
.3133
.3389
.3621
.3830
.4015
.4177
.4319
.4441
.4545
.4633
.4706
.4767
.4817
.4857
.4890
.4916
.4936
.4952
.4964
.4974
.4981
.4986
.4990 *
Practice finding some areas (probabilities) for various z scores.
Again remind students that these areas will be from the mean to the particular z score. Drawing the correct graph and shaded area is very helpful in learning this concept. *

47. To find: z Score
the distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of Table A-2
Area
the region under the curve; refer to the values in the body of Table A-2
page 231 of text
It will be important to distinguish between finding a probability (area) given a particular z score from that of finding a z score given a particular area.
Being able to recognize the correct method for each of these processes will be critical.

48. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees.
page 231 of text

49. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees.
P ( 0 < x < 1.58 ) =

50. * * z Table A-2 Standard Normal (z) Distribution
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
.0000
.0398
.0793
.1179
.1554
.1915
.2257
.2580
.2881
.3159
.3413
.3643
.3849
.4032
.4192
.4332
.4452
.4554
.4641
.4713
.4772
.4821
.4861
.4893
.4918
.4938
.4953
.4965
.4974
.4981
.4987
.0040
.0438
.0832
.1217
.1591
.1950
.2291
.2611
.2910
.3186
.3438
.3665
.3869
.4049
.4207
.4345
.4463
.4564
.4649
.4719
.4778
.4826
.4864
.4896
.4920
.4940
.4955
.4966
.4975
.4982
.4987
.0080
.0478
.0871
.1255
.1628
.1985
.2324
.2642
.2939
.3212
.3461
.3686
.3888
.4066
.4222
.4357
.4474
.4573
.4656
.4726
.4783
.4830
.4868
.4898
.4922
.4941
.4956
.4967
.4976
.4982
.4987
.0120
.0517
.0910
.1293
.1664
.2019
.2357
.2673
.2967
.3238
.3485
.3708
.3907
.4082
.4236
.4370
.4484
.4582
.4664
.4732
.4788
.4834
.4871
.4901
.4925
.4943
.4957
.4968
.4977
.4983
.4988
.0160
.0557
.0948
.1331
.1700
.2054
.2389
.2704
.2995
.3264
.3508
.3729
.3925
.4099
.4251
.4382
.4495
.4591
.4671
.4738
.4793
.4838
.4875
.4904
.4927
.4945
.4959
.4969
.4977
.4984
.4988
.0199
.0596
.0987
.1368
.1736
.2088
.2422
.2734
.3023
.3289
.3531
.3749
.3944
.4115
.4265
.4394
.4505
.4599
.4678
.4744
.4798
.4842
.4878
.4906
.4929
.4946
.4960
.4970
.4978
.4984
.4989
.0239
.0636
.1026
.1406
.1772
.2123
.2454
.2764
.3051
.3315
.3554
.3770
.3962
.4131
.4279
.4406
.4515
.4608
.4686
.4750
.4803
.4846
.4881
.4909
.4931
.4948
.4961
.4971
.4979
.4985
.4989
.0279
.0675
.1064
.1443
.1808
.2157
.2486
.2794
.3078
.3340
.3577
.3790
.3980
.4147
.4292
.4418
.4525
.4616
.4693
.4756
.4808
.4850
.4884
.4911
.4932
.4949
.4962
.4972
.4979
.4985
.4989
.0319
.0714
.1103
.1480
.1844
.2190
.2517
.2823
.3106
.3365
.3599
.3810
.3997
.4162
.4306
.4429
.4535
.4625
.4699
.4761
.4812
.4854
.4887
.4913
.4934
.4951
.4963
.4973
.4980
.4986
.4990
.0359
.0753
.1141
.1517
.1879
.2224
.2549
.2852
.3133
.3389
.3621
.3830
.4015
.4177
.4319
.4441
.4545
.4633
.4706
.4767
.4817
.4857
.4890
.4916
.4936
.4952
.4964
.4974
.4981
.4986
.4990 *
Using the z- distribution to find the area from the mean (=0) to the z score of 1.58. *

51. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees.
Area =
P ( 0 < x < 1.58 ) =

52. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees.
Area =
P ( 0 < x < 1.58 ) =
Interpretation of numerical answer.
The probability that the chosen thermometer will measure freezing water between 0 and 1.58 degrees is

53. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees.
Area =
P ( 0 < x < 1.58 ) =
Another interpretation that will be important to discuss.
There is 44.29% of the thermometers with readings between 0 and 1.58 degrees.

54. Using Symmetry to Find the Area to the Left of the Mean
Because of symmetry, these areas are equal.
Figure 5-7
(a)
(b)
0.4925
0.4925
page of text
Students will need to be reminded why probability (and area) can never be negative.
z =
z = 2.43
Equal distance away from 0
NOTE: Although a z score can be negative, the area under the curve (or the corresponding probability) can never be negative.

55. Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads freezing water between degrees and 0 degrees.
Area =
P ( < x < 0 ) =
Example on page 232 of text
Again emphasize that the answer - a probability (and area) - will not be negative, even though the z score was negative.
The probability that the chosen thermometer will measure freezing water between and 0 degrees is

56. Standard Normal Distribution: µ = 0 and  = 1
The Empirical Rule
Standard Normal Distribution: µ = 0 and  = 1
Some student have difficulty understand the idea of ‘within one standard deviation of the mean’. Emphasize that this means the interval from one standard deviation below the mean to one standard deviation above the mean.

57. Standard Normal Distribution: µ = 0 and  = 1
The Empirical Rule
Standard Normal Distribution: µ = 0 and  = 1
68% within
1 standard deviation
Some student have difficulty understand the idea of ‘within one standard deviation of the mean’. Emphasize that this means the interval from one standard deviation below the mean to one standard deviation above the mean.
34%
34%
x - s x
x + s

58. Standard Normal Distribution: µ = 0 and  = 1
The Empirical Rule
Standard Normal Distribution: µ = 0 and  = 1
95% within
2 standard deviations
68% within
1 standard deviation
34%
34%
13.5%
13.5%
x - 2s
x - s x
x + s
x + 2s

59. The Empirical Rule Standard Normal Distribution: µ = 0 and  = 1
99.7% of data are within 3 standard deviations of the mean
95% within
2 standard deviations
68% within
1 standard deviation
These percentages will be verified by the concepts learned in Chapter 5.
Emphasize the Empirical Rule is appropriate for data that is in a BELL-SHAPED distribution.
34%
34%
2.4%
2.4%
0.1%
0.1%
13.5%
13.5%
x - 3s
x - 2s
x - s x
x + s
x + 2s
x + 3s

60. Probability of Half of a Distribution
0.5
Students will need this explanation. It will not be immediately intuitive for some. Because the normal curve is ‘symmetric’ and the total probability (area) is equal to 1, from the mean to one side (or the other) of the curve will have a probability (area) equal to 0.5.
This concept is necessary to complete the other types of probability situations that can occur.

61. Finding the Area to the Right of z = 1.27
Value found
in Table A-2
0.3980
This area is =
Example on page 233 of text
z = 1.27

62. Finding the Area Between z = 1.20 and z = 2.30
(from Table A-2 with z = 2.30)
Area A is =
0.3849 A
Example on page 234 of text
z = 1.20
z = 2.30

63. denotes the probability that the z score is greater than a
Notation
P(a < z < b)
denotes the probability that the z score is between a and b
P(z > a)
denotes the probability that the z score is greater than a
P (z < a)
denotes the probability that the z score is less than a
page 235 of text
A review of inequality signs that their meanings will most likely be necessary.

64. Figure 5-10 Interpreting Area Correctly

65. Interpreting Area Correctly
Subtract
from
0.5
‘greater than x’
‘at least x’
‘more than x’
‘not less than x’
Add to
0.5
0.5 x x
More emphasis on different ways to word various inequality statements.
Also the different ways the shadings could be drawn.

66. Interpreting Area Correctly
Subtract
from
0.5
‘greater than x’
‘at least x’
‘more than x’
‘not less than x’
Add to
0.5
0.5 x x
‘less than x’
‘at most x’
‘no more than x’
‘not greater than x’
Add to
0.5
Subtract
from
0.5
0.5 x x

67. Interpreting Area Correctly
Subtract
from
0.5
‘greater than x’
‘at least x’
‘more than x’
‘not less than x’
Add to
0.5
0.5 x x
‘less than x’
‘at most x’
‘no more than x’
‘not greater than x’
Add to
0.5
Subtract
from
0.5
0.5
Remind students that the bottom right example drawing could have just as well been drawn to the right of the mean.
Reminding students that the horizontal line of the normal curve is a number line that increases from left to right and that the given problem values should be place correctly in relation to the mean and each other is vital to obtaining the correct drawing, and thus, the correct final answer. x x
Add C
Use
A = C - B
‘between x1 and x2’ A B x1 x2 x1 x2

68. Finding a z - score when given a probability Using Table A-2
1. Draw a bell-shaped curve, draw the centerline, and identify the region under the curve that corresponds to the given probability. If that region is not bounded by the centerline, work with a known region that is bounded by the centerline.
2. Using the probability representing the area bounded by the centerline, locate the closest probability in the body of Table A-2 and identify the corresponding z score.
3. If the z score is positioned to the left of the centerline, make it a negative.
page 236 of text
Some students get confused with this process. It is actually the ‘reverse’ process of that discussed in the first part of the section.
The most common mistake students make with this process is to not make z-scores to the left of the mean negative.
Checking the final answer for reasonableness - reminding students that the horizontal line is a ‘number line’ - is important.

69. Finding z Scores when Given Probabilities
95% 5%
5% or 0.05
0.45
0.50
Drawing for example on page 237 of text z
( z score will be positive )
Finding the 95th Percentile

70. Finding z Scores when Given Probabilities
95% 5%
5% or 0.05
0.45
0.50
1.645
(z score will be positive)
Finding the 95th Percentile

71. Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
Drawing for example on page 238 of text
0.10
0.40 z
(z score will be negative)
Finding the 10th Percentile

72. Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
Student will need to be reminded that Table A-2 will not indicate that the correct z score for this problem will be negative. This is something that the student will have to remember to do when writing the correct answer to the problem.
Emphasize the number line aspect again and that numbers below 0 will be negative.
0.10
0.40
-1.28
(z score will be negative)
Finding the 10th Percentile

73. Other Normal Distributions
If   0 or   1 (or both), we will convert values to standard scores using Formula 6-7, then procedures for working with all normal distributions are the same as those for the standard normal distribution.
x - µ 
z =
The z score formula was first presented in Chapter 2, Section 2-6, Measures of Position, page 85.

74. Converting to Standard Normal DistributionP  x
(a)

75. Converting to Standard Normal Distribution
x -  
z =
Tell students to round z score answers to two decimal places if using Table A-2 from the text. P P  x z
(a)
(b)

76. Probability of Weight between 143 pounds and 201 pounds
z =
= 2.00
x = 143 29
σ =29
page 242 of text
Weight z

77. Probability of Weight between 143 pounds and 201 pounds
Value found
in Table A-2
x = 143
σ =29
Emphasize that the area in question is the area that Table A-2 will give - from the mean to the score.
Weight z

78. Probability of Weight between 143 pounds and 201 pounds
0.4772
x = 143
σ =29
Looking 2.00 up in Table A-2 produces an area of
Weight z

79. Probability of Weight between 143 pounds and 201 pounds
There is a probability of randomly selecting a woman with a weight between 143 and 201 lbs.
x = 143
σ =29
Interpretation of answer should be emphasized.
Weight z

80. Probability of Weight between 143 pounds and 201 pounds
OR % of women have weights between 143 lb and 201 lb.
x = 143
σ =29
Alternative way of interpreting answer.
Weight z

81. Cautions to keep in mind
1. Don’t confuse z scores and areas.
    Z scores are distances along the horizontal  scale, but areas are regions under the  normal curve. Table A-2 lists z scores in the left column and across the top row, but  areas are found in the body of the table.
2. Choose the correct (right/left) side of the graph.
3. A z score must be negative whenever it is located
to the left of the centerline of 0.
page 249 of text

82. Finding z Scores when Given Probabilities
95% 5%
5% or 0.05
0.45
0.50
1.645
(z score will be positive)
Finding the 95th Percentile

83. Finding z Scores when Given Probabilities
90%
10%
Bottom 10%
Student will need to be reminded that Table A-2 will not indicate that the correct z score for this problem will be negative. This is something that the student will have to remember to do when writing the correct answer to the problem.
Emphasize the number line aspect again and that numbers below 0 will be negative.
0.10
0.40
-1.28
(z score will be negative)
Finding the 10th Percentile

84. Procedure for Finding Values Using Table A-2 and Formula 6-7
1. Sketch a normal distribution curve, enter the given probability or percentage in the appropriate region of the graph, and identify the x value(s) being sought.
2. Use Table A-2 to find the z score corresponding to the region bounded by x and the centerline of Cautions:
    Refer to the BODY of Table A-2 to find the closest area, then identify the corresponding z score.
    Make the z score negative if it is located to the left of the centerline.
3. Using Formula 5-2, enter the values for µ, , and the z score found in step 2, then solve for x.
x = µ + (z • ) (Another form of Formula 6-7)
4. Refer to the sketch of the curve to verify that the solution makes sense in the context of the graph and the context of the problem.
page of text

85. Finding P10 for Weights of Women
10%
90%
40%
50%
Drawing for example on page 250 of text
Weight
x = ?
143

86. Finding P10 for Weights of Women
0.10
0.40
0.50
Emphasize z score will need to be made negative since it is below the mean.
Weight
x = ?
143
-1.28

87. Finding P10 for Weights of Women
x = (-1.28 • 29) =
0.10
0.40
0.50
Using the alternative form of the z score formula, find the P10 value.
Weight
x = ?
143
-1.28

88. Finding P10 for Weights of Women
The weight of 106 lb (rounded) separates the lowest 10% from the highest 90%.
0.10
0.40
0.50
Interpretation should be emphasized.
Weight
x = 106
143
-1.28

89. Forgot to make z score negative???
x = (1.28 • 29) = 180
0.10
0.40
0.50
Results if one forgets to make the z score negative.
page 251 of text
Weight
x = ?
143
1.28

90. Forgot to make z score negative???
x = (1.28 • 29) = 180
0.10
0.40
0.50
Remind students that the horizontal line is a number line where the numbers increase from left to right.
180 is an unreasonable answer for the position of the normal curve on this number line.
Weight
x = 180
143
1.28

91. Forgot to make z score negative???
UNREASONABLE ANSWER!
0.10
0.40
0.50
Weight
x = 180
143
1.28

92. REMEMBER!
Make the z score negative if the value is located to the left (below) the mean. Otherwise, the z score will be positive.

93. A pizza company advertises that it puts
A pizza company advertises that it puts .5 lb of cheese on its medium pizzas. In reality, the amount of cheese is normally distributed with a mean value of .5 lbs with a standard deviation of .025 lbs.
What is the probability that the amount of cheese on a medium pizza is between .525 and .550 lbs?
What is the probability that a medium pizza has less than .47 lbs of cheese?
If the pizza has more cheese than 95% of the other pizzas, how many pounds of cheese does it have?
=.1359 a.

94. A pizza company advertises that it puts
A pizza company advertises that it puts .5 lb of cheese on its medium pizzas. In reality, the amount of cheese is normally distributed with a mean value of .5 lbs with a standard deviation of .025 lbs.
What is the probability that the amount of cheese on a medium pizza is between .525 and .550 lbs?
What is the probability that a medium pizza has less than .47 lbs of cheese?
If the pizza has more cheese than 95% of the other pizzas, how many pounds of cheese does it have? b.
=.1150

95. A pizza company advertises that it puts
A pizza company advertises that it puts .5 lb of cheese on its medium pizzas. In reality, the amount of cheese is normally distributed with a mean value of .5 lbs with a standard deviation of .025 lbs.
What is the probability that the amount of cheese on a medium pizza is between .525 and .550 lbs?
What is the probability that a medium pizza has less than .47 lbs of cheese?
If the pizza has more cheese than 95% of the other pizzas, how many pounds of cheese does it have?
x = μ + zσ
z = 1.645
x = .5 + (1.645)(.025) c.
x =.5411

96. A car’s gas mileage averages 30 mi/gal with a standard deviation of 1
A car’s gas mileage averages 30 mi/gal with a standard deviation of 1.2 mi/gal
What is the probability that the mileage of a randomly selected car is between 31.2 mi/gal and 33 mi/gal?
What is the probability that a car selected at random exceeds 32 mi/gal?
What is the miles per gallon if the car is in the bottom 20%?
=.1525 a.

97. A car’s gas mileage averages 30 mi/gal with a standard deviation of 1
A car’s gas mileage averages 30 mi/gal with a standard deviation of 1.2 mi/gal
What is the probability that the mileage of a randomly selected car is between 31.2 mi/gal and 33 mi/gal?
What is the probability that a car selected at random exceeds 32 mi/gal?
What is the miles per gallon if the car is in the bottom 20%?
= .0475 b.

98. A car’s gas mileage averages 30 mi/gal with a standard deviation of 1
A car’s gas mileage averages 30 mi/gal with a standard deviation of 1.2 mi/gal
What is the probability that the mileage of a randomly selected car is between 31.2 mi/gal and 33 mi/gal?
What is the probability that a car selected at random exceeds 32 mi/gal?
What is the miles per gallon if the car is in the bottom 20%?
x = μ + zσ
x = 30 – (.84)(1.2) = 28.99 c.