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ELEMENTARY
Chapter 5
STATISTICS
Normal Probability Distributions
MARIO F. TRIOLA
EIGHTH
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
EDITION1
Measures of Position
(Section 2.6)
 z Score
(or standard score)
the number of standard deviations
that a given value x is above or below
the mean
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
2
Measures of Position
z score
Sample
x
x
z= s
Population
x
µ
z=

Round to 2 decimal places
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
3
FIGURE 2-16
Interpreting Z Scores
Unusual
Values
Ordinary
Values
-3
-2
-1
0
Unusual
Values
1
2
3
Z
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
4
Chapter 5
Normal Probability Distributions
5-1
Overview
5-2
The Standard Normal Distribution
5-3
Normal Distributions: Finding Probabilities
5-4
Normal Distributions: Finding Values
5-5
The Central Limit Theorem
5-6
Normal Distribution as Approximation to
Binomial Distribution (skipped)
5-7
Determining Normality
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
5
Normal Distribution
 Continuous random variable
 Graph is symmetric and bell shaped
 Area under the curve is equal to 1,
therefore there is correspondence
between area and probability.
µ
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
6
Heights of Adult Men and Women
•Normal distributions (bell shaped) are a
family of distributions that have the same
general shape.
•The mean, mu, controls the center
•Standard deviation, sigma, controls the
spread.
Women:
µ = 63.6
 = 2.5
Figure 5-4
Men:
µ = 69.0
 = 2.8
63.6
69.0
Height (inches)
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
7
5-2
The Standard Normal
Distribution
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
8
Definition
Standard Normal Distribution
a normal probability distribution that has a
mean of 0 and a standard deviation of 1
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
9
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
34%
34%
2.4%
2.4%
0.1%
0.1%
13.5%
-3
-2
13.5%
-1
0
1
2
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
3
z
10
Example:
Using your knowledge of the Emperical
rule, what is the probability that a value falls between the
mean and 1 standard deviation above the mean?
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
11
Example:
Using your knowledge of the Emperical
rule, what is the probability that a value falls between the
mean and 1 standard deviation above the mean?
34%
0
1
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
z
12
What if we need to find the
probability of a value that does not
fall on a standard deviation?
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
13
Finding Probabilities from Standard(Z) scores
Step 1: Draw a normal curve with the centerline
labeled 0 on the x-axis.
Step 2: Label given value(s) on the appropriate
location on the x-axis. Draw vertical lines on
the normal curve above the given values.
Step 3: Shade the area of the region in question.
Step 4: Use calculator function normalcdf() to
find the area (probability) in question.
normalcdf(left boundary, right boundary)
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
14
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. Since  = 0 and  =1, the values
are standard or z-scores.
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
15
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. Since  = 0 and  =1, the values
are standard or z-scores.
0
1.58
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
16
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. Since  = 0 and  =1, the values
are standard or z-scores.
P ( 0 < x < 1.58 ) = ?
0
1.58
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
17
Table A-2 Standard Normal (z) Distribution
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
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
*
*
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
18
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. Since  = 0 and  =1, the values
are standard or z-scores.
P ( 0 < x < 1.58 ) = Normalcdf(0, 1.58) = .443
0
1.58
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
19
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. Since  = 0 and  =1, the values
are standard or z-scores.
Area = 0.443
P ( 0 < x < 1.58 ) = 0.443
0
1.58
The probability that the chosen
thermometer will measure freezing water
between 0 and 1.58 degrees is 0.443.
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
20
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. Since  = 0 and  =1, the values
are standard or z-scores.
Area = 0.443
P ( 0 < x < 1.58 ) = 0.443
0
1.58
There is 44.3% of the thermometers with
readings between 0 and 1.58 degrees.
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
21
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 -2.43 degrees and 0 degrees.
-2.43
0
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
22
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 -2.43 degrees and 0 degrees.
Area = 0.493
-2.43
P ( -2.43 < x < 0 ) =
Normalcdf(-2.43, 0)= 0.493
0
NOTE: Although a z score can be negative, the area
under the curve (or the corresponding probability)
can never be negative.
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
23
Notation
P(a < z < b) → Normalcdf(a,b)
denotes the probability that the z score is between a
and b
P(z > a) → Normalcdf(a,9999)
denotes the probability that the z score is greater
than a
P (z < a) → Normalcdf(-9999,a)
denotes the probability that the z score is less than a
Note: 9999 is used to represent infinity
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
24
Probability of Half of a Distribution
Remember the graph represents the entire
distribution with its area of 1.
0.5
0.5
=0
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
25
Finding the Area to the Right of z = 1.27
P(z>1.27) = normalcdf(1.27, 9999) = 0.1020
0
z = 1.27
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
26
Finding the Area Between z = 1.20 and z = 2.30
P(1.20 <z< 2.30) = normalcdf(1.20, 2.30) = 0.104
Area = .104
0
z = 1.20 z = 2.30
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
27
Interpreting Area Correctly
Figure 5-10
‘greater than
‘at least
x’
x’
‘more than
x’
‘not less than
x’
x
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
x
28
Interpreting Area Correctly
‘greater than
‘at least
x’
x’
‘more than
x’
‘not less than
x’
x
‘less than
‘at most
x
x’
x’
‘no more than
x’
‘not greater than
x’
x
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
x
29
Finding a z - score when given a probability
1. Draw a bell-shaped curve, draw the centerline, and shade the
region under the curve that corresponds to the given
probability. The line(s) that bound the region denotes the zscore(s) you are trying to find.
2. Using the probability representing the area bounded by the
z-score, convert it to area to the left of the z-score and enter
that value into invNorm() function in your calculator.
z = invNorm(area to the left) This command finds the z-score that has area to the
left of the boundary under the normal curve. Since the area represent probability the
argument must be between 0 and 1 inclusive.
3. If the z score is positioned to the left of the centerline, make
sure it’s negative.
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
30
Finding z Scores when Given Probabilities
Find the z-score that denotes the 95th Percentile.
95%
0.95
0
z = invNorm(.95) =1.65
z
( z score will be positive )
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
31
Finding z Scores when Given Probabilities
Find the z-score that separate the bottom 10% and upper 90%.
90%
10%
Bottom 10%
0.90
0.10
z
0
z = invNorm(.10) = -1.28
(z score will be negative)
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
32
Finding z Scores when Given Probabilities
Find the z-scores that denotes the middle 60%
20%
20%
60%
Bottom 20%
0.60
0.20
0.20
0
z = invNorm(.20) = -0.84
z = invNorm(.80) = 0.84
Chapter 5. Section 5-1 and 5-2. Triola, Elementary Statistics, Eighth Edition. Copyright 2001. Addison Wesley Longman
33