Transcript Case Study: Chest Sizes of Scottish Militiamen (page 352)

Computing Probabilities From the
Standard Normal Distribution
Anthony J Greene
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APPENDIX B
T AB L E B.1
Statistical Tables
THE UNIT NORMAL TABLE*
—————————————————————
Table
B.1
p. 687
*Column A lists z-score values. A vertical line drawn through a normal distribution at a z-score location divides the distribution into two sections.
Column B identifies the proportion in the larger section, called the body.
Column C identifies the proportion in the smaller section, called the tail.
Column D identifies the proportion between the mean and the z-score.
Note: Because the normal distribution is symmetrical, the proportions for negative z-scores are the same as those for positive z-scores.
(A)
z
(B)
PROPORTION
IN BODY
(C)
PROPORTION
IN TAIL
(D)
PROPORTION
BETWEEN MEAN AND z
(A)
z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.11
0.12
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.20
0.21
0.22
0.23
0.24
.5000
.5040
.5080
.5120
.5160
.5199
.5239
.5279
.5319
.5359
.5398
.5438
.5478
.5517
.5557
.5596
.5636
.5675
.5714
.5753
.5793
.5832
.5871
.5910
.5948
.5000
.4960
.4920
.4880
.4840
.4801
.4761
.4721
.4681
.4641
.4602
.4562
.4522
.4483
.4443
.4404
.4364
.4325
.4286
.4247
.4207
.4168
.4129
.4090
.4052
.0000
.0040
.0080
.0120
.0160
.0199
.0239
.0279
.0319
.0359
.0398
.0438
.0478
.0517
.0557
.0596
.0636
.0675
.0714
.0753
.0793
.0832
.0871
.0910
.0948
0.25
0.26
0.27
0.28
0.29
0.30
0.31
0.32
0.33
0.34
0.35
0.36
0.37
0.38
0.39
0.40
0.41
0.42
0.43
0.44
0.45
0.46
0.47
0.48
0.49
(B)
(C)
(D)
PROPORTION PROPORTIONIN
PROPORTION
IN BODY
TAIL
BETWEENMEAN AND z
.5987
.6026
.6064
.6103
.6141
.6179
.6217
.6255
.6293
.6331
.6368
.6406
.6443
.6480
.6517
.6554
.6591
.6628
.6664
.6700
.6736
.6772
.6808
.6844
.6879
.4013
.3974
.3936
.3897
.3859
.3821
.3783
.3745
.3707
.3669
.3632
.3594
.3557
.3520
.3483
.3446
.3409
.3372
.3336
.3300
.3264
.3228
.3192
.3156
.3121
.0987
.1026
.1064
.1103
.1141
.1179
.1217
.1255
.1293
.1331
.1368
.1406
.1443
.1480
.1517
.1554
.1591
.1628
.1664
.1700
.1736
.1772
.1808
.1844
.1879
Table
B.1
A
Closer
Look
(A)
z
(B)
(C)
PROPORTION PROPORTION
IN BODY
0.17
0.18
0.19
0.20
0.21
0.22
0.23
0.24
.5675
.5714
.5753
.5793
.5832
.5871
.5910
.5948
IN TAIL
.4325
.4286
.4247
.4207
.4168
.4129
.4090
.4052
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(D)
PROPORTION
BETWEEN MEAN AND z
.0675
.0714
.0753
.0793
.0832
.0871
.0910
.0948
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The Normal Distribution:
why use a table?
P
x2

x1
1
2
2
e
 ( X   ) 2 / 2 2
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d
dx
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From x or z to P
To determine a percentage or
probability for a normally
distributed variable
Step 1 Sketch the normal curve associated with the variable
Step 2 Shade the region of interest and mark the delimiting xvalues
Step 3 Compute the z-scores for the delimiting x-values found
in Step 2
Step 4 Use Table B.1 to obtain the area under the standard
normal curve delimited by the z-scores found in Step 3
Use Geometry and remember that the total area under
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the curve is always 1.00.
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From x or z to P
Finding percentages for a normally
distributed variable from areas under
the standard normal curve
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Finding percentages for a normally
distributed variable from areas under the
standard normal curve
1.
,  are given.
2. a and b are any two values of the variable x.
3. Compute z-scores for a and b.
4. Consult table B-1
5. Use geometry to find desired area.
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Given that a quiz has a mean score of 14
and an s.d. of 3, what proportion of the
class will score between 9 & 16?
1.
 = 14 and  = 3.
2. a = 9 and b = 16.
3. za = -5/3 = -1.67, zb = 2/5 = 0.4.
4. In table B.1, we see that the area to the left of a is 0.0475
and that the area to the right of b is 0.3446.
5. The area between a and b is therefore
1 – (0.0475 + 0.3446) = 0.6079 or 60.79%
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Finding the area under the standard
normal curve to the left of z = 1.23
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What if you start with x
instead of z?
z = 1.50: Use Column C; Anthony
P = 0.0668
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What is the
probability of
selecting a
random
student who
scored above
650 on the
SAT?
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Finding the area under the standard
normal curve to the right of z = 0.76
The easiest way would be to use Column C, but lets use
Column B instead
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Finding the area under the standard
normal curve that lies between
z = –0.68 and z = 1.82
P = 1 – 0.0344 – 0.2483
= 0.7173
One Strategy: Start with the area to the left of 1.82, then
subtract the area to the right of -0.68.
Second Strategy: Start with 1.00 and subtract off the two
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tails
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Determination of the percentage of
people having IQs between 115 and
140
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From x or z to P
Review of Table B.1 thus far
Using Table B.1 to find the area under the standard normal
curve that lies
(a) to the left of a specified z-score,
(b) to the right of a specified z-score,
(c) between two specified z-scores
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Then if x is asked for, convert
from
z to x
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From P to z or x
Now the other way around
To determine the observations
corresponding to a specified
percentage or probability for a
normally distributed variable
Step 1 Sketch the normal curve associated the the variable
Step 2 Shade the region of interest (given as a probability or area
Step 3 Use Table B.1 to obtain the z-scores delimiting the region
in Step 2
Step 4 Obtain the x-values having the z-scores found in Step 3
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From P to z or x
Finding z- or x-scores corresponding
x
to a given region.
z
Finding the z-score having area 0.04 to its left
x=σ×z+μ

x  z  
If μ is 242 σ is 100, then
x = 100 × -1.75 + 242
x = 67
Use Column C:
The z corresponding to 0.04 Anthony
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left tail is -1.75
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The z
Notation
The symbol zα is used to denote the
z-score having area α (alpha) to its
right under the standard normal
curve. We read “zα” as “z sub α” or
more simply as “z α.”
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The z notation : P(X>x) = α
P(X>x)= α
This is the z-score that
demarks an area under the
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curve with P(X>x)= α
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The z notation : P(X<x) = α
P(X<x)= α
Z
This is the z-score that
demarks an area under the
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curve with P(X<x)= α
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The z notation : P(|X|>|x|) = α
P(|X|>|x|)= α
α/2
1- α
This is the z-score that
demarks an area under the
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curve with P(|X|>|x|)= α
α/2
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Finding z 0.025
Use Column C:
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The z corresponding to 0.025Anthony
in the
right tail is 1.96
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Finding z 0.05
Use Column C:
The z corresponding to 0.05 Anthony
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right tail is 1.64
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Finding the two z-scores dividing the
area under the standard normal curve
into a middle 0.95 area and two outside
0.025 areas
Use Column C:
Anthony
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The z corresponding to 0.025
in both
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Finding the 90th percentile for IQs
z0.10 = 1.28
z = (x-μ)/σ
1.28 = (x – 100)/16
120.48 = x
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What you should be able to do
1. Start with z-or x-scores and compute regions
2. Start with regions and compute z- or x-scores
z
x

x  z  
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