Section 6 – 3B:
Finding probabilities involving Z scores
The probability that a number in the z distribution is less than a given z value
is the area under the z curve that is LEFT of the z value
The area under the z curve that is to the left of a given Z value represents the probability of
selecting one number from the z distribution and having that number be less than the Z value.
Find P( z < –2.1 )
If we want to know the probability of selecting one number from the z distribution and having that
number be less than the negative z value of –2.1 we need to find the area to the left of z = –2.1
The yellow area to
the left of z = –2.1
represents
P( z < –2.1)
z = –2.1
0
Find P( z < 2.7 )
If we want to know the probability of selecting one number from the z distribution and having that
number be less than the positive Z value of 2.7 we need to find the area to the left of z = 2.7
The yellow area to
the left of z = 2.7
represents
P( z < 2.7)
0
z= 2.7
We use two different tables to help find the area to the left of a given z value or z score.
The Negative Z Scores Table
is used to find the area that is
to the left of a negative z value
z = – 2.1
The Positive Z Scores Table
is used to to find the area that is
to the left of a positive z value
0
Section 6 – 3B Lecture
0
Page 1 of 16
z= 2.7
© 2012 Eitel
The Negative Z Scores Table
The Negative Z Scores Table is used to find the area that is to the LEFT of a negative z value.
z = – 2.1
0
The vertical line in the middle of the graph divides the total area of 1 in half. The area to the left of this
line is .5000. All the shaded areas to the left of a negative Z value will be less than .5000
The 2 decimal place Z score is the negative 1 decimal place z value
from the left column (in red)
with an additional decimal place from the row on top (in red)
The 4 decimal place number in yellow at the
intersection of the left column z value (in red) and top row z value (in red)
is the area to the LEFT of that given negative z score
The Z Table only gives areas to the left of a Z value.
Negative Z Scores
Standard Normal (Z) Distribution: Area to the LEFT of Z
Z
–3.4
–3.3
–3.2
–3.1
–3.0
–2.9
–2.8
–2.7
–2.6
–2.5
–2.4
–2.3
–2.2
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0003
0.0003
0.0003
0.0003
0.0003
0.0003
0.0003
0.0003
0.0003
0.0002
0.0005
0.0005
0.0005
0.0004
0.0004
0.0004
0.0004
0.0004
0.0004
0.0003
0.0007
0.0007
0.0006
0.0006
0.0006
0.0006
0.0006
0.0005
0.0005
0.0005
0.0010
0.0009
0.0009
0.0009
0.0008
0.0008
0.0008
0.0008
0.0007
0.0007
0.0013
0.0019
0.0026
0.0035
0.0047
0.0062
0.0082
0.0107
0.0139
0.0013
0.0018
0.0025
0.0034
0.0045
0.0060
0.0080
0.0104
0.0136
0.0013
0.0018
0.0024
0.0033
0.0044
0.0059
0.0078
0.0102
0.0132
0.0012
0.0017
0.0023
0.0032
0.0043
0.0057
0.0075
0.0099
0.0129
0.0012
0.0016
0.0023
0.0031
0.0041
0.0055
0.0073
0.0096
0.0125
0.0011
0.0016
0.0022
0.0030
0.0040
0.0054
0.0071
0.0094
0.0122
0.0011
0.0015
0.0021
0.0029
0.0039
0.0052
0.0069
0.0091
0.0119
0.0011
0.0015
0.0021
0.0028
0.0038
0.0051
0.0068
0.0089
0.0116
0.0010
0.0014
0.0020
0.0027
0.0037
0.0049
0.0066
0.0087
0.0113
0.0010
0.0014
0.0019
0.0026
0.0036
0.0048
0.0064
0.0084
0.0110
This is only a portion of the entire Negative z Score Table
Section 6 – 3B Lecture
Page 2 of 16
© 2012 Eitel
Example 1A
Finding the area to the left of a negative Z score
Find P( z < – 1.86 )
The area under the z curve that is to the left of a given z value represents the probability of
selecting one number from the z distribution and having that number be less than the z value.
If we want to find the probability of selecting one number from the z distribution and having that
number be less than the z value of – 1.86 we need to find the area to the LEFT of z = – 1.86
The number at the intersection of the
–1.8
row and the
0.06
column is 0.0314
Negative Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–1.8
0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307
0.08
0.09
0.0301
0.0294
This means that the area to the left of z = – 1.86 is 0.0314
the area to the left of
z = – 1.86 is .0314
left tail area
= .0314
z = – 1.86 0
negative
value
If the area to the left of z = – 1.86 is .0314 then
P( z < – 1.86 ) = 0.0314
Section 6 – 3B Lecture
Page 3 of 16
© 2012 Eitel
Example 1B
Finding the area to the Right of a negative Z score
Find P( z > – 1.86 )
The area under the z curve that is to the right of a given z value represents the probability of
selecting one number from the z distribution and having that number be more than the z value.
If we want to find the probability of selecting one number from the z distribution and having that
number be more than the z value of – 1.86 we need to find the area to the RIGHT of z = – 1.86
The number at the intersection of the
–1.8
row and the
0.06
column is 0.0314
Negative Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–1.8
0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307
0.08
0.09
0.0301
0.0294
This means that the area to the LEFT of z = – 1.86 is 0.0314
The vertical line drawn at the value of z divides the area under the z curve into two areas. The yellow
area is to the left of the z value and the white area is to the right of the z value.
The total of the yellow and white areas is 1
the area to the left of z = – 1.86
is .0314
left tail area
= .0314
the area to the right of z = – 1.86
is 1– .0314 = .9686
right tail area
= .9686
z = – 1.86 0
so
If the area to the Left of Z = – 1.86 is .0314
then the area to the Right of Z = – 1.86 is 1 – .0314 = .9686
P( z > – 1.86 ) = 1 – 0.0314 = .9686
Section 6 – 3B Lecture
Page 4 of 16
© 2012 Eitel
Example 2A
Find the area to the left of a negative Z score
Find P( z < – 1.02 )
The area under the z curve that is to the left of a given z value represents the probability of
selecting one number from the z distribution and having that number be less than the z value.
If we want to find the probability of selecting one number from the z distribution and having that
number be less than the z value of – 1.02 we need to find the area to the LEFT of z = – 1.02
The number at the intersection of the
–1.0
row and the
0.02
column is 0.1539
Negative Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–1.0
0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423
0.08
0.09
0.1401
0.1379
This means that the area to the left of z = –1.02 is 0.1539
the area to the left of
z = – 1.02 is .1539
left tail area
= .1539
z = – 1.02
negative
value
0
If the area to the left of z = – 1.86 is .1539 then
P( z < – 1.02 ) = 0.1539
Section 6 – 3B Lecture
Page 5 of 16
© 2012 Eitel
Example 2B
Finding the area to the Right of a negative Z score
Find P( z > – 1.02 )
The area under the z curve that is to the right of a given z value represents the probability of
selecting one number from the z distribution and having that number be more than the z value.
If we want to find the probability of selecting one number from the z distribution and having that
number be more than the z value of – 1.02 we need to find the area to the RIGHT of z = – 1.02
the area to the left of z = – 1.02
is .1539
left tail area
= .1539
the area to the right of z = – 1.02
is 1– .1539 = .8461
right tail area
= .8461
z = – 1.02
0
If the area to the Left of Z = – 1.02 is 0.1539
then the area to the Right of Z = – 1.02 is 1 – .1539
P( z > – 1.02 ) = 1 – 0.1539 = .8461
Section 6 – 3B Lecture
Page 6 of 16
© 2012 Eitel
Example 3
Find the area to the left of a negative Z score
A Special Case for z = – 2.575
P( z < – 2.575 )
If we want to find the probability of selecting one number from the z distribution and having that
number be less than the z value of – 2.575 we need to find the area to the left of z = – 2.575
Two negative Z scores with 3 decimal places are used in this course. These two 3 decimal place
values and the areas to the left of them are listed separately at the bottom of the table.
The cells at the bottom of the negative z scores table show that
the area to the left of Z = –2.575 is 0.0050
Negative Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–2.5
0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051
Z scores of –3.5 or less use .0001
AREA
Z Score
0.0500 –1.645
AREA
0.08
0.09
0.0049
0.0048
Z Score
0.0050 –2.575
This means that the area to the LEFT of z = – 1.86 is 0.0050
P( z < – 2.575 ) = 0.0050
Finding the area to the Right of a negative Z score
P( z > – 2.575 )
If we want to find the probability of selecting one number from the z distribution and having that
number be greater than the z value of – 2.575 we need to find the area to the right of z = – 2.575
If the area to the Left of Z = – 1.02 is 0.0050
then the area to the Right of Z = – 2.575 is 1 – .0050 = .9950
P( z > – 2.575 ) = 1 – 0.0050 = .9950
Section 6 – 3B Lecture
Page 7 of 16
© 2012 Eitel
Example 4
Find the area to the left of a negative Z score
A Special Case for z = – 1.645
P( z < – 1.654 )
If we want to find the probability of selecting one number from the z distribution and having that
number be less than the z value of – 1.654 we need to find the area to the left of z = – 1.654
The cells at the bottom of the negative z scores table show that
the area to the left of Z = –1.645 is 0.0500
Negative Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–2.5
0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051
Z scores of –3.5 or less use .0001
AREA
Z Score
0.0500 –1.645
AREA
0.08
0.09
0.0049
0.0048
Z Score
0.0050 –2.575
This means that the area to the left of z = –1.645 is 0.0500
P( z < – 1.654 ) = 0.0500
Finding the area to the Right of a negative Z score
P( z > – 1.654 )
If we want to find the probability of selecting one number from the z distribution and having that
number be greater than the z value of – 1.654 we need to find the area to the right of z = – 1.654
If the area to the Left of Z = – 1.645 is 0.0500
then the area to the Right of Z = – 1.645 is 1 – .0500 = .9500
P( z > – 1.654 ) 1 – 0.0500 = .9500
Section 6 – 3B Lecture
Page 8 of 16
© 2012 Eitel
The Positive Z Scores Table
The Positive Z Scores Table is used to find the area that is to the left of a positive z value.
z = 2.1
The vertical line in the middle of the graph divides the total area of 1 in half. The area to the left of this
line is .5000. All the shaded areas to the left of a positive z value will be more than .5000
The 2 decimal place z score
is the positive 1 decimal place z value from the left column (in red)
with an additional decimal place from the row on top (in red)
The 4 decimal place number in yellow at the
intersection of the left column z value (in red) and top row z value (in red)
stands for the area to the LEFT of that given z score
The Z Table only gives areas to the left of a Z value.
Positive Z Scores
Standard Normal (Z) Distribution: Area to the LEFT of Z
Z
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
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.5000
0.5040
0.5080
0.5120
0.5160
0.5199
0.5239
0.5279
0.5319
0.5359
0.5398
0.5438
0.5478
0.5517
0.5557
0.5596
0.5636
0.5675
0.5714
0.5753
0.5793
0.5832
0.5871
0.5910
0.5948
0.5987
0.6026
0.6064
0.6103
0.6141
0.6179
0.6217
0.6255
0.6293
0.6331
0.6368
0.6406
0.6443
0.6480
0.6517
0.6554
0.6915
0.7257
0.7580
0.7881
0.8159
0.8413
0.8643
0.8849
0.6591
0.6950
0.7291
0.7611
0.7910
0.8186
0.8438
0.8665
0.8869
0.6628
0.6985
0.7324
0.7642
0.7939
0.8212
0.8461
0.8686
0.8888
0.6664
0.7019
0.7357
0.7673
0.7967
0.8238
0.8485
0.8708
0.8907
0.6700
0.7054
0.7389
0.7704
0.7995
0.8264
0.8508
0.8729
0.8925
0.6736
0.7088
0.7422
0.7734
0.8023
0.8289
0.8531
0.8749
0.8944
0.6772
0.7123
0.7454
0.7764
0.8051
0.8315
0.8554
0.8770
0.8962
0.6808
0.7157
0.7486
0.7794
0.8078
0.8340
0.8577
0.8790
0.8980
0.6844
0.7190
0.7517
0.7823
0.8106
0.8365
0.8599
0.8810
0.8997
0.6879
0.7224
0.7549
0.7852
0.8133
0.8389
0.8621
0.8830
0.9015
This is only a portion of the entire Positive z Score table
Section 6 – 3B Lecture
Page 9 of 16
© 2012 Eitel
Example 5
Finding the area to the left of a positive Z score
P( z < 1.41 )
If we want to find the probability of selecting one number from the z distribution and having that
number be less than the z value of 1.41 we need to find the area to the LEFT of z = 1.41
The number at the intersection of the
1.4
row and the
0.01
column is 0.9207
Positive Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
1.4
0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292
0.08
0.09
0.9306
0.9319
This means that the area to the left of z = 1.41 is 0.9207
the area to the left of
z = 1.41 is .9207
left tail area
= .1539left tail area
= .9207
0
z =1.41
positive
value
P( z < 1.41 ) = 0.9207
Finding the area to the Right of a positive Z score
P( z > 1.41 )
If we want to find the probability of selecting one number from the z distribution and having that
number be more than the z value of 1.41 we need to find the area to the RIGHT of z = 1.41
The total of the yellow and white areas is 1 so the right area (in white) = 1 – the left area
If the area to the Left of Z = 1.41 is 0.9207
then the area to the Right of Z = 1.41 is 1 – .9207 = .0793
P( z > 1.41 ) = 1 – 0.9207 = .0793
Example 6
Section 6 – 3B Lecture
Page 10 of 16
© 2012 Eitel
Finding the area to the left of a positive Z score
P( z < 2.33 )
If we want to find the probability of selecting one number from the z distribution and having that
number be less than the z value of 2.33 we need to find the area to the LEFT of z = 2.33
The number at the intersection of the
2.3
row and the
0.03
column is 0.9901
Positive Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
2.3
0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911
0.08
0.09
0.9913
0.9916
This means that the area to the left of z = 2.33 is 0.9901
the area to the left of
z = 2.33 is .9901
left tail area
= .1539left tail area
= .9901
0
z = 2.33
positive
value
P( z < 2.33 )= 0.9901
Finding the area to the Right of a positive Z score
P( z > 2.33 )
If we want to find the probability of selecting one number from the z distribution and having that
number be more than the z value of 2.33 we need to find the area to the RIGHT of z = 2.33
If the area to the Left of Z = 2.33 is .9901
then the area to the Right of Z = 2.33 is 1 – .9901= .0099
P( z > 2.33 ) = 1 – 0.9901 = .0099
Section 6 – 3B Lecture
Page 11 of 16
© 2012 Eitel
Example 7
Finding the area to the left of a positive Z score
A Special Case for z = 1.645
P( z < 1.645 )
If we want to find the probability of selecting one number from the z distribution and having that
number be less than the z value of 1.645 we need to find the area to the LEFT of z = 1.645
The cells at the bottom of the positive z scores table show that
the area to the left of Z = 1.645 is 0.9500
Positive Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
1.6
0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525
Z scores of 3.5 or more use .9999
AREA
0.9500
Z Score
1.645
AREA
0.08
0.09
0.9535
0.9545
Z Score
0.9950
2.575
This means that the area to the left of z = 1.645 is 0.9500
left tail area = .9500
0
right tail area
= .0500
z= 1.645
P( z < 1.645 ) = .9500
Finding the area to the Right of a positive Z score
P( z > 1.645 )
If the area to the Left of Z = 1.645 is .9500
then the area to the Right of Z = 1.645 is 1 – .9500 = .0500
P( z > 1.645 ) = 1 – 0.9500 = .0500
Section 6 – 3B Lecture
Page 12 of 16
© 2012 Eitel
Finding the area in yellow between two Z scores
P( z 1 < Z < z 2)
z1
z2
subtract the
blue area to
the left of z1
from the
yellow area to
the left of z2
z1
z2
subtract the
blue area to
the left of z1
from the
z2
z1
yellow area to
the left of z2
z1
z2
The area in yellow between Two Z scores z 1 and z 2
P( z 1 < Z < z 2) =
the Area to the left of z 2 – the area to the left of z 1
Section 6 – 3B Lecture
Page 13 of 16
© 2012 Eitel
Example 8
Find the area between z = –2.48 and z = 2.87
P( –2.48 < z < 2.87 )
If we want to find the probability of selecting one number from the z distribution and having that
number be BETWEEN the z value of –2.48 the z value of 2.87 we need to find the area
BETWEEN z = 2.87 and z = 2.87
z = –2.48
z =2.87
Positive Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
2.8
0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979
0.08
0.09
0.9980
0.9981
0.08
0.09
0.0066
0.0064
Positive Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–2.4
0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068
If the area to the left of Z = 2.87 is 0.9979
and the area to the left of Z = –2.48 is 0.0066
then the area BETWEEN Z = 2.87 and Z = –2.48 is
.9979 – .0066 = .9913
P( –2.48 < z < 2.87 ) = .9913
Section 6 – 3B Lecture
Page 14 of 16
© 2012 Eitel
Example 9
Find the area between Z = –1.75 and Z = 1.88
P( –1.75 < z < 1.88 )
If we want to find the probability of selecting one number from the z distribution and having that
number be BETWEEN the z value of –1.75 the z value of 1.88 we need to find the area
BETWEEN z = 1.75 and z = 1.88
z = –1.75
z =1.88
Positive Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
1.8
0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693
0.08
0.09
0.9699
0.9706
0.08
0.09
0.0375
0.0367
Negative Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–1.7
0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384
If the area to the left of Z = 2.87 is 0.9699
and the area to the left of Z = –2.48 is 0.0401
then the area BETWEEN Z = –1.75 and Z = 1.88 is
.9699 – .0401 = .9298
P( –1.75 < z < 1.88 ) = .9298
Section 6 – 3B Lecture
Page 15 of 16
© 2012 Eitel
Example 10
Find the area between Z = –2.51 and Z = 2.575
P( –2.51 < z < 2.575
If we want to find the probability of selecting one number from the z distribution and having that
number be BETWEEN the z value of –2.51 the z value of 2.575 we need to find the area
BETWEEN z = 2.51 and z = 2.575
z = –2.51
z = 2.575
the area to the left of Z = 2.575 is .9950
Positive Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z scores of 3.5 or more use .9999
AREA
0.9500
Z Score
1.645
AREA
0.9950
Z Score
2.575
the area to the left of Z = –2.51 is .0060
Negative Z Scores
Standard Normal (Z) Distribution: Cumulative Area to the LEFT of Z
Z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
–2.5
0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051
0.08
0.09
0.0049
0.0048
If the area to the left of Z = 2.575 is 0.9950
and the area to the left of Z = –2.51 is 0.0060
then the area BETWEEN Z = 2.575 and Z = –2.51 is
.9950 – .0060= .9890
P( –2.51 < z < 2.575 ) = .9890
Section 6 – 3B Lecture
Page 16 of 16
© 2012 Eitel