NEGATIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 -3.50 and lower .0001 .0003 .0003 0003 .0003 .0004 -3.4 .0003 .0003 .0003 .0003 .0003 .0002 -3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0003 -3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005 -3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007 -3.0 .0013 .0013 .0013 .0012 .0012 .001 .001 .001 .0010 .0010 -2.9 .0019 .O018 .0018 .0017 .0016 .0016 .O015 .0015 .0014 .0014 -2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 -2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026 -2.6 ,0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 ,0036 -2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 * .0049 .0048 -2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064 -2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084 -2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110 -2.1 .0179 0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143 -2.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188 .0183 -1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233 -1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294 -1.7 .0446 .0436 .0427 .0418 0409 .0401 .0392 .0384 .0375 .0367 Print Done
NEGATIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 -3.50 and lower .0001 .0003 .0003 0003 .0003 .0004 -3.4 .0003 .0003 .0003 .0003 .0003 .0002 -3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0003 -3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005 -3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 .0007 .0007 -3.0 .0013 .0013 .0013 .0012 .0012 .001 .001 .001 .0010 .0010 -2.9 .0019 .O018 .0018 .0017 .0016 .0016 .O015 .0015 .0014 .0014 -2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 -2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026 -2.6 ,0047 .0045 .0044 .0043 .0041 .0040 .0039 .0038 .0037 ,0036 -2.5 .0062 .0060 .0059 .0057 .0055 .0054 .0052 .0051 * .0049 .0048 -2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064 -2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084 -2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110 -2.1 .0179 0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143 -2.0 .0228 .0222 .0217 .0212 .0207 .0202 .0197 .0192 .0188 .0183 -1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233 -1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294 -1.7 .0446 .0436 .0427 .0418 0409 .0401 .0392 .0384 .0375 .0367 Print Done
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Concept explainers
Contingency Table
A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.
Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
Topic Video
Question
2
![tage
esult
NEGATIVE z Scores
Standard Normal (z) Distribution: Cumulative Area from the LEFT
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
-3.50
and
lower
.001
-3.4
.0003
.0003
.0003
.0003
.0003
.0003
.0003
.0003
.0003
.0002
-3.3
.0005
.0005
.0005
.0004
.0004
.0004
.0004
.0004
.0004
.0003
.0005
.0007
-3.2
,0007
.0007
.0006
.0006
.0006
.0006
.0006
.0005
.0005
-3.1
.0010
.0009
.0009
.0009
.0008
.0008
.0008
.0008
.0007
-3.0
.0013
.o013
.0013
.0012
.0012
.001
.001
.001
.0010
.0010
-2.9
.0019
.0018
.0018
.0017
.0016
.0016
.0015
.0015
.0014
.0014
-2.8
.0026
.0025
.0024
.0023
.0023
.0022
.0021
.0021
.0020
.0019
-2.7
.0035
.0034
.0033
.0032
.0031
.0030
.0029
.0028
.0027
.0026
-2.6
.0047
.0045
.0044
.0043
.0041
.0040
.0039
.0038
.0037
.0036
-2.5
.0062
.0060
.0059
.0057
.0055
.0054
.0052
.0051
* .0049
.0048
-2.4
.0082
.0080
.0078
.0075
.0073
.0071
.0069
.0068
.0066
.0064
-2.3
.0107
.0104
.0102
.0099
.0096
.0094
.0091
.0089
.0087
.0084
-2.2
.0139
.0136
.0132
.0129
.0125
.0122
.0119
.0116
0113
.0110
-2.1
.0179
0174
.0170
.0166
0162
.0158
.0154
.0150
.0146
.0143
-2.0
.0228
.0222
.0217
.0212
.0207
.0202
.0197
.0192
.0188
.0183
-1.9
.0287
.0281
.0274
.0268
.0262
.0256
.0250
.0244
.0239
.0233
-1.8
.0359
.0351
.0344
.0336
.0329
.0322
.0314
.0307
.0301
.0294
-1.7
.0446
.0436
.0427
.0418
0409
.0401
.0392
.0384
.0375
.0367
Print
Done](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa3d4fb4b-44f5-4612-bf88-fe843821e62a%2F4d1bc8df-e7c6-42dd-8cd5-3e67a940a08b%2Fcxwxkj9_processed.png&w=3840&q=75)
Transcribed Image Text:tage
esult
NEGATIVE z Scores
Standard Normal (z) Distribution: Cumulative Area from the LEFT
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
-3.50
and
lower
.001
-3.4
.0003
.0003
.0003
.0003
.0003
.0003
.0003
.0003
.0003
.0002
-3.3
.0005
.0005
.0005
.0004
.0004
.0004
.0004
.0004
.0004
.0003
.0005
.0007
-3.2
,0007
.0007
.0006
.0006
.0006
.0006
.0006
.0005
.0005
-3.1
.0010
.0009
.0009
.0009
.0008
.0008
.0008
.0008
.0007
-3.0
.0013
.o013
.0013
.0012
.0012
.001
.001
.001
.0010
.0010
-2.9
.0019
.0018
.0018
.0017
.0016
.0016
.0015
.0015
.0014
.0014
-2.8
.0026
.0025
.0024
.0023
.0023
.0022
.0021
.0021
.0020
.0019
-2.7
.0035
.0034
.0033
.0032
.0031
.0030
.0029
.0028
.0027
.0026
-2.6
.0047
.0045
.0044
.0043
.0041
.0040
.0039
.0038
.0037
.0036
-2.5
.0062
.0060
.0059
.0057
.0055
.0054
.0052
.0051
* .0049
.0048
-2.4
.0082
.0080
.0078
.0075
.0073
.0071
.0069
.0068
.0066
.0064
-2.3
.0107
.0104
.0102
.0099
.0096
.0094
.0091
.0089
.0087
.0084
-2.2
.0139
.0136
.0132
.0129
.0125
.0122
.0119
.0116
0113
.0110
-2.1
.0179
0174
.0170
.0166
0162
.0158
.0154
.0150
.0146
.0143
-2.0
.0228
.0222
.0217
.0212
.0207
.0202
.0197
.0192
.0188
.0183
-1.9
.0287
.0281
.0274
.0268
.0262
.0256
.0250
.0244
.0239
.0233
-1.8
.0359
.0351
.0344
.0336
.0329
.0322
.0314
.0307
.0301
.0294
-1.7
.0446
.0436
.0427
.0418
0409
.0401
.0392
.0384
.0375
.0367
Print
Done
![Assume that human body temperatures are normally distributed with a mean of 98.19°F and a standard deviation of 0.64°F.
a. A hospital uses 100.6°F as the lowest temperature considered to be a fever. What percentage of normal and healthy persons would be considered to have a fever? Does this percentage suggest that a cutoff of 100.6°F is appropriate?
b. Physicians want to select a minimum temperature for requiring further medical tests. What should that temperature be, if we want only 5.0% of healthy people to exceed it? (Such a result is a false positive, meaning that the test result is
positive, but the subject is not really sick.)
Click to view page 1 of the table. Click to view page 2 of the table.
a. The percentage of normal and healthy persons considered to have a fever is
%.
(Round to two decimal places as needed.)
Does this percentage suggest that a cutoff of 100.6°F is appropriate?
O A. Yes, because there is a large probability that a normal and healthy person would be considered to have a fever.
O B. Yes, because there is a small probability that a normal and healthy person would be considered to have a fever.
C. No, because there is a small probability that a normal and healthy person would be considered to have a fever.
D. No, because there is a large probability that a normal and healthy person would be considered to have a fever.
b. The minimum temperature for requiring further medical tests should be °F if we want only 5.0% of healthy people to exceed it.
(Round to two decimal places as needed.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa3d4fb4b-44f5-4612-bf88-fe843821e62a%2F4d1bc8df-e7c6-42dd-8cd5-3e67a940a08b%2F5oakomc_processed.png&w=3840&q=75)
Transcribed Image Text:Assume that human body temperatures are normally distributed with a mean of 98.19°F and a standard deviation of 0.64°F.
a. A hospital uses 100.6°F as the lowest temperature considered to be a fever. What percentage of normal and healthy persons would be considered to have a fever? Does this percentage suggest that a cutoff of 100.6°F is appropriate?
b. Physicians want to select a minimum temperature for requiring further medical tests. What should that temperature be, if we want only 5.0% of healthy people to exceed it? (Such a result is a false positive, meaning that the test result is
positive, but the subject is not really sick.)
Click to view page 1 of the table. Click to view page 2 of the table.
a. The percentage of normal and healthy persons considered to have a fever is
%.
(Round to two decimal places as needed.)
Does this percentage suggest that a cutoff of 100.6°F is appropriate?
O A. Yes, because there is a large probability that a normal and healthy person would be considered to have a fever.
O B. Yes, because there is a small probability that a normal and healthy person would be considered to have a fever.
C. No, because there is a small probability that a normal and healthy person would be considered to have a fever.
D. No, because there is a large probability that a normal and healthy person would be considered to have a fever.
b. The minimum temperature for requiring further medical tests should be °F if we want only 5.0% of healthy people to exceed it.
(Round to two decimal places as needed.)
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