In a survey of time taken by students to reach Dubai from UD by bus, the researcher uses a sample of 36 clients. The published data shows that the population is normally distributed with a mean (p) equals 1.5 hours and a population standard deviation is 1 hour. What is the probability that the travel time of the sample (X) is between 1 and 1.25 hours? Note: Use standard error X-H)(X-p) X±Za2 S X+la/2 Standard Normal Probabilities Table entry Table entry for is the area under the standard normal cunve to the left of z 00 .01 02 .03 .04 06 07 08 09 0003 0004 0006 .0008 .0011 .0003 0002 .0004 .0005 .0007 0010 0003 .0003 .0003 0003 0003 L0005 L0007 .0010 .0003 0005 .0006 .0009 .0013 -3.4 0003 0004 .0003 0004 0006 0008 0004 .0005 0008 .0005 .0004 -3.3 -3.2 -3.1 -3.0 .0006 .0008 .0012 .0016 .0005 .0007 0010 0014 0019 0007 0009 .0013 0018 0025 .0034 0045 .0006 .0009 .0012 .0017 .0023 .0032 0043 .0057 0075 0099 0136 0132 0129 0166 0011 0011 0016 .0022 0030 0040 0054 0071 0094 0122 0158 0202 0256 0322 0401 0495 .0606 .0735 .0013 0019 .0019 0026 .0035 0047 0014 .0015 0021 0015 0021 2.9 -2.8 2.7 -2.6 .0018 0024 .0033 .0020 0027 .0023 .0031 0041 .0055 .0073 L0096 .0125 0162 0028 0038 0026 .0029 0039 0052 .0069 0091 0119 .0036 0048 .0037 0044 0051 0068 0089 .0049 .0066 .0087 0116 .0113 .0146 0188 0239 .0301 .0375 0465 0571 .0694 -2.5 -2.4 23 -2.2 -2.1 .0062 .0082 .0107 .0060 .0080 0104 0059 .0078 0102 .0064 0084 0110 0139 0179 -2.0 0228 .0287 0359 .0446 .0548 0668 .0808 0143 0183 .0233 .0294 0367 .0150 .0154 0197 .0250 .0314 .0392 0485 .0594 .0721 0170 0174 .0222 10281 0192 .0207 0262 0329 0409 0217 0212 0244 .0268 ,0336 -19 0274 0307 0384 0475 0582 0708 0344 <-1.8 -1.7 -1.6 S15 .0351 0436 O537 0655 0793 .0427 0526 0643 .0418 L0516 .0630 .0455 0559 .0681 0823 OSOS 0618 .0764 0918 -1.4 0778 .0749 n901 n960 13
In a survey of time taken by students to reach Dubai from UD by bus, the researcher uses a sample of 36 clients. The published data shows that the population is normally distributed with a mean (p) equals 1.5 hours and a population standard deviation is 1 hour. What is the probability that the travel time of the sample (X) is between 1 and 1.25 hours? Note: Use standard error X-H)(X-p) X±Za2 S X+la/2 Standard Normal Probabilities Table entry Table entry for is the area under the standard normal cunve to the left of z 00 .01 02 .03 .04 06 07 08 09 0003 0004 0006 .0008 .0011 .0003 0002 .0004 .0005 .0007 0010 0003 .0003 .0003 0003 0003 L0005 L0007 .0010 .0003 0005 .0006 .0009 .0013 -3.4 0003 0004 .0003 0004 0006 0008 0004 .0005 0008 .0005 .0004 -3.3 -3.2 -3.1 -3.0 .0006 .0008 .0012 .0016 .0005 .0007 0010 0014 0019 0007 0009 .0013 0018 0025 .0034 0045 .0006 .0009 .0012 .0017 .0023 .0032 0043 .0057 0075 0099 0136 0132 0129 0166 0011 0011 0016 .0022 0030 0040 0054 0071 0094 0122 0158 0202 0256 0322 0401 0495 .0606 .0735 .0013 0019 .0019 0026 .0035 0047 0014 .0015 0021 0015 0021 2.9 -2.8 2.7 -2.6 .0018 0024 .0033 .0020 0027 .0023 .0031 0041 .0055 .0073 L0096 .0125 0162 0028 0038 0026 .0029 0039 0052 .0069 0091 0119 .0036 0048 .0037 0044 0051 0068 0089 .0049 .0066 .0087 0116 .0113 .0146 0188 0239 .0301 .0375 0465 0571 .0694 -2.5 -2.4 23 -2.2 -2.1 .0062 .0082 .0107 .0060 .0080 0104 0059 .0078 0102 .0064 0084 0110 0139 0179 -2.0 0228 .0287 0359 .0446 .0548 0668 .0808 0143 0183 .0233 .0294 0367 .0150 .0154 0197 .0250 .0314 .0392 0485 .0594 .0721 0170 0174 .0222 10281 0192 .0207 0262 0329 0409 0217 0212 0244 .0268 ,0336 -19 0274 0307 0384 0475 0582 0708 0344 <-1.8 -1.7 -1.6 S15 .0351 0436 O537 0655 0793 .0427 0526 0643 .0418 L0516 .0630 .0455 0559 .0681 0823 OSOS 0618 .0764 0918 -1.4 0778 .0749 n901 n960 13
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
Topic Video
Question
100%
![-1
In a survey of time taken by students to reach Dubai from UD by bus, the researcher uses a sample of 36 clients. The published data
shows that the population is normally distributed with a mean (u) equals 1.5 hours and a population standard deviation is 1 hour. What
is the probability that the travel time of the sample (X) is between 1 and 1.25 hours? Note: Use standard error
ed
out of
X-) (X-u)
S.
X+la/2 Jn
In
question
Standard Normal Probabilities
Table entry,
Table entry for : is the area under the standard normal curve
to the left of z.
1.
.00
.01
.02
.03
.04
.05
.06
07
.08
.09
.0003
- .0003
Lo005
0003
.0004
0006
0009
* 0002
-3.4
-3.3
-3.2
-3.1 0010
-3.0
-2.9
-2.8
-2.7
-2.6
25
0003
.0003
.0004
.0006
.0008
.0011
0003
.0003
.0003
.0004
.0006
.0008
.0012
0003
.0004
.0005
.0008
0005
0005
.0006
.0009
0013
.0018
.0024
0033
0044
0004
0006
.0008
.0011
.0015 0015
.0021
0029
0039
.0004
0005
.0007
0011 .0010
0014
0020
.0027
.0003
0005
.0007
0010
.0014
.0019
0026
0036
.0048
.0064
0084
0110
.0007
0007
0009
.0013
0018
.0013
.0019
.0026
.0035
.0047
0062
L0012
.0017
0023
.0032
0043
.0057
0075
,0099
.0129
0166
0212
0268
.0025
0034
.0016
.0023
.0031
.0016
.0022
.0030
0040
0021
.0028
0038
.0051
0045
.0041
.0037
.0060
.0080
0104
0136
0174
.0222
.0281
.0351
.0052
.0069
.0091
.0119
0154
0197
0250
.0314
.0392
0485
L0594
.0721
.0049
.0059
0078
L0102
.0054
0071
L0094
L0122
0158
.0055
-2.4
-2.3
22
.0082
0107
0139
0179
.0228
0287
0359
0446
0548
0668
.0808
.0066
.0087
.0113
0146
.0188
.0239
0301
.0375
.0465
.0571
0073
0068
0089
0116
.0150
0192
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
0132
0170
0217
.0274
0344
.0427
0526
L0643
0778
.0096
0125
0162
0207
0262
.0329
10409
OSOS
L0618
0749
0202
0256
0322
L0401
0495
0606
0143
0183
.0233
0294
0367
L0244
0307
0336
0418
.0436
0537
0655
LOS16
0630
.0764
.0384
0475
LOS82
.0708
.0455
0559
.0681
0823
.0793
L0735
.0694
13
VO](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffef7f554-66ab-474b-9b63-0f0c8d649fe0%2Fd06551ea-8638-4823-85f1-e6156189d02f%2Fppjr1mw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:-1
In a survey of time taken by students to reach Dubai from UD by bus, the researcher uses a sample of 36 clients. The published data
shows that the population is normally distributed with a mean (u) equals 1.5 hours and a population standard deviation is 1 hour. What
is the probability that the travel time of the sample (X) is between 1 and 1.25 hours? Note: Use standard error
ed
out of
X-) (X-u)
S.
X+la/2 Jn
In
question
Standard Normal Probabilities
Table entry,
Table entry for : is the area under the standard normal curve
to the left of z.
1.
.00
.01
.02
.03
.04
.05
.06
07
.08
.09
.0003
- .0003
Lo005
0003
.0004
0006
0009
* 0002
-3.4
-3.3
-3.2
-3.1 0010
-3.0
-2.9
-2.8
-2.7
-2.6
25
0003
.0003
.0004
.0006
.0008
.0011
0003
.0003
.0003
.0004
.0006
.0008
.0012
0003
.0004
.0005
.0008
0005
0005
.0006
.0009
0013
.0018
.0024
0033
0044
0004
0006
.0008
.0011
.0015 0015
.0021
0029
0039
.0004
0005
.0007
0011 .0010
0014
0020
.0027
.0003
0005
.0007
0010
.0014
.0019
0026
0036
.0048
.0064
0084
0110
.0007
0007
0009
.0013
0018
.0013
.0019
.0026
.0035
.0047
0062
L0012
.0017
0023
.0032
0043
.0057
0075
,0099
.0129
0166
0212
0268
.0025
0034
.0016
.0023
.0031
.0016
.0022
.0030
0040
0021
.0028
0038
.0051
0045
.0041
.0037
.0060
.0080
0104
0136
0174
.0222
.0281
.0351
.0052
.0069
.0091
.0119
0154
0197
0250
.0314
.0392
0485
L0594
.0721
.0049
.0059
0078
L0102
.0054
0071
L0094
L0122
0158
.0055
-2.4
-2.3
22
.0082
0107
0139
0179
.0228
0287
0359
0446
0548
0668
.0808
.0066
.0087
.0113
0146
.0188
.0239
0301
.0375
.0465
.0571
0073
0068
0089
0116
.0150
0192
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
0132
0170
0217
.0274
0344
.0427
0526
L0643
0778
.0096
0125
0162
0207
0262
.0329
10409
OSOS
L0618
0749
0202
0256
0322
L0401
0495
0606
0143
0183
.0233
0294
0367
L0244
0307
0336
0418
.0436
0537
0655
LOS16
0630
.0764
.0384
0475
LOS82
.0708
.0455
0559
.0681
0823
.0793
L0735
.0694
13
VO
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