Assume that a procedure yields a binomial distribution with n = 2 trials and a probability of success of p = 0.05. Use a binomial probability table to find the probability that the number of successes x is exactly 1. Click on the icon to view the binomial probabilities table. P(1) = (Round to three decimal places as needed.)

MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
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Binomial Probabilities Table
Binomial Probabilities
.01
.05
.10
.20
.30
.40
.50
60
.70
80
.90
.95
.99
2
.980
.902
.810
.640
.490
.360
.250
.160
090
040
010
002
0+
2
1
020
.095
.180
.320
.420
.480
500
A80
420
.180
095
.020
1
2
0+
.002
.010
.040
.090
.160
.250
360
490
640
810
902
.980
2
3
.970
.857
.729
.512
.343
.216
.125
064
027
008
001
0+
0+
3
1
029
.135
.243
.384
.441
.432
.375
208
.189
095
027
007
0+
2
0+
.007
.027
.096
.189
.288
.375
432
441
384
243
.135
.029
2
3
0+
0+
.001
.008
.027
.064
.125
216
343
512
.729
857
.970
3
.961
.815
.656
410
.240
.130
.062
026
008
002
0+
0+
0+
1
039
.171
.292
410
.412
.346
.250
.154
076
026
004
0+
0+
1
2
001
.014
.049
.154
.265
.346
.375
346
265
.154
049
014
.001
2
3
0+
0+
.004
.026
.076
.154
.250
346
412
410
292
.171
.039
3
4
0+
0+
0+
.002
.008
.026
.062
.130
240
410
656
815
.961
4.
.951
.774
.590
.328
.168
.078
.031
010
002
0+
0+
0+
0+
5
1
048
.204
.328
410
.360
.259
.156
077
028
006
0+
0+
0+
2
001
.021
.073
.205
.309
.346
.312
230
.132
051
008
001
0+
3
0+
.001
.008
.051
.132
.230
.312
346
309
205
073
021
.001
3
4
0+
0+
0+
.006
.028
.077
.156
259
360
410
328
204
.048
5
0+
0+
0+
0+
.002
.010
.031
078
.168
328
590
.774
.951
5
.941
.735
.531
.262
.118
.047
.016
004
001
0+
0+
0+
1
057
.232
.354
.393
.303
.187
.094
037
010
002
0+
0+
0+
2
001
.031
.098
246
.324
.311
.234
.138
060
015
001
0+
0+
3
0+
.002
.015
.082
.185
.276
.312
276
.185
082
015
002
0+
3
4
0+
0+
.001
.015
.060
.138
.234
311
324
246
098
031
.001
4
Print
Done
Transcribed Image Text:Binomial Probabilities Table Binomial Probabilities .01 .05 .10 .20 .30 .40 .50 60 .70 80 .90 .95 .99 2 .980 .902 .810 .640 .490 .360 .250 .160 090 040 010 002 0+ 2 1 020 .095 .180 .320 .420 .480 500 A80 420 .180 095 .020 1 2 0+ .002 .010 .040 .090 .160 .250 360 490 640 810 902 .980 2 3 .970 .857 .729 .512 .343 .216 .125 064 027 008 001 0+ 0+ 3 1 029 .135 .243 .384 .441 .432 .375 208 .189 095 027 007 0+ 2 0+ .007 .027 .096 .189 .288 .375 432 441 384 243 .135 .029 2 3 0+ 0+ .001 .008 .027 .064 .125 216 343 512 .729 857 .970 3 .961 .815 .656 410 .240 .130 .062 026 008 002 0+ 0+ 0+ 1 039 .171 .292 410 .412 .346 .250 .154 076 026 004 0+ 0+ 1 2 001 .014 .049 .154 .265 .346 .375 346 265 .154 049 014 .001 2 3 0+ 0+ .004 .026 .076 .154 .250 346 412 410 292 .171 .039 3 4 0+ 0+ 0+ .002 .008 .026 .062 .130 240 410 656 815 .961 4. .951 .774 .590 .328 .168 .078 .031 010 002 0+ 0+ 0+ 0+ 5 1 048 .204 .328 410 .360 .259 .156 077 028 006 0+ 0+ 0+ 2 001 .021 .073 .205 .309 .346 .312 230 .132 051 008 001 0+ 3 0+ .001 .008 .051 .132 .230 .312 346 309 205 073 021 .001 3 4 0+ 0+ 0+ .006 .028 .077 .156 259 360 410 328 204 .048 5 0+ 0+ 0+ 0+ .002 .010 .031 078 .168 328 590 .774 .951 5 .941 .735 .531 .262 .118 .047 .016 004 001 0+ 0+ 0+ 1 057 .232 .354 .393 .303 .187 .094 037 010 002 0+ 0+ 0+ 2 001 .031 .098 246 .324 .311 .234 .138 060 015 001 0+ 0+ 3 0+ .002 .015 .082 .185 .276 .312 276 .185 082 015 002 0+ 3 4 0+ 0+ .001 .015 .060 .138 .234 311 324 246 098 031 .001 4 Print Done
Assume that
procedure yields a binomial distribution with n =2 trials and a probability of success of p = 0.05. Use a binomial probability table to find the probability
that the number of successes x is exactly 1.
E Click on the icon to view the binomial probabilities table.
P(1) = (Round to three decimal places as needed.)
Transcribed Image Text:Assume that procedure yields a binomial distribution with n =2 trials and a probability of success of p = 0.05. Use a binomial probability table to find the probability that the number of successes x is exactly 1. E Click on the icon to view the binomial probabilities table. P(1) = (Round to three decimal places as needed.)
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