Assume that a procedure yields a binomial distribution with n =33 trials and a p
ID: 3073704 • Letter: A
Question
Assume that a procedure yields a binomial distribution with n =33 trials and a probability of success of p=0.10. Use a binomial probability table to find the probability that the number of successes x is exactly 11
Binomial Probabilities
p
n x .01 .05 .10 .20 .30 .40 .50 .60 .70 .80 .90 .95 .99 x n
2 0 .980 .902 .810 .640 .490 .360 .250 .160 .090 .040 .010 .002 0+ 0 2
1 .020 .095 .180 .320 .420 .480 .500 .480 .420 .320 .180 .095 .020 1
2 0+ .002 .010 .040 .090 .160 .250 .360 .490 .640 .810 .902 .980 2
3 0 .970 .857 .729 .512 .343 .216 .125 .064 .027 .008 .001 0+ 0+ 0 3
1 .029 .135 .243 .384 .441 .432 .375 .288 .189 .096 .027 .007 0+ 1
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
4 0 .961 .815 .656 .410 .240 .130 .062 .026 .008 .002 0+ 0+ 0+ 0 4
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
5 0 .951 .774 .590 .328 .168 .078 .031 .010 .002 0+ 0+ 0+ 0+ 0 5
1 .048 .204 .328 .410 .360 .259 .156 .077 .028 .006 0+ 0+ 0+ 1
2 .001 .021 .073 .205 .309 .346 .312 .230 .132 .051 .008 .001 0+ 2
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 4
5 0+ 0+ 0+ 0+ .002 .010 .031 .078 .168 .328 .590 .774 .951 5
6 0 .941 .735 .531 .262 .118 .047 .016 .004 .001 0+ 0+ 0+ 0+ 0 6
1 .057 .232 .354 .393 .303 .187 .094 .037 .010 .002 0+ 0+ 0+ 1
2 .001 .031 .098 .246 .324 .311 .234 .138 .060 .015 .001 0+ 0+ 2
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
5 0+ 0+ 0+ .002 .010 .037 .094 .187 .303 .393 .354 .232 .057 5
6 0+ 0+ 0+ 0+ .001 .004 .016 .047 .118 .262 .531 .735 .941 6
7 0 .932 .698 .478 .210 .082 .028 .008 .002 0+ 0+ 0+ 0+ 0+ 0 7
1 .066 .257 .372 .367 .247 .131 .055 .017 .004 0+ 0+ 0+ 0+ 1
2 .002 .041 .124 .275 .318 .261 .164 .077 .025 .004 0+ 0+ 0+ 2
3 0+ .004 .023 .115 .227 .290 .273 .194 .097 .029 .003 0+ 0+ 3
4 0+ 0+ .003 .029 .097 .194 .273 .290 .227 .115 .023 .004 0+ 4
5 0+ 0+ 0+ .004 .025 .077 .164 .261 .318 .275 .124 .041 .002 5
6 0+ 0+ 0+ 0+ .004 .017 .055 .131 .247 .367 .372 .257 .066 6
7 0+ 0+ 0+ 0+ 0+ .002 .008 .028 .082 .210 .478 .698 .932 7
8 0 .923 .663 .430 .168 .058 .017 .004 .001 0+ 0+ 0+ 0+ 0+ 0 8
1 .075 .279 .383 .336 .198 .090 .031 .008 .001 0+ 0+ 0+ 0+ 1
2 .003 .051 .149 .294 .296 .209 .109 .041 .010 .001 0+ 0+ 0+ 2
3 0+ .005 .033 .147 .254 .279 .219 .124 .047 .009 0+ 0+ 0+ 3
4 0+ 0+ .005 .046 .136 .232 .273 .232 .136 .046 .005 0+ 0+ 4
5 0+ 0+ 0+ .009 .047 .124 .219 .279 .254 .147 .033 .005 0+ 5
6 0+ 0+ 0+ .001 .010 .041 .109 .209 .296 .294 .149 .051 .003 6
7 0+ 0+ 0+ 0+ .001 .008 .031 .090 .198 .336 .383 .279 .075 7
8 0+ 0+ 0+ 0+ 0+ .001 .004 .017 .058 .168 .430 .663 .923 8
n x .01 .05 .10 .20 .30 .40 .50 .60 .70 .80 .90 .95 .99 x n
p
NOTE: 0+ represents a positive probability less than 0.0005.
P(1)=_____
(Round to three decimal places as needed.)
Explanation / Answer
Here, X~Bin(33,0.1) i.e. n=33 and p=0.1
From the binomial table we get the cumulative probabilties i.e. F(x) = P(X<=x).
So, for P(X=11) = P(X<=11)-P(X<11) = P(X<=11) - P(X<=10) = F(11) - F(10) = 0.999953-0.999762 = 0.000191 ~ 0.000(round off to 3 decimal places)
and P(X=1) = P(X<=1)-P(X<1)=P(X<=1)-P(X=0)=0.144215-0.030903 = 0.113312 ~ 0.113(Round off to 3 decimal places)
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