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How to Manipulate formula for Binomial Distribution
How can I change binomial distribution formula to do:
successes of at least 2 (this equates to 1 - f(0) - (f1) number of trials 6 probability of both f((0) and f(1) individually equal .23 |
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#2
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How to Manipulate formula for Binomial Distribution
You have not specified the binomial p, but I will guess that p=0.4 from your
ambiguously worded "probability of both f(0) and f(1) individually equal 0.23" =1-BINOMDIST(1,6,0.4,TRUE) returns 0.76672. Jerry "GH" wrote: How can I change binomial distribution formula to do: successes of at least 2 (this equates to 1 - f(0) - (f1) number of trials 6 probability of both f((0) and f(1) individually equal .23 |
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#3
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How to Manipulate formula for Binomial Distribution
Thanks Jerry for your response. The problem I am trying to solve reads "In a
sample of 6 International travelers, what is the probability that AT LEAST 2 will remain with the tour group?" The binomial dist formula computes exact probability not the "AT LEAST 2" which requires the formula be adjusted to this. This is what I am trying to determine. HELP GH "Jerry W. Lewis" wrote: You have not specified the binomial p, but I will guess that p=0.4 from your ambiguously worded "probability of both f(0) and f(1) individually equal 0.23" =1-BINOMDIST(1,6,0.4,TRUE) returns 0.76672. Jerry "GH" wrote: How can I change binomial distribution formula to do: successes of at least 2 (this equates to 1 - f(0) - (f1) number of trials 6 probability of both f((0) and f(1) individually equal .23 |
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#4
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How to Manipulate formula for Binomial Distribution
Yes, that is what you asked and I answered the first time. See Help for
BINOMDIST, paying particular attention to the final argument, to understand the solution. Jerry "GH" wrote: Thanks Jerry for your response. The problem I am trying to solve reads "In a sample of 6 International travelers, what is the probability that AT LEAST 2 will remain with the tour group?" The binomial dist formula computes exact probability not the "AT LEAST 2" which requires the formula be adjusted to this. This is what I am trying to determine. HELP GH "Jerry W. Lewis" wrote: You have not specified the binomial p, but I will guess that p=0.4 from your ambiguously worded "probability of both f(0) and f(1) individually equal 0.23" =1-BINOMDIST(1,6,0.4,TRUE) returns 0.76672. Jerry "GH" wrote: How can I change binomial distribution formula to do: successes of at least 2 (this equates to 1 - f(0) - (f1) number of trials 6 probability of both f((0) and f(1) individually equal .23 |
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