N & P of Binomial Distribution with Mean 12 & SD 2.683

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To determine N (number of trials) and P (probability of success) for a binomial distribution with a mean of 12 and a standard deviation of 2.683, the relationships between these parameters must be utilized. The mean is calculated as N * P, while the standard deviation is given by the formula sqrt(N * P * (1 - P)). By substituting the known values into these equations, one can derive the values for N and P. The discussion emphasizes the importance of demonstrating effort in problem-solving before seeking help. Understanding the definitions and relationships in binomial distributions is crucial for accurately solving such problems.
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A binomial distribution has a mean of 12 and a standard deviation of 2.683, what are N and P?

Thanks
 
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First, let's see the $...

Is this homework?

Show some effort then ask the question again. At least write out the definition of a binomial dist. and suggest how the problem can be approached (you started out that way but need further advice) or cannot be approached (because you tried and didn't work).
 

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