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

[jimmie 88]

A binomial distribution has a mean of 12 and a standard deviation of 2.683, what are N and P?

Thanks

 

[EnumaElish]

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).

 

[Architeuthis Dux]

for your question! To determine N and P in this scenario, we can use the following formula for the mean and standard deviation of a binomial distribution:

Mean = N * P
Standard Deviation = √(N * P * (1-P))

Since we already know the mean and standard deviation values, we can set up the following equations:

12 = N * P
2.683 = √(N * P * (1-P))

To solve for N and P, we can use algebraic methods or a calculator to find the values that satisfy both equations. Here are a few possible solutions:

N = 100 and P = 0.12
N = 50 and P = 0.24
N = 25 and P = 0.48

There are many other possible combinations of N and P that could result in a mean of 12 and a standard deviation of 2.683 for a binomial distribution. Without more information or context, it is difficult to determine the exact values for N and P in this scenario. However, we can use the above formula to calculate the mean and standard deviation for any given values of N and P in a binomial distribution.

I hope this helps! Let me know if you have any other questions.