Standard Deviation Binomial Dist: n=50, p=0.3, Prob (x > 13 & 13 < x < 17)

  • Thread starter Thread starter jim1174
  • Start date Start date
AI Thread Summary
To find the standard deviation for a binomial distribution with n=50 and p=0.3, the formula used is √(n * p * (1 - p)), resulting in a standard deviation of approximately 3.87. The normal approximation can be applied to calculate probabilities for the specified ranges: for (a) P(x > 13) and (b) P(13 < x < 17). Using the normal distribution, these probabilities can be estimated using z-scores derived from the mean and standard deviation. The discussion emphasizes the importance of showing work to receive targeted assistance. The thread is categorized under homework due to the nature of the inquiries.
jim1174
Messages
79
Reaction score
0
find the standard deviation for the binomial distribution which has stated values of n=50 and p=0.3. use normal distribution to approximate the binomial distribution and find probability of (a) x> 13 and (b) 13<x 17.
 
Physics news on Phys.org
? Any reason why we should?

1) This looks like homework, not something you are just interested in.

2) Everything you ask here can be done by using simple formulas. What are they?

Show what you have done yourself so we will know where you need help.

I am moving this to the homework section.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top