Statistics help?

Question

All, please see the below for your second assignment. Please be sure to include all work. Show your equations and step through the work in a clear and comprehensive way.

Find the mean for the binomial distribution which has the following values of n and p. Round answer to nearest tenth: n=37, p=0.2
Find the standard deviation for the binomial distribution which has the following values of n and p. Round answer to nearest hundredth: n=22, p= 3/5
Use the given values of n and p to find the minimum and maximum usual values: n=97, p=0.24

 

Answer 1

To do these problems you need to know the formulas for the mean and standard deviation of a binomial distribution which are 

 

mean = n*p 

standard deviation = sqrt( n*p*(1-p) ) 

 

If n = 37 and p = 0.2 then the mean = 37*0.2 = 7.4

 

If n = 22 and p = 3/5 then the standard deviation = sqrt( 22*(3/5)*(2/5) ) = sort(5.28) = 2.30 

 

If n = 97 and p = 0.24 then 

mean = 97*0.24 = 23.28

standard deviation = sqrt(97*0.24*0.76)) = sqrt(17.6928) = 4.21 

 

I don't know how your class defines minimum and maximum usual values, but if they are ±3 standard deviations from the mean then

the minimum usual value would be 23.28 - 3*4.21 = 10.65

and the maximum usual value would be 23.28 + 3*4.21 = 35.91