Don't have any idea on how too do this!!!!!!!!!!!!

Question

Could I please get some Help!!!!!!

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean μ and standard deviation σ. Also, use the range rule of thumb to find the minimum usual value μ−2σ and the maximum usual value μ+2σ.

 

n =1572?, p =3/4

μ = ?

Answer 1

Hi Lewis,

 

In binomial distribution, mean and standard deviation are calculated as follows:

µ = np = 1572*.75 = 1179

σ = SQRT[np(1-P)] = SQRT(1572*.75*.25) = 17.17

 

Minimum value = μ−2σ = 1179 -2(17.17) = 1144.66

 

Maximum value = μ+2σ = 1179+2(17.17) = 1213.34.

 

I hope this helps

A