Part B: You can also find the standard deviation for the binomial probability distribution of a specific outcome in a binomial experiment. Use the formula to find the standard deviation. You've already identified n and p in Part A. Show your work, and round your answer to two decimal places.

Mean = np + n(n-1)p²

Standard Deviation= √ σ²   = √npq

Step-by-step explanation:

The m.g.f of the binomial probability distribution b(x;n,p) is derived as below.

M₀t = E (e)^tx

        = ∑(e)^tx (nCx) (p)^x q^(n-x)            { x varies from 0 to n}

        = ∑(e)^tx (nCx) (pe^t)^x q^(n-x)

          = (q +pe^t)^n

The expansion of this binomial is purely algebraic and need not to be interpreted in terms of probabilities.

WE get the moments by differentiating M₀(t)  once, twice, etc. with respect to t and putting t=0

Thus

μ₁`  = E(X) = [ d/dt (q +pe^t)^n]    t=0

     = [ npe^t (q +pe^t)^n-1]  t=0  

     = np

And

μ₂`  = E(X²) = [ d²/dt² (q +pe^t)^n]    t=0

      = [ npe^t (q +pe^t)^n-1]    +  [ n(n-1)p²e^²t (q +pe^t)^n-2]    t=0

= np + n(n-1)p²

Variance= μ₂= μ₂`  - μ₁`²

             σ² =E(X²)-  E(X)²

              σ² =np + n(n-1)p²- (np )²

             σ²   =npq

Standard Deviation= √ σ²   = √npq

yes

step-by-step explanation: