Answer:
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
