Binomial Mean and Standard Deviation For X ∼ B ( n , p ) {X \sim B(n, p)} X ∼ B ( n , p ) :
σ X 2 = n p ( 1 − p ) \sigma_X^2 = np(1-p) σ X 2 = n p ( 1 − p )
σ X = n p ( 1 − p ) \sigma_X = \sqrt{np(1-p)} σ X = n p ( 1 − p ) These formulas emerge directly from the general formulas for sums of independent Bernoulli variables.
Bernoulli trial
For a Bernoulli variable Y {Y} Y : μ Y = p {\mu_Y = p} μ Y = p , σ Y 2 = p ( 1 − p ) {\sigma_Y^2 = p(1-p)} σ Y 2 = p ( 1 − p )
The binomial variable
X = Y 1 + Y 2 + ⋯ + Y n {X = Y_1 + Y_2 + \cdots + Y_n} X = Y 1 + Y 2 + ⋯ + Y n (sum of
n {n} n independent Bernoullis):
μ X = ∑ μ Y i = n p \mu_X = \sum \mu_{Y_i} = np μ X = ∑ μ Y i = n p
σ X 2 = ∑ σ Y i 2 = n ⋅ p ( 1 − p ) \sigma_X^2 = \sum \sigma_{Y_i}^2 = n \cdot p(1-p) σ X 2 = ∑ σ Y i 2 = n ⋅ p ( 1 − p ) Example: X ∼ B ( 50 , 0.6 ) {X \sim B(50, 0.6)} X ∼ B ( 50 , 0.6 ) :
μ X = 50 ( 0.6 ) = 30 {\mu_X = 50(0.6) = 30} μ X = 50 ( 0.6 ) = 30 , σ X = 50 ( 0.6 ) ( 0.4 ) = 12 ≈ 3.46 {\sigma_X = \sqrt{50(0.6)(0.4)} = \sqrt{12} \approx 3.46} σ X = 50 ( 0.6 ) ( 0.4 ) = 12 ≈ 3.46