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Mathematics · Ch 15 — Binomial Distribution

Mean and Variance of Binomial Distribution ( Formulae without proof )

15.3

Mean and Variance of Binomial Distribution ( Formulae without proof )

Mean and Variance of Binomial Distribution ( Formulae without proof )

These results are stated here without proof. If X∼B(n,p)X\sim B(n,p), then the mean (expected value) of XX is denoted μ\mu or E(X)E(X) and given by

μ=E(X)=np.\mu = E(X) = np.

The variance is denoted Var(X)\text{Var}(X) and given by

Var(X)=npq.\text{Var}(X) = npq.

The standard deviation is denoted SD(X)SD(X) or σX\sigma_X, and is the square root of the variance:

SD(X)=σX=Var(X)=npq.SD(X) = \sigma_X = \sqrt{\text{Var}(X)} = \sqrt{npq}.

In-text example. If X∼B(10,0.4)X\sim B(10, 0.4), here n=10n=10, p=0.4p=0.4, so q=1−0.4=0.6q=1-0.4=0.6. Then E(X)=np=10×0.4=4E(X)=np=10\times0.4=4 and Var(X)=npq=10×0.4×0.6=2.4\text{Var}(X)=npq=10\times0.4\times0.6=2.4.

Worked examples.

Ex.1 - The p.m.f. P(X=x)=4Cx(59)x(49)4−xP(X=x)={}^{4}C_x\left(\tfrac59\right)^x\left(\tfrac49\right)^{4-x}, x=0,1,2,3,4x=0,1,2,3,4, is a binomial distribution with n=4n=4, p=59p=\tfrac59, q=49q=\tfrac49 (matching the standard nCxpxqn−x{}^{n}C_xp^xq^{n-x} pattern). So E(X)=np=4×59=209E(X)=np=4\times\tfrac59=\tfrac{20}{9} and Var(X)=npq=4×59×49=8081\text{Var}(X)=npq=4\times\tfrac59\times\tfrac49=\tfrac{80}{81}. …

Misc InEx1In-text example: X~B(10, 0.4), find E(X) and Var(X)

Worked out. A short in-text illustration applying the mean and variance formulas directly with n=10, p=0.4, q=0.6, giving E(X)=4 and Var(X)=2.4 - the simplest 'forward' use of the two formulas before the harder reverse-direction worked examples that follow. …

Misc Ex1p.m.f. given as 4Cx(5/9)^x(4/9)^(4-x) - find E(X) and Var(X)

Worked out. Recognises a printed probability function as a binomial distribution by matching it to the nCx p^x q^(n-x) pattern, reads off n=4, p=5/9, q=4/9 from the given formula, then applies E(X)=np and Var(X)=npq to get 20/9 and 80/81. …

Misc Ex2E(X)=6 and Var(X)=4.2 given - find n and p

Worked out. Works the mean/variance formulas in reverse: dividing Var(X)=npq by E(X)=np cancels the unknown n and isolates q=Var(X)/E(X)=0.7, then p=1-q=0.3, and finally n=E(X)/p=20, recovering both parameters from the two summary numbers alone. …