Binomial Distribution Elements of Binomial Distribution (Bernouli Process) n Identical Trials 2 Outcomes: Success and Failure P(S) = p P(F) = q Constant Probability of Success Independent Trials ( ½ + ½)2 = 1/4 + 2/4 + 1/4 ( ½ + ½ )3 = 1/8 + 3/8 + 3/8 + 1/8

Binomial Probability Formula: Example: S = Bus Major p = .20 n = 4 q = .80 P4(0) = 4C0(.2)0(.8)4 = P4(1) = 4C1(.2)1(.8)3 = P4(2) = 4C2(.2)2(.8)2 = P4(3) = 4C3(.2)3(.8)1 = P4(4) = 4C4(.2)4(.8)0 =

Example: 5 Coin Toss p = ½ n = 5 q = ½ P5(0) = 5C0(1/2)0(1/2)5 = P5(1) = 5C1(1/2)1(1/2)4 = P5(2) = 5C2(1/2)2(1/2)3 = P5(3) = 5C3(1/2)3(1/2)2 = P5(4) = 5C4(1/2)4(1/2)1 = P5(5) = 5C5(1/2)5(1/2)0 =

Mean for Binomial Distribution - µ = n•p Variance for Binomial Distribution – σ2 = n•p•q S – Bus Student n = 20 p = .2 q = .8 P20(5) = 20C5(.2)5(.8)15 = S – Female Student n = 20 p = .6 q = .4 P20(6) = 20C6(.6)6(.4)14 = Binomial Probability Table -

S – Business Major n = 10 p = .2 q = .8 P10(x = 2) = P10(x > 2) = 1 – P10(x≤2) = P10(x ≤ 2) = P(0) + P(1) + P(2) P10(4 ≤ x ≤ 7) = P(4) + P(5) + P(6) + P(7)

Poisson Distribution – QC Defect Rate Inventory Withdrawals Service Center Arrivals