Mean & Variance for the Binomial Distribution

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Presentation transcript:

Mean & Variance for the Binomial Distribution We can calculate the probability distribution for a binomial situation If we look at tossing a coin 4 times and finding the number of heads. We have n=4 π = 0.5 so X~B(4,0.5) We therefore have the probability distribution values from tables μ=E(X)=0x0.0625+1x0.2500+2x0.3750+3x0.2500+4x0.0625 = 2 =4x0.5 =nπ x 1 2 3 4 P(X=x) 0.0625 0.2500 0.3750

σ2 = E(X2)- μ2 =(02x0.0625+12x0.25+22x0.375+32x0.25+42x0.0625)-4 = 5-4 = 1 =4x0.5x0.5 = n π(1- π) We could repeat with other cases to show that μ= n π and σ2 = n π(1- π) σ = √nπ(1- π) eg A machine produces goods which have a 1% chance of being faulty. Find the mean number of faulty goods per batch of 1000, and the standard deviation in the number of faulty goods. X~B(1000,0.01) μ= n π = 1000x0.01 =10 σ = √nπ(1- π) = √1000x0.01x0.99 = √9.9 = 3.146