A firm employing 100 workers has an average absenteeism rate of 4%. On a given day, what is the probability of
(a) no workers,
(b) one worker,
(c) more than six workers being absent?
Given X~Binomial(n=100, p=0.04)
P(X=x)=xC100*(0.04^x)*(0.96^(100-x))
Answer:
(a) no workers,
P(X=0)=0.96^100=0.01687032
(b) one worker,
P(X=1)=1C100*(0.04^1)*(0.96^(100-1))= 0.070293
(c) more than six workers being absent?
P(X>6)=1-P(X=0)-..-P(X=6)=0.1063923
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