A manufacturer produces fuses. The percentage of non defective fuses is 0.96. A sample of 100 fuses was selected. Use normal approximation to binomial distribution to find the probability of selecting more than 6 defective fuses
X ~ Bin(100, 0.04). According to the normal approximation, X ~ N(100*0.04, 100*0.04*0.96)=N(4, 3.84)
"P(X>6)=P(N(4,3.84)>6)=P(4+1.96N(0,1)>6)=P(N(0,1)>1.02)=1-P(N(0,1)<1.02)=1-0.84614=0.15386"
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