Question

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

Answer 1

X ~ Bin(100, 0.04). According to the normal approximation, X ~ N(100*0.04, 100*0.04*0.96)=N(4, 3.84)

$P(X>6)=P(N(4,3.84)>6)=P(4+1.96N(0,1)>6)=P(N(0,1)>1.02)=1-P(N(0,1)<1.02)=1-0.84614=0.15386$