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Suppose that $0.3 \%$ of bolts made by a machine are defective, the defectives occurring at random during production. If the bolts are packaged in boxes of 100 , what is the Poisson approximation that a given box will contain $x$ defectives? Suppose you buy 8 boxes of bolts. What is the distribution of the number of boxes with no defective bolts? What is the expected number of boxes with no defective bolts?
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$D$ is $\operatorname{Bin}(100,0.003)$ which is approximately Poisson with parameter $100 \times 0.003=0.3 .$ Hence $P(D=$ $x) \approx e^{-0.3}(0.3)^{x} / x !, x=0,1, \ldots$ In particular, $P(D=0) \approx 0.7408 .$ Finally, $N$, the no. of boxes with no defectives is $\operatorname{Bin}(8,0.7408)$ and so $E(N)=8 \times 0.7408=5.926$.
by Diamond (74,866 points)

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