A company manufactures fuses. The percentage of non-defective fuses is 95.4%. A sample of 9 fuse was selected. Calculate the probability of selecting at least 3 defective fuses.
The Poisson distribution is
P(X=x)=e−λλxx!P\left(X=x\right)=\frac{e^{-\lambda }\lambda ^x}{x!}P(X=x)=x!e−λλx
Let xxx denote the defective fuse
P=100%−95.4%=4.6%=0.046P=100\%-95.4\%=4.6\%=0.046P=100%−95.4%=4.6%=0.046
n=9n=9n=9
mean=λ=np=9×0.046=0.414mean=\lambda =np=9\times 0.046=0.414mean=λ=np=9×0.046=0.414
P(at least 3 defective fuses)=∑x≥3e−0.4140.41433=0.0156P\left(at\,least\:3\:defective\:fuses\right)=\sum_{x \ge 3} \frac{e^{-0.414}0.414^3}{3}=0.0156P(atleast3defectivefuses)=∑x≥33e−0.4140.4143=0.0156
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