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Youngster
 3 years ago
Wegman claims that their checkout scanners scan correctly 99.8% of the items sold. How many would you expect to buy, on average, to find one that scans incorrectly?
Youngster
 3 years ago
Wegman claims that their checkout scanners scan correctly 99.8% of the items sold. How many would you expect to buy, on average, to find one that scans incorrectly?

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Youngster
 3 years ago
Best ResponseYou've already chosen the best response.0i don't think it is that. You have to use the equation\[P(n=r)=nCr*q^{r1}*p^{1}\] It is a binomial model

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0\[P(correct)=p=0.998\]\[P(incorrect)=q=0.002\] We want to find an n so that \[\binom{n}{0}(0.998)^n(0.002)^0<\binom{n}{1}(0.998)^{n1}(0.002)^1\]

Youngster
 3 years ago
Best ResponseYou've already chosen the best response.0wait, what does\[\left(\begin{matrix}n \\ 0\end{matrix}\right)\] mean?

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0It means nC0 \[\binom{n}{0}=1,\binom{n}{1}=1\] \[0.998^n<n0.998^{n1}0.002\] \[\frac{0.998}{0.002}<n\] =499

Youngster
 3 years ago
Best ResponseYou've already chosen the best response.0oh, that makes sense. thanks so much! im going to post another question in a couple of minutes, can u help me with that?

Youngster
 3 years ago
Best ResponseYou've already chosen the best response.0wait, actually ur supposed to use expected value for binormial... i think you get the same answer, but its less confusing. Thanks anyways

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.1Probability of an incorrect scan = 1.000  0.998 = 0.002 Let the expected quantity required to get one incorrect scan = n E(X) = 1 = 0.002n \[n=\frac{1}{0.002}=?\]

Youngster
 3 years ago
Best ResponseYou've already chosen the best response.0@kropot72 500.. so that's the answer?

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.1@Youngster Yes, that's the answer.
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