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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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YoungsterBest ResponseYou've already chosen the best response.0
i 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
 one year ago

henpenBest 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 \[\binom{n}{0}(0.998)^n(0.002)^0<\binom{n}{1}(0.998)^{n1}(0.002)^1\]
 one year ago

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

henpenBest ResponseYou've already chosen the best response.0
It 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
 one year ago

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

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

kropot72Best ResponseYou've already chosen the best response.1
Probability 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 \[n=\frac{1}{0.002}=?\]
 one year ago

YoungsterBest ResponseYou've already chosen the best response.0
@kropot72 500.. so that's the answer?
 one year ago

kropot72Best ResponseYou've already chosen the best response.1
@Youngster Yes, that's the answer.
 one year ago
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