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Youngster Group Title

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
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  1. koalamon Group Title
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    99

    • one year ago
  2. koalamon Group Title
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    wait no

    • one year ago
  3. koalamon Group Title
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    999

    • one year ago
  4. koalamon Group Title
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    yeah

    • one year ago
  5. Youngster Group Title
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    i don't think it is that. You have to use the equation\[P(n=r)=nCr*q^{r-1}*p^{1}\] It is a binomial model

    • one year ago
  6. henpen Group Title
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    \[P(correct)=p=0.998\]\[P(incorrect)=q=0.002\] We want to find an n so that \[\binom{n}{0}(0.998)^n(0.002)^0<\binom{n}{1}(0.998)^{n-1}(0.002)^1\]

    • one year ago
  7. Youngster Group Title
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    wait, what does\[\left(\begin{matrix}n \\ 0\end{matrix}\right)\] mean?

    • one year ago
  8. henpen Group Title
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    It means nC0 \[\binom{n}{0}=1,\binom{n}{1}=1\] \[0.998^n<n0.998^{n-1}0.002\] \[\frac{0.998}{0.002}<n\] =499

    • one year ago
  9. henpen Group Title
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    \[\binom{a}{b}=aCb\]

    • one year ago
  10. Youngster Group Title
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    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
  11. Youngster Group Title
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    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
  12. kropot72 Group Title
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    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
  13. Youngster Group Title
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    @kropot72 500.. so that's the answer?

    • one year ago
  14. kropot72 Group Title
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    @Youngster Yes, that's the answer.

    • one year ago
  15. Youngster Group Title
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    thanks!!

    • one year ago
  16. kropot72 Group Title
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    You're welcome :)

    • one year ago
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