Youngster
  • Youngster
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?
Statistics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
99
anonymous
  • anonymous
wait no
anonymous
  • anonymous
999

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
yeah
Youngster
  • Youngster
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
anonymous
  • anonymous
\[P(correct)=p=0.998\]\[P(incorrect)=q=0.002\] We want to find an n so that \[\binom{n}{0}(0.998)^n(0.002)^0<\binom{n}{1}(0.998)^{n-1}(0.002)^1\]
Youngster
  • Youngster
wait, what does\[\left(\begin{matrix}n \\ 0\end{matrix}\right)\] mean?
anonymous
  • anonymous
It means nC0 \[\binom{n}{0}=1,\binom{n}{1}=1\] \[0.998^n
anonymous
  • anonymous
\[\binom{a}{b}=aCb\]
Youngster
  • Youngster
oh, that makes sense. thanks so much! im going to post another question in a couple of minutes, can u help me with that?
Youngster
  • Youngster
wait, actually ur supposed to use expected value for binormial... i think you get the same answer, but its less confusing. Thanks anyways
kropot72
  • kropot72
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 \[n=\frac{1}{0.002}=?\]
Youngster
  • Youngster
@kropot72 500.. so that's the answer?
kropot72
  • kropot72
@Youngster Yes, that's the answer.
Youngster
  • Youngster
thanks!!
kropot72
  • kropot72
You're welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.