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xEnOnn

Each box is packed with 40 units of components. An inspector will randomly select 5 units from a box to test if the components are in good condition. A box will be rejected if 2 or more tested units are defective. A shipment includes 400 boxes of the components. Each box will be tested by the inspector. The whole shipment will be returned if more than 90 boxes of components are rejected. The ordering records in the past showed that 20% of the boxes were rejected. What is the probability that a given shipment will be returned?

  • one year ago
  • one year ago

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  1. SmoothMath
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    Is there enough information given in this problem? My initial impression is no.

    • one year ago
  2. xEnOnn
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    But this seems like all the question has for me to answer it. And because this is an exam question, I think there should have enough information provided to solve the problem.

    • one year ago
  3. Zarkon
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    looks binomial

    • one year ago
  4. Chlorophyll
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    @Zarkon, also seems like hypergeomtric!

    • one year ago
  5. Zarkon
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    but that part doesn't give you any information

    • one year ago
  6. xEnOnn
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    My initial intuition was binomial too. But if this is binomial, it means that the selection of the boxes for the inspection has replacement, which I don't think it has.

    • one year ago
  7. xEnOnn
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    hmm..but wouldn't there be a chance that one of the selected boxes for inspection is a repeated box? ie, a box that has already been inspected earlier?

    • one year ago
  8. Zarkon
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    each box is being tested...the box is either good or bad...the probability it is bad is 20%

    • one year ago
  9. Chlorophyll
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    Of course without replacement!

    • one year ago
  10. Chlorophyll
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    20% 400 = 80

    • one year ago
  11. SmoothMath
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    ...Doing it as binomial seems like a serious pain. You'd have to add up P(89) + P(88)+ P(87) +... + P(2) + P(1)

    • one year ago
  12. Zarkon
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    you could use a normal approximation

    • one year ago
  13. Zarkon
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    I'm all about the normal approximation today ;)

    • one year ago
  14. SmoothMath
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    Tell me more, Zarkon. How could we determine the SD?

    • one year ago
  15. xEnOnn
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    Could the numbers be used as a hypergeometric case? I was trying to figure out the numbers from this question for the the hypergeometric formula's parameters but doesn't look possible.

    • one year ago
  16. xEnOnn
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    haha...normal approximation today...

    • one year ago
  17. Zarkon
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    if it is binomial with n trials and p prob of success...then the sd is \[\sqrt{n\cdot p\cdot (1-p)}\]

    • one year ago
  18. Zarkon
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    brb

    • one year ago
  19. SmoothMath
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    Oh really? That's pretty cool. That does it for you then. You have the mean will clearly be 80. That SD gives you a z score for 90. Problem's doneso.

    • one year ago
  20. SmoothMath
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    Normal approximation was the first thing I thought of, but I figured you needed a given SD. It makes sense that you can calculate it though.

    • one year ago
  21. xEnOnn
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    This is assuming that there is replacement. If there is replacement, it will follow binomial distribution. However, there is a possibility that we picked out another box which we had tested earlier.

    • one year ago
  22. SmoothMath
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    I don't think so, EnOnn. It says that they simply test all 400 boxes. There's no chance of testing a box twice.

    • one year ago
  23. Zarkon
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    you don't...you are looking at all the boxes...and checking each one once.

    • one year ago
  24. Zarkon
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    all you need is independence between boxes

    • one year ago
  25. xEnOnn
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    ohhhh yea you are right! I was looking at the units! It is the units of components that are only randomly picked.

    • one year ago
  26. xEnOnn
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    Thanks a lot!! :D

    • one year ago
  27. xEnOnn
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    oh yes zarkon, in this case, they are all indepent. Thanks!! :D

    • one year ago
  28. xEnOnn
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    independent*

    • one year ago
  29. SmoothMath
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    Mean: 80 SD: sqrt(400*0.2*(.8)) = 8 90 is a z-score of 1.25

    • one year ago
  30. SmoothMath
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    |dw:1335139433649:dw|

    • one year ago
  31. xEnOnn
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    yea....thanks! one more condition which makes this approximation possible is also that n(1-p) and np are both greater than 15. Since this condition is met, the approximation is pretty close.

    • one year ago
  32. Zarkon
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    For a slightly better answer one should use a continuity correction. again...i feel like a broken record ;)

    • one year ago
  33. Zarkon
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    most texts want n(1-p) and np to be larger than 5

    • one year ago
  34. xEnOnn
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    oh yea you are right. since i want 90 to be in as well, the value for the x-bar in the z-score would be 85.5.

    • one year ago
  35. xEnOnn
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    i mean 89.5

    • one year ago
  36. Zarkon
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    again this is a CLT result...since the binomial distribution is the sum of independent Bernoulli random variables

    • one year ago
  37. xEnOnn
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    yea...the CLT again like the previous one. haha... thanks so much!

    • one year ago
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