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xEnOnn

  • 2 years ago

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?

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

  2. xEnOnn
    • 2 years ago
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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.

  3. Zarkon
    • 2 years ago
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    looks binomial

  4. Chlorophyll
    • 2 years ago
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    @Zarkon, also seems like hypergeomtric!

  5. Zarkon
    • 2 years ago
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    but that part doesn't give you any information

  6. xEnOnn
    • 2 years ago
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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.

  7. xEnOnn
    • 2 years ago
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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?

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

  9. Chlorophyll
    • 2 years ago
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    Of course without replacement!

  10. Chlorophyll
    • 2 years ago
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    20% 400 = 80

  11. SmoothMath
    • 2 years ago
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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)

  12. Zarkon
    • 2 years ago
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    you could use a normal approximation

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

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

  15. xEnOnn
    • 2 years ago
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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.

  16. xEnOnn
    • 2 years ago
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    haha...normal approximation today...

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

  18. Zarkon
    • 2 years ago
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    brb

  19. SmoothMath
    • 2 years ago
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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.

  20. SmoothMath
    • 2 years ago
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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.

  21. xEnOnn
    • 2 years ago
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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.

  22. SmoothMath
    • 2 years ago
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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.

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

  24. Zarkon
    • 2 years ago
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    all you need is independence between boxes

  25. xEnOnn
    • 2 years ago
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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.

  26. xEnOnn
    • 2 years ago
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    Thanks a lot!! :D

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

  28. xEnOnn
    • 2 years ago
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    independent*

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

  30. SmoothMath
    • 2 years ago
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    |dw:1335139433649:dw|

  31. xEnOnn
    • 2 years ago
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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.

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

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

  34. xEnOnn
    • 2 years ago
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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.

  35. xEnOnn
    • 2 years ago
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    i mean 89.5

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

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

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