anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Is there enough information given in this problem? My initial impression is no.
anonymous
  • anonymous
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.
Zarkon
  • Zarkon
looks binomial

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anonymous
  • anonymous
@Zarkon, also seems like hypergeomtric!
Zarkon
  • Zarkon
but that part doesn't give you any information
anonymous
  • anonymous
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.
anonymous
  • anonymous
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?
Zarkon
  • Zarkon
each box is being tested...the box is either good or bad...the probability it is bad is 20%
anonymous
  • anonymous
Of course without replacement!
anonymous
  • anonymous
20% 400 = 80
anonymous
  • anonymous
...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)
Zarkon
  • Zarkon
you could use a normal approximation
Zarkon
  • Zarkon
I'm all about the normal approximation today ;)
anonymous
  • anonymous
Tell me more, Zarkon. How could we determine the SD?
anonymous
  • anonymous
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.
anonymous
  • anonymous
haha...normal approximation today...
Zarkon
  • Zarkon
if it is binomial with n trials and p prob of success...then the sd is \[\sqrt{n\cdot p\cdot (1-p)}\]
Zarkon
  • Zarkon
brb
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
I don't think so, EnOnn. It says that they simply test all 400 boxes. There's no chance of testing a box twice.
Zarkon
  • Zarkon
you don't...you are looking at all the boxes...and checking each one once.
Zarkon
  • Zarkon
all you need is independence between boxes
anonymous
  • anonymous
ohhhh yea you are right! I was looking at the units! It is the units of components that are only randomly picked.
anonymous
  • anonymous
Thanks a lot!! :D
anonymous
  • anonymous
oh yes zarkon, in this case, they are all indepent. Thanks!! :D
anonymous
  • anonymous
independent*
anonymous
  • anonymous
Mean: 80 SD: sqrt(400*0.2*(.8)) = 8 90 is a z-score of 1.25
anonymous
  • anonymous
|dw:1335139433649:dw|
anonymous
  • anonymous
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.
Zarkon
  • Zarkon
For a slightly better answer one should use a continuity correction. again...i feel like a broken record ;)
Zarkon
  • Zarkon
most texts want n(1-p) and np to be larger than 5
anonymous
  • anonymous
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.
anonymous
  • anonymous
i mean 89.5
Zarkon
  • Zarkon
again this is a CLT result...since the binomial distribution is the sum of independent Bernoulli random variables
anonymous
  • anonymous
yea...the CLT again like the previous one. haha... thanks so much!

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