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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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Is there enough information given in this problem? My initial impression is no.
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.
looks binomial

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Other answers:

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

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