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?

- anonymous

- katieb

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- anonymous

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

- 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

looks binomial

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- anonymous

@Zarkon, also seems like hypergeomtric!

- Zarkon

but that part doesn't give you any information

- 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

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

each box is being tested...the box is either good or bad...the probability it is bad is 20%

- anonymous

Of course without replacement!

- anonymous

20% 400 = 80

- 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

you could use a normal approximation

- Zarkon

I'm all about the normal approximation today ;)

- anonymous

Tell me more, Zarkon. How could we determine the SD?

- 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

haha...normal approximation today...

- 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

brb

- 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

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

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

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

you don't...you are looking at all the boxes...and checking each one once.

- Zarkon

all you need is independence between boxes

- anonymous

ohhhh yea you are right! I was looking at the units! It is the units of components that are only randomly picked.

- anonymous

Thanks a lot!! :D

- anonymous

oh yes zarkon, in this case, they are all indepent. Thanks!! :D

- anonymous

independent*

- anonymous

Mean: 80
SD: sqrt(400*0.2*(.8)) = 8
90 is a z-score of 1.25

- anonymous

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- 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

For a slightly better answer one should use a continuity correction. again...i feel like a broken record ;)

- Zarkon

most texts want n(1-p) and np to be larger than 5

- 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

i mean 89.5

- Zarkon

again this is a CLT result...since the binomial distribution is the sum of independent Bernoulli random variables

- anonymous

yea...the CLT again like the previous one. haha... thanks so much!

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