Question

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?

Answer 1

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

Answer 2

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.

Answer 3

looks binomial

Answer 4

@Zarkon, also seems like hypergeomtric!

Answer 5

but that part doesn't give you any information

Answer 6

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.

Answer 7

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?

Answer 8

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

Answer 9

Of course without replacement!

Answer 10

20% 400 = 80

Answer 11

...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)

Answer 12

you could use a normal approximation

Answer 13

I'm all about the normal approximation today ;)

Answer 14

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

Answer 15

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.

Answer 16

haha...normal approximation today...

Answer 17

if it is binomial with n trials and p prob of success...then the sd is

\[\sqrt{n\cdot p\cdot (1-p)}\]

Answer 18

brb

Answer 19

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.

Answer 20

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.

Answer 21

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.

Answer 22

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

Answer 23

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

Answer 24

all you need is independence between boxes

Answer 25

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

Answer 26

Thanks a lot!! :D

Answer 27

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

Answer 28

independent*

Answer 29

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