A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 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?
anonymous
 4 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?

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is there enough information given in this problem? My initial impression is no.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Zarkon, also seems like hypergeomtric!

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2but that part doesn't give you any information

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My 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
 4 years ago
Best ResponseYou've already chosen the best response.0hmm..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
 4 years ago
Best ResponseYou've already chosen the best response.2each box is being tested...the box is either good or bad...the probability it is bad is 20%

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Of course without replacement!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0...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
 4 years ago
Best ResponseYou've already chosen the best response.2you could use a normal approximation

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2I'm all about the normal approximation today ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Tell me more, Zarkon. How could we determine the SD?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Could 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
 4 years ago
Best ResponseYou've already chosen the best response.0haha...normal approximation today...

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2if it is binomial with n trials and p prob of success...then the sd is \[\sqrt{n\cdot p\cdot (1p)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh 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
 4 years ago
Best ResponseYou've already chosen the best response.0Normal 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
 4 years ago
Best ResponseYou've already chosen the best response.0This 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
 4 years ago
Best ResponseYou've already chosen the best response.0I don't think so, EnOnn. It says that they simply test all 400 boxes. There's no chance of testing a box twice.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2you don't...you are looking at all the boxes...and checking each one once.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2all you need is independence between boxes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhhh yea you are right! I was looking at the units! It is the units of components that are only randomly picked.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yes zarkon, in this case, they are all indepent. Thanks!! :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mean: 80 SD: sqrt(400*0.2*(.8)) = 8 90 is a zscore of 1.25

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1335139433649:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea....thanks! one more condition which makes this approximation possible is also that n(1p) and np are both greater than 15. Since this condition is met, the approximation is pretty close.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2For a slightly better answer one should use a continuity correction. again...i feel like a broken record ;)

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2most texts want n(1p) and np to be larger than 5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh yea you are right. since i want 90 to be in as well, the value for the xbar in the zscore would be 85.5.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2again this is a CLT result...since the binomial distribution is the sum of independent Bernoulli random variables

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea...the CLT again like the previous one. haha... thanks so much!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.