Thread: year 12 prob assignment

1. year 12 prob assignment

cant get my head around some of these problems..excuse me if they are confusing.

A company manufactures bolts which are packed in batches of 10,000. The manufacturer operates a simple sampling scheme whereby a random sample of 10 is taken from each batch. If the manufacturer finds that there are fewer than 3 faulty bolts the batch is allowed to be shipped out. Otherwise, the whole batch is rejected and reprocessed

Show that if 100p% of all bolts produced are known to be defective, then p(batch is accepted) = (1-p)^8(1 + 8 p + 36 p ^ 2), 0 <(or equal to) p <(or equal to) 1

ANSWERED! Thank you Mr. Fantastic =D

Using a graphical calculator, sketch the graph of p('batch is accepted') versus p.

Would I just enter the above quadratic in graph mode of a casio graphics calculator? Or...?

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The quality control for the manufactoring of screws is carried out by taking a random sample of 15 screws from a batch of 10000. Empirical data shows that 10% of screws are defective. If it is found that there are three or more defective screws in the sample, that particular batch is rejected.

What is the probability that the manufacturer rejects a batch?

It costs the manufacturer $20.00 to process a batch of 10000 screws. Each batch is sold for$38.00, otherwise it is sold for scrap at $5.00 per batch. What profit can the manufacturer expect to make on each batch? The answer to a) is 0.1841 and b)$9.45

Theres one more, but I think once I read your responses to these 2, I can answer the last one myself. I use a Casio Graphics calculator, and we have used the normal distribution and binomial functions. Thanks for any help at all.

EDIT: a) is done, but I still do not get b)

2. Originally Posted by proddy
[snip]
A company manufactures bolts which are packed in batches of 10,000. The manufacturer operates a simple sampling scheme whereby a random sample of 10 is taken from each batch. If the manufacturer finds that there are fewer than 3 faulty bolts the batch is allowed to be shipped out. Otherwise, the whole batch is rejected and reprocessed

Show that if 100p% of all bolts produced are known to be defective, then p(batch is accepted) = (1-p)^8(1 + 8 p + 36 p ^ 2), 0 <(or equal to) p <(or equal to) 1

Using a graphical calculator, sketch the graph of p('batch is accepted') versus p.

Would I just enter the above quadratic in graph mode of a casio graphics calculator? Mr F says: Yes.
Or...?
[snip]
Let X be the random variable number of defective bolts in batch.

Then X ~ Binomial(n = 10, p = p)

Pr(Accept batch) = Pr(X < 3) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2).

$\displaystyle \Pr(X = 0) = {10 \choose 0} p^0 (1 - p)^{10 - 0} = (1 - p)^{10}$.

$\displaystyle \Pr(X = 1) = {10 \choose 1} p^1 (1 - p)^{10 - 1} = 10 p (1 - p)^9$.

$\displaystyle \Pr(X = 2) = {10 \choose 2} p^2 (1 - p)^{10 - 2} = 45 p^2 (1 - p)^8$.

Therefore Pr(Accept batch) = $\displaystyle (1 - p)^{10} + 10 p (1 - p)^9 + 45 p^2 (1 - p)^8 = (1 - p)^8 [(1 - p)^2 + 10 p (1 - p) + 45 p^2] = ....$.

3. Originally Posted by proddy
[snip]
The quality control for the manufactoring of screws is carried out by taking a random sample of 15 screws from a batch of 10000. Empirical data shows that 10% of screws are defective. If it is found that there are three or more defective screws in the sample, that particular batch is rejected.

What is the probability that the manufacturer rejects a batch?

Mr F says: Follow the previous question, but use n = 15 and p = 0.1, to get Pr(Accept). The Pr(Reject) = 1 - Pr(Accept).

It costs the manufacturer $20.00 to process a batch of 10000 screws. Each batch is sold for$38.00, otherwise it is sold for scrap at $5.00 per batch. What profit can the manufacturer expect to make on each batch? The answer to a) is 0.1841 and b)$9.45

[snip]
Pr(Reject) = 0.1841. Pr(Accept) = 0.8159.

Expected profit = (0.1841)(5.00 - 20.00) + (0.8159)(38.00 - 20) = $11.92. So either the given answer to (a) (which I've used) is wrong, or the answer to (b) is wrong. 4. Originally Posted by mr fantastic Pr(Reject) = 0.1841. Pr(Accept) = 0.8159. Expected profit = (0.1841)(5.00 - 20.00) + (0.8159)(38.00 - 20) =$11.92.

So either the given answer to (a) (which I've used) is wrong, or the answer to (b) is wrong.
The given answer to (a) is correct. So the given answer to (b) is wrong ..... It should be \$11.92.

5. Yeah, I've noticed quite a few questions have the given answer wrong. Its very confusing, because I try to redo the question every other way I can think of, and I get the same result. I'll have to talk with my teacher.

Thanks so much Mr Fantastic.