ecmt1010 ..QUiZ next week? (1 Viewer)

[sarevok]

zzzzzzzzzzzzz

 

[stazi]

im tackling 11, after i finish dinner

 

[Komit]

Hallelujah!!! I got one right. Thanks sarevok - you've been a big help (fuck q11). 1time4thppl - thanks also. Let us pray that there is no probability tomorrow and that the monday tutes cop it.

 

[Lainee]

any one feel up to tackling question 11?

i understand that the prior probability is just p(a), p(a) being the probability a defect occurs. using the revised question, drawing a tree diagram:

Machine in good condition P(0.95) ---> Defect P(0.02)
---> Normal P(0.98)

Machine in bad condition P(0.05) ---> Defect P(0.15)
---> Normal P(0.75)

So, the probability of p(a) is 0.95 * 0.02 + 0.05 * 0.15 = 53/2000

can any one confirm if this is right?

That's what I actually did for 11(a) because I figured that that wasn't using Bayes' Theorem like they suggested... but now I'm stuck for 11(b), so... you might be right after all.

 

[stazi]

wait for 11a. Didnt we get a really similar problem on the overheads put up? ALso about factory defects

 

[s.m.i.t.h]

is al the stuff we need to know just in these practise questions?
i.e if it isnt covered then we dont have to learn it for the quiz i.e geo mean?

 

[sarevok]

zzzzzzzzzzzz

 

[stazi]

FOUND IT!!
Suppose we're interested in the condition of a machine that produces a particular item. From experience, it is known that the machine is in good condition 90% of the time
When the machine is in good condition, 1 % of the time an item will be defective.
When the machine is bad, 10% of the time the item will be defective.
One item is randomly selected, checked and found to be defective.

With this info what's the probability the machine is in good condition?

However 11a asks what does the manufacturer believe the prior probability of a defect is?
I really don't get that q

 

[Lainee]

...I think Q11 is still written incorrectly. Like the example in the textbook 154-155.

 

[stazi]

any one feel up to tackling question 11?

i understand that the prior probability is just p(a), p(a) being the probability a defect occurs. using the revised question, drawing a tree diagram:

Machine in good condition P(0.95) ---> Defect P(0.02)
---> Normal P(0.98)

Machine in bad condition P(0.05) ---> Defect P(0.15)
---> Normal P(0.75)

So, the probability of p(a) is 0.95 * 0.02 + 0.05 * 0.15 = 53/2000

can any one confirm if this is right?

Can u explain what the Normal P is ?

 

[sarevok]

by normal i mean not defective =)

 

[stazi]

ok using the example on the overhead but relating it to 11a.
P(S-atisfactory)=0.95 P(D-efect|S)=0.02
P(S')=0.05 P(D|S')=0.15

We want ??? can someone help me out here.
In the example for whether its in good condition was P(S|D), P(S'|D)

 

[sarevok]

i think lainee is right, the question must still be written incorrectly.

because what does 'revise the probability of defects' mean? using bayes theorem, we can work out, if we get a defect, the probability the machine is in good condition or bad condition, but nothing else as far as i can see.

 

[Lainee]

Would anyone have Tig's email on hand?

 

[stazi]

k ive officially given up on 11. Even 11a. Not sure where all the numbers are coming from. I got an answer different to the one posted here. I tabulated it as per the overhead, yet the one here was strange.
Oh well, on to 12 hope that's easy.
I'm just going to have to go in tomorrow hoping q11 won't be in there

 

[stazi]

can someone tell me if i have these right?
a) 0.5
b) 0.72
c) 0.5
d) 0.225
and could someone please run me through 12e again. The formula and how I would expand it to fit this data?

 

[sarevok]

What's the answer to the overhead question about Bayes' Theorem?

 

[stazi]

sarevok; why are you deleting all youve said?

 

[sarevok]

i figure it's not needed anymore and i don't really want randoms coming through and taking my answers.

 

[Sarah168]

Would anyone have Tig's email on hand?

http://www.econ.usyd.edu.au/content.php?pageid=970