Probability and Statistics brainiacs.

[sanantoniosooner]

I need some assistance.

Let me know if you are up to some long distance tutoring.

[Vaevictis]

I'm not a brainiac, but I'm willing to give it a shot. I know waaaay more about probability than I do about statistics though.

[Frozen Sooner]

How advanced of statistics are we talking here?

[Ike]

whatchuwanna know?

[sanantoniosooner]

This isn't my instructor, but it is the same course.

BTW, I'm a math major, and typically I'm good with math, but I carried a lot of baggage into this semester and I'm playing catchup. I'd like to do some email corresondance, and if you don't dig it, it's all right.

http://faculty.business.utsa.edu/vdeolive/STA3513/

[sooner_born_1960]

Out.

[Howzit]

I can give you the statistic on the number of people on my ignore list:

1

[olevetonahill]

planning on joining the professional gambler circuit huh ?

[Beef]

I always ran out of my statistics tests crying.

[yermom]

heh

that sounds like fun SAS, i was a math major for about a year :O

[crawfish]

There is a 70-40 chance I can help you out.

[Ike]

This isn't my instructor, but it is the same course.

BTW, I'm a math major, and typically I'm good with math, but I carried a lot of baggage into this semester and I'm playing catchup. I'd like to do some email corresondance, and if you don't dig it, it's all right.

http://faculty.business.utsa.edu/vdeolive/STA3513/

OK. so whats giving you trouble then?

[skycat]

Just use this thread to post your particular questions. I'm sure I don't remember everything you need to know, but I bet between all of the geekanerds on this board we could come up with something for most of your questions.

[Vaevictis]

Actually, I'm taking a class right now where the syllabus is significantly similar to yours. We might be able to actually help each other ;)

[sanantoniosooner]

In a certain region of the country it is known from past experience that the probability of selecting an adult over 40 year of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a pesron with cancer as having the disease is 0.78 and the probability of incorrectly diagnosing a person without cancer as having the disease is 0.06.

a-What is the probability that a person is diagnosed as having cancer?

b-What is the probability that a person diagnosed as having cancer actuall has the disease?

THERE IS NO POINT IN JUST POSTING ANSWERS. FOR ME TO UNDERSTAND THIS CRAP, THERE HAS TO BE SOME EXPLANATION

[Frozen Sooner]

OK. The probablility that someone has cancer is .05. So, if doctors have a unity chance of correctly diagnosing the disease, then the chance that someone will be diagnosed with cancer is .05.

However, doctors will only diagnose .78 of the people with the disease as having it. So there is only a .78*.05 probability that someone will both have the disease and be diagnosed with it.

For answer a, you need to recognize that doctors will both diagnose people with cancer who DO have cancer and some who don't.

To find the proportion of those who don't who are given a false positive, take the remainder of the population who don't have cancer (1-.05=.95) and multiply that by the chance of a false positive (.06). The answer to A is (.05)(.78)+(.95)(.06)

[Frozen Sooner]

Now, to find the chance that someone who is diagnosed actually has the disease...

Say that you have a population of 1000 people.

Of those 1000 people, 50 will have cancer, and .78 of those 50 (or 39) will be diagnosed with it.
Of those 1000 people, 60 will be given a false positive.

So 39 out of 99 will be given a true positive out of all positives, or roughly .394.

[Frozen Sooner]

I ****IN' RULE!

[MiccoMacey]

Do I have to start with MY explanations on math again? :D

[Frozen Sooner]

Mike>MiccoMacey

That's all the math you need, son.

;)

[Ike]

In a certain region of the country it is known from past experience that the probability of selecting an adult over 40 year of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a pesron with cancer as having the disease is 0.78 and the probability of incorrectly diagnosing a person without cancer as having the disease is 0.06.

a-What is the probability that a person is diagnosed as having cancer?

b-What is the probability that a person diagnosed as having cancer actuall has the disease?

THERE IS NO POINT IN JUST POSTING ANSWERS. FOR ME TO UNDERSTAND THIS CRAP, THERE HAS TO BE SOME EXPLANATION

I am assuming that by "person" they mean "person over 40"

The probablilty that any person would be diagnosed as having cancer would be P(diagnosed)=P(having cancer)*P(correct Diagnosis) + P(not having cancer)*P(false positive)

which is P(diagnosed)=0.05*0.78 + 0.95*0.06 = 0.096.
(in other words, 1 in 10 people would be diagnosed with cancer...which is double the number of people that actually have cancer.)

The probability then of being diagnosed and actually having cancer would then be the fraction of people that are correctly diagnosed as having cancer out of all the people that were diagnosed:
P(diagnosed and having cancer)=P(having cancer)*P(correct Diagnosis)/P(diagnosed)
P(diagnosed and having cancer)=0.05*0.78/0.096 = 0.40625

In other words, if you are diagnosed with cancer, There is a 40% chance that you actually have cancer.

[Frozen Sooner]

Hah! I beat Ike! And his answer has rounding errors!

[MiccoMacey]

Hah! I beat Ike! And his answer has rounding errors!

Is that some kind of weird math nerd smack? :D

[sanantoniosooner]

An undergraduate committee of 4 students is to be randomly selected from a group of 5 freshman, 4 sophomores 5 juniors and 1 senior.

a-What is the probability that the committee consists of all Sophomores?

b-What is the probability that the committee contains the Senior?

c-What is the probability that the committee consists of a person from each class?

[Ike]

Hah! I beat Ike! And his answer has rounding errors!

no...yours has addition errors :)

[OCUDad]

Here's a slightly different way to look at it.

For the sake of simplicity, assume the “certain region of the country” has 1000 people over 40. Actual numbers don’t matter because we’re talking percentages.

So, if .05 (5%) of the population has cancer, that’s 50 people with cancer and 950 without cancer.

Of the 50 with cancer, .78 (78%), or 39, will be correctly diagnosed as having cancer.

Of the 950 without cancer, .06 (6%), or 57, will be incorrectly diagnosed as having cancer.

That’s 39+57 people (96 people) diagnosed as having cancer.

The probability of a cancer diagnosis is therefore 96/1000, which is 9.6%

Of the 96 diagnoses of cancer, 39 are accurate, so the probability of an accurate diagnosis is 39/96, or 41%

[Frozen Sooner]

a. Your initial selection is 4 out of 15. Your next selection is 3 out of 14. Your next is 2 out of 13. Next is one out of 12. Probability of all four selections being Sophomores is 4/15*3/14*2/13*1/12.

b. Initial selection is 1 out of 15, then 1 out of 14, then 1 out of 13, then 1 out of 12. Probability of one selection being the senior is 1/15+1/14+1/13+1/12

[Frozen Sooner]

no...yours has addition errors :)

D'oh. You're correct. I forgot to remove the people with cancer from the population that could get a false positive.

[sooner_born_1960]

I didn't know they let you surf the innerweb while taking the ACT. This looks like cheating to me.

[Ike]

An undergraduate committee of 4 students is to be randomly selected from a group of 5 freshman, 4 sophomores 5 juniors and 1 senior.

a-What is the probability that the committee consists of all Sophomores?

b-What is the probability that the committee contains the Senior?

c-What is the probability that the committee consists of a person from each class?

so you just fill each slot accordingly. there are 5+4+5+1=15 students to choose from.

a) the answer is the probability that a soph is chosen for each seat...so for the first seat, thats 4/15. for the second seat, thats 3/14 (cause the first sophmore has to be gone from the pool). for the third, its 2/13 and for the 4th its 1/12....so the answer is (4*3*2*1)/(15*14*13*12) = really farkin small
(7.33*10^-4)

b) we calculate the probability that each seat gets filled by a senior and add them all up. 1/15 + 1/14 + 1/13 + 1/12 = 0.30 (roughly)

for the last part, its just like the second part, but the numerator changes.

[12]

I don't think I've ever seen a more fitting avatar.

[MiccoMacey]

Pretty simple stuff, actually.

If you can visualize, all numbers and powers of all finite and infinite vectors can be, at some point, both systematically as well as geometrically joined by their opposite counterpoint on a "linear" line graph.

To wit: (and this juxtaposition is where it makes it's money):

Any representation of a valid and joined system can and will be a dual host for both members in any equal or bilateral subset of either vector on a geographical plane.

Start there and work backwards. Let me know what you come up with.

[OCUDad]

Ike = Vizzini
MiccoMacey = Fezzik

[MiccoMacey]

It's inconceivable that I'm Fezzik.

[Jerk]

OK. The probablility that someone has cancer is .05. So, if doctors have a unity chance of correctly diagnosing the disease, then the chance that someone will be diagnosed with cancer is .05.

However, doctors will only diagnose .78 of the people with the disease as having it. So there is only a .78*.05 probability that someone will both have the disease and be diagnosed with it.

For answer a, you need to recognize that doctors will both diagnose people with cancer who DO have cancer and some who don't.

To find the proportion of those who don't who are given a false positive, take the remainder of the population who don't have cancer (1-.05=.95) and multiply that by the chance of a false positive (.06). The answer to A is (.05)(.78)+(.95)(.06)

Ok, just give me a minute here to figure this shiznit out...

[OCUDad]

It's inconceivable that I'm Fezzik.You keep using that word. I do not think it means what you think it means. :rolleyes:

[MiccoMacey]

Isn't that what he said all the time?

[MiccoMacey]

My bad...I had the wrong guy as Fezzik.

[MiccoMacey]

By the way...I am inconceivable. I had the "V" in June. :D

[OCUDad]

Vizzini was the "smart" one. Fezzik was the "big dumb" one.

That's OK, though. The smart one died. :D

[lil'duck]

I don't think H.O.U.S's really exist. (Horns of unusual size)

[Jerk]

I don't know what you mixed into that drink, but you need to tell me what it was...please.

(and can I still pass a urine test if I consume it? Muchos gracious (that's mexican for "thanks!"))

[OCUDad]

You couldn't pass a urine test if you studied for a week. :rolleyes:

[lil'duck]

vodkaX meets grapefruitY at an intersection going Zmiles per hour....

oh, you know the rest....a F1(Duck) meets F2(Sooner) and levels F3(Horn). :D

[MiccoMacey]

vodkaX meets grapefruitY at an intersection going Zimas per hour....

NTTAWWT

[Frozen Sooner]

Ok, just give me a minute here to figure this shiznit out...

It'd probably be easier if I did it right. But I forgot to remove the fifty people with cancer from the population that can get a false positive.

[sanantoniosooner]

Get your own thread simpletons.

[Al Gore]

Too complex for me....
Please wait while the page is being loaded. If this message is shown forever, the page did not load. So try again...

[GottaHavePride]

You keep using that word. I do not think it means what you think it means. :rolleyes:

Could it be that you are mad at something else, and looking to take it out on me?

http://www.garnersclassics.com/pics/amigos/jefe.jpg



*And yes, I know I switched movies. The lines are similar!

[OCUDad]

Can you get me one of those hats?

[Al Gore]

Too complex for me....
Please wait while the page is being loaded. If this message is shown forever, the page did not load. So try again...What the heck is this red message on my post??? Is it the man watching me????

[OKC-SLC]

If Carmen does not open her flower to me, I will kill her.