Elementary Statistics: Could Use Some Help

[HappyDude20]

Googling Math Lab or Math Help only resulted in drug addiction sites; aside, realized a Math community with a thriving group of people doesn't really exist.

I have my 2nd math exam this coming Thursday. In fact, the exam was today, however figured i'd use my "being held up @ gun-point" story to my advantage and the professor gave me the extra 2 days. (I told him I could take the exam, just that admittedly throughout the weekend the incident was more on my mind than studying for math)

Anyways..as I said, the exam is on Thursday so may come back here with a question or two asking for some help.

Essentially with me, if I know how to do a math problem, i'll be golden and be able to move on.

My 1st exam a month ago I got 100%, however since getting that perfect score got a bit cocky and didn't really open the book (Elemen. Stats by Triola 10th Ed.)

Here's the one current question on my mind I can't solve:

Ch. 4.4 Probability - Multiplication Rule

#12: It is common for public opinion polls to have a "confidence level" of 95%, meaning that there is a 0.95 probability that the poll results are accurate within the claimed margins of error. If six different organizations conduct independent polls, what is the probability that all six of them are accurate within the claimed margins of error? Does the result suggest that with a confidence level of 95%, we can expect that almost all polls will be within the claimed margin of error?



Btw, I'm using a TI-83.

I'm not really asking for you to solve the problem for me...I just wanna understand it. Hell, yes, gimme the answer too; but I wanna know how to get there.

 

[RedTomato]

Eh its been a long time since I did maths, but that seems quite an easy question. *No calculator needed!*

As you know, the art with wordy questions like that is to strip out the verbiage and reduce it to something simple to grasp.

The first sentence seems to mean just:

'x has a .95 probability of being right.'

It doesn't seem to matter what the margins of error are or anything like that.

So the first question is just 'if x is done 6 times in a row, what is the probability of x being right every time?'

There's a huge clue in the title of the question: "Probability - Multiplication Rule."

This isn't hard at all - I was doing this stuff when I was 14. Maybe I missed something in the question.

 

[Rodimus Prime]

Eh its been a long time since I did maths, but that seems quite an easy question. *No calculator needed!*

As you know, the art with wordy questions like that is to strip out the verbiage and reduce it to something simple to grasp.

The first sentence seems to mean just:

'x has a .95 probability of being right.'

It doesn't seem to matter what the margins of error are or anything like that.

So the first question is just 'if x is done 6 times in a row, what is the probability of x being right every time?'

There's a huge clue in the title of the question: "Probability - Multiplication Rule."

This isn't hard at all - I was doing this stuff when I was 14. Maybe I missed something in the question.

No the 95% probability for being right is not what you explain. It has something to do with the standard deviation I believe.

It has been a long time since I took statistics so I do not know exactly what it means but I know it is not what you explain. I know it has something to do with error correcting due to sample sizes and deviation and what not. I remember that being in some of the formals I used in that class. But that is higher level.

As for the answer to the question my best guess is 19^6/20^6 but it might be even different than that. Like I said it has been a long time since I did that class.

 

[OutThere]

Googling Math Lab or Math Help only resulted in drug addiction sites; aside, realized a Math community with a thriving group of people doesn't really exist.

I have my 2nd math exam this coming Thursday. In fact, the exam was today, however figured i'd use my "being held up @ gun-point" story to my advantage and the professor gave me the extra 2 days. (I told him I could take the exam, just that admittedly throughout the weekend the incident was more on my mind than studying for math)

Anyways..as I said, the exam is on Thursday so may come back here with a question or two asking for some help.

Essentially with me, if I know how to do a math problem, i'll be golden and be able to move on.

My 1st exam a month ago I got 100%, however since getting that perfect score got a bit cocky and didn't really open the book (Elemen. Stats by Triola 10th Ed.)

Here's the one current question on my mind I can't solve:

Ch. 4.4 Probability - Multiplication Rule

#12: It is common for public opinion polls to have a "confidence level" of 95%, meaning that there is a 0.95 probability that the poll results are accurate within the claimed margins of error. If six different organizations conduct independent polls, what is the probability that all six of them are accurate within the claimed margins of error? Does the result suggest that with a confidence level of 95%, we can expect that almost all polls will be within the claimed margin of error?



Btw, I'm using a TI-83.

I'm not really asking for you to solve the problem for me...I just wanna understand it. Hell, yes, gimme the answer too; but I wanna know how to get there.

The multiplication rule of probability tells us that the probability of two independent events occurring together is the product of the probability of each event occurring individually, multiplied together.

So the probability of rolling a 3 and then a 5 on a 6 sided die is (1/6)*(1/6). For this it would just mean .95^6, or a .735 probability that 6 independent polls with 95% confidence intervals would all have the answer pegged within the claimed margin of error.

 

[mstrze]

Aren't we looking for the total probability of the polls being correct if each independently has a 95% probability of being correct?

I seem to recall the key to something like this is to take the probility of each being wrong and then multiply it by each additional probability of the others being wrong. Then the actual chance of all polls being accurate would be 100% minus the percentage you came up with of them NOT being accurate.

EDIT: I think what I did was find the percentage that ANY of the 6 will be accurate, not ALL of them...so disregard.

 

[HappyDude20]

Thanks for the responses, however in all honesty admit I'm still confused on that problem and have since moved on.

Here's something easier I think I need answered:

#13: Recent developments appear to make it possible for couples to dramatically increase the likelihood that they will conceive a child with the gender of their choice. In a test of a gender slection method, 12 couples try to have baby girls. If this gender selection method has no effect, what is the probability that the 12 babies will be all girls?

I answered this by: p(girl) x p(girl) x p(girl) ... up to 12 total...which I got on my calculator as: 0.0002441

The answer on the back of the book states the answer is: 1/4096

When I put "1/4096" on my calculator, the answer it gives me is 2.441e-4, which essentially is 0.0002441.

I'm wondering how I get the answer (on my calculator) 1/4096, from 2.441e-4

anyone?

...anyone?

 

[Rodimus Prime]

Thanks for the responses, however in all honesty admit I'm still confused on that problem and have since moved on.

Here's something easier I think I need answered:

#13: Recent developments appear to make it possible for couples to dramatically increase the likelihood that they will conceive a child with the gender of their choice. In a test of a gender slection method, 12 couples try to have baby girls. If this gender selection method has no effect, what is the probability that the 12 babies will be all girls?

I answered this by: p(girl) x p(girl) x p(girl) ... up to 12 total...which I got on my calculator as: 0.0002441

The answer on the back of the book states the answer is: 1/4096

When I put "1/4096" on my calculator, the answer it gives me is 2.441e-4, which essentially is 0.0002441.

I'm wondering how I get the answer (on my calculator) 1/4096, from 2.441e-4

anyone?

...anyone?

Well since it has no effect having a girl is 1/2 for the chances. Same as a coin flip

1/2*1/2.... (Total of 12 times) = 1/4096=.00024414...

Same as 0.5^12

 

[HappyDude20]

Well since it has no effect having a girl is 1/2 for the chances. Same as a coin flip

1/2*1/2.... (Total of 12 times) = 1/4096=.00024414...

Same as 0.5^12

Hmm, however the calculator is what gives me .00024414, while it seems the book (& professor) wants the "1/4096" answer, of which I don't know how to get, considering the only information my calculator will give me during tomorrow's exam will be .00024414

-

Aside from that, here's another one:

#15: The principle redundancy is used when system reliability is improved through redundant or backup components. Assume your alarm clock has a 0.975 probability of working on any given morning.

With one alarm clock, you have a 0.975 probability of being awakened. What is the probability of being awakened if you use two alarm clocks?

The answer is: 0.999375 - Though I've no idea how that answer came to be.

(btw, I'm currently at the library, switching between MBP and Math book, all in the effort of forwarding my homework.)

If I understand my homework, i'll understand my exam tomorrow.

Again, all help is appreciated; including you as well Rodimus Prime.

 

[OutThere]

Thanks for the responses, however in all honesty admit I'm still confused on that problem and have since moved on.

Here's something easier I think I need answered:

#13: Recent developments appear to make it possible for couples to dramatically increase the likelihood that they will conceive a child with the gender of their choice. In a test of a gender slection method, 12 couples try to have baby girls. If this gender selection method has no effect, what is the probability that the 12 babies will be all girls?

I answered this by: p(girl) x p(girl) x p(girl) ... up to 12 total...which I got on my calculator as: 0.0002441

The answer on the back of the book states the answer is: 1/4096

When I put "1/4096" on my calculator, the answer it gives me is 2.441e-4, which essentially is 0.0002441.

I'm wondering how I get the answer (on my calculator) 1/4096, from 2.441e-4

anyone?

...anyone?

Your method of multiplying the probabilities in the fertility question is the same method you'd use to solve the first question you posted about the polls. As for the fraction vs. decimal problem: the simplest answer is that you have taken (1/2)^12, or (1^12)/(2^12), which is 1/4096.

You can convert a decimal to a fraction on the calculator, but I believe the function is quite limited on a TI83. I got a TI89 in high school because it could basically do my homework for me.

 

[plinden]

Hmm, however the calculator is what gives me .00024414, while it seems the book (& professor) wants the "1/4096" answer, of which I don't know how to get, considering the only information my calculator will give me during tomorrow's exam will be .00024414

1/2*1/2 = 1/4
1/2*1/2*1/2 = 1/8
1/2*1/2*1/2*1/2 = 1/16
...
1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2 = 1/4096

Inverting:
2*2*2*2*2*2*2*2*2*2*2*2 = 4096

 

[Rodimus Prime]

Hmm, however the calculator is what gives me .00024414, while it seems the book (& professor) wants the "1/4096" answer, of which I don't know how to get, considering the only information my calculator will give me during tomorrow's exam will be .00024414

-

Aside from that, here's another one:

#15: The principle redundancy is used when system reliability is improved through redundant or backup components. Assume your alarm clock has a 0.975 probability of working on any given morning.

With one alarm clock, you have a 0.975 probability of being awakened. What is the probability of being awakened if you use two alarm clocks?

The answer is: 0.999375 - Though I've no idea how that answer came to be.

(btw, I'm currently at the library, switching between MBP and Math book, all in the effort of forwarding my homework.)

If I understand my homework, i'll understand my exam tomorrow.

Again, all help is appreciated; including you as well Rodimus Prime.

Well you are relaying to much on your calculator and not learning the principle.

As for quetion 15.
Take the probably of the alarm clock not waking you up (1-.975=0.025.

Now that you have that the problity of them both failing is 0.025*0.025=0.000625
So you anwser is 1-0.000625=0.999375

Posted from my blackberry

 

[HappyDude20]

1/2*1/2 = 1/4
1/2*1/2*1/2 = 1/8
1/2*1/2*1/2*1/2 = 1/16
...
1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2 = 1/4096

I just tried this on mine, however gave me the same 2.441e-4 answer.

You can convert a decimal to a fraction on the calculator, but I believe the function is quite limited on a TI83. I got a TI89 in high school because it could basically do my homework for me.

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Hopefully my professor will allow the decimal answers in lieu instead.

 

[HappyDude20]

Well you are relaying to much on your calculator and not learning the principle.

As for quetion 15.
Take the probably of the alarm clock not waking you up (1-.975=0.025.

Now that you have that the problity of them both failing is 0.025*0.025=0.000625
So you anwser is 1-0.000625=0.999375

Posted from my blackberry

OH man thank you.

I do agree i'm relying on my calculator too much, however this semester the professor is rather easy compared to previous math professors and considering this will be the last math class I ever take, I just wanna pass the class and get the hell out of there.

/justification.

 

[CalBoy]

#15: The principle redundancy is used when system reliability is improved through redundant or backup components. Assume your alarm clock has a 0.975 probability of working on any given morning.

With one alarm clock, you have a 0.975 probability of being awakened. What is the probability of being awakened if you use two alarm clocks?

The answer is: 0.999375 - Though I've no idea how that answer came to be.

The probability that one alarm clock will work is .975, meaning that the probability that one alarm clock won't work is .025 (1-.975). In order to find the odds of two alarm clocks working, you take the probability that both won't work (.025^2) and subtract that figure from 1. Thus,

.025*.025=.000625

1-.000625=.999375

Thanks for the responses, however in all honesty admit I'm still confused on that problem and have since moved on.

Your original question actually has nothing to do with confidence intervals, which is why I think you are confused.

The question is asking the odds of all 6 polls being correct if each has a 95% chance of capturing the correct range. This is a much simpler problem than worrying about confidence intervals and standard deviations.

.95^6=.735, meaning that there is a 73.5% chance that all six polls are within the range.

 

[plinden]

1/2*1/2 = 1/4
1/2*1/2*1/2 = 1/8
1/2*1/2*1/2*1/2 = 1/16
...
1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2*1/2 = 1/4096

I just tried this on mine, however gave me the same 2.441e-4 answer.

Hmm, yes, you are relying too much on your calculator. You missed my point.

If you look at what else I wrote:

2*2*2*2*2*2*2*2*2*2*2*2 = 4096

Inverting that is the same as multiplying all those 0.5, so you get 1/4096

 

[CalBoy]

I just tried this on mine, however gave me the same 2.441e-4 answer.

Hopefully my professor will allow the decimal answers in lieu instead.

If you need to find the denominator of a fraction with 1 as the numerator, just inverse the number (x^-1).

Most TIs also have a function that allows you to find the fraction (or nearest fraction) for any decimal. It's been many years since I've had to use one, but I know that it's either a primary key or one operated by a shift function.

 

[HappyDude20]

Here's another one , simple, yet pissing me off I can't get it.

Find the probability of a couple having a baby girl when their third child is born, given that the first two children were both girls. Is the result the same as the probability of getting three girls among three children?

The answer is 0.5, while the answer to the question at the end is no.

I'm feeling dumb with the lack of being able to answer these simple sounding questions.

This question is now part of 4.5 - Probability, Multiplication Rule: Complements & Conditional Probability

Admittedly, I'm moving along my homework nicely, though I do have about 3 more sections after this one. Then it's just read and review to feel good about walking into the exam for tomorrow....Again, thanks.

 

[plinden]

Maybe you can relate these to a coin toss, which is a 1 in 2 probability like gender of a baby.

What's the probability of getting three heads in a row when tossing a coin?

 

[HappyDude20]

Maybe you can relate these to a coin toss, which is a 1 in 2 probability like gender of a baby.

What's the probability of getting three heads in a row when tossing a coin?

Well, I'm thinking, 1/2 x 1/2 x 1/2, but that equals .125

 

[plinden]

Yes. That's right. And 1/8 is not equal to 1/2

The probability of getting three girls out three is not equal to the probability of the third child being a girl since, like a coin toss, the probability of getting a girl at any pregnancy doesn't depend on what's happened previously (edit: ie 1 in 2 or a half)

It's like what I've seen people talk about here on MacRumors - "I've had three broken Macs in a row, the odds of getting a broken one now must be astronomical" - which is wrong. The odds are exactly the same as if the person hadn't had any broken Macs.

Or it's related to the "gambler's fallacy". Some one losing at roulette will tell themselves that their luck is bound to change if they've lost 20 times in a row, but in reality, the odds of losing are the same no matter how many times they've played and lost (or won - and assuming the wheel isn't fixed).

 

[HappyDude20]

Yes. That's right. And 1/8 is not equal to 1/2

I'm sorry but i'm still confused. The answer for the problem (#13) is 0.5, of which I;m not sure how to get.

I've always been dumb when it comes to math so do please excuse me if i'm missing the obvious.

 

[plinden]

OK, I'll try again.

Find the probability of a couple having a baby girl when their third child is born, given that the first two children were both girls.

The second clause "given that the first two children were both girls" has no bearing on the probability that the third child will be a girl. Any pregnancy is 1/2 chance of a girl.

Is the result the same as the probability of getting three girls among three children?

No is correct. As you've calculated, it's 1/8.

 

[HappyDude20]

OK, I'll try again.



The second clause "given that the first two children were both girls" has no bearing on the probability that the third child will be a girl. Any pregnancy is 1/2 chance of a girl.


No is correct. As you've calculated, it's 1/8.

I do understand the 1/8 part; which is the same as .125.

What I don't understand is why the answer proposed is 0.5, and how to get it.

 

[Rodimus Prime]

Well, I'm thinking, 1/2 x 1/2 x 1/2, but that equals .125

Well for example the girl boy thing. Each event is separate and independent. The results of the previous event have no baring on the result of the current event.

A none independent example is I have a draw full of 12 socks. 6 white and 6 black and I am pulling out 3 socks. What is the probably of me pulling out a white sock given that the first 2 are were black. (Answer is 6/10)
Now if you want to find the probulity that all 3 are going to be white the answer is 1/2* 5/11*4/10.

 

[Rodimus Prime]

I do understand the 1/8 part; which is the same as .125.

What I don't understand is why the answer proposed is 0.5, and how to get it.

Separate and independent actions.

first 2 being girls has no bearing on the 3 child being a girl. Sock example from my previous post should help explain it.