Anyone good with Statistics and Probabilities?

[Deleted User]

I am stumped! Been trying to understand this probability stuff better and searched all over the net for help. I have a question that I do not know how to answer.

The percentage of customers who enter a restaurant and ask to be seated in a smoking section is 15%. Suppose that 100 people enter the restaurant.
a) what is the probability that 15 people request a smoking table?
b) what is the standard deviation of the number of requests for a smoking table?

Anyone good with probability theories would be greatly appreciated!

Halen

unc2701

a)Exactly 15 or at least 15?
b)Can you use the normal approximation?

[Deleted User]

If 100 people were to enter the restaurant, what are the chances there are exactly 15 people that smokes. Anything would be very helpful.

stebesplace

Regarding question 1, your P(A) would be 0.15, and your P(A') would be 0.85.

Regarding question 2, it goes something like this, and anyone can correct me if i am wrong:

You need to calculate the mean, so in this case it would be 1+2+3 ... 14+15 = 120

Divide 120 by the number of entries, in this case, 15. Answer = 8

Find your deviation list, which would be each number in your original list minus the mean. So for example, 1-8 = -7, 2-8 = -6, etc. You need to then take the squares of each, and add them all together. The final value here would be 280.

You then divide that by the (mean - 1), so in this case, 14, which gets you 20.

The final calculation is the sqrt, or square root of that final value (20). Your standard deviation would then be 4.472 (approx).

This was a quick reply, so double check to make sure my math was right.

unc2701

a) Exactly 15 would be (100 C 15) * (.15^15) * (.85^85) for clarity, exactly 14 would be:
(100 C 14) * (.15^14) * (.85^86)

b)I have no idea stebesplace is doing above... you gotta do x*p(x) for x=0 to 100 if you were gonna do it the long way. The normal approximation gives you sqrt(n*theta*(1-theta)) or sqrt(100*.15*.85).

stebesplace

I should have put a disclaimer that I am not a math expert.

*DISCLAIMER*
I am not a math expert!
*END DISCLAIMER*

:rolleyes:

[Deleted User]

unc2701, could you please elaborate on the answer? I am not able to understand the formula you wrote. thanks

unc2701

Which part?

[Deleted User]

stebesplace, so I just add up from and divide to get the mean? sum of x's divide by total number of x's?

[Deleted User]

(100 C 15) and the normal approximation sqrt(100*.15*.85). if solved what would be the answers?

unc2701

(100 C 15) is a combination- the thing w/ all the factorials. Since no one does these things for fun, you really need to learn how to do that. Might come up on a test. The second one is just the square root of (100*.15*.85) When you have a big enough sample size, the binomial looks a lot like the normal, so you can just use that instead of doing some **** up factorials and summations.

BAD ASP

If you were in Illinois it would be simple.... there's no smoking in restaurants so your probability would be zero!:D:D:D

stebesplace

BAD ASP wrote: »

If you were in Illinois it would be simple.... there's no smoking in restaurants so your probability would be zero!:D:D:D

HA, well put.

stebesplace

halenhoang wrote: »

stebesplace, so I just add up from and divide to get the mean? sum of x's divide by total number of x's?

Thats how you find the mean, yes.

[Deleted User]

thank you.

unc2701

stebesplace wrote: »

Thats how you find the mean, yes.

Er. sadly, no. What you're looking for in that formula is the expected value, which is summation of 0 to 100 of x*prob(x)

Prob(x) for each of those is (100 C X) *(.15^x)* (.85^(100-x). However, if you work through the math, it'll simplify down to n*theta or 100*.15=15

For the SD, you can just use the formula I gave. Square root (100 * .15 * .85) I just remembered that it's not an approximation, it really simplifies down to that formula (when you go to do inference, using that as the SD, it becomes an approximation).

cdn5003

This is actually an advanced problem in probability and statistics involving the binomial distribution. The difficulty is in the EXACTLY part of the question. The formula is nCr x (p)^r x (1-p)^(n-r). terms defined as n= number of trials, r= number of events you wish to obtain, p= probability the event will occur. nCr is the combination operator n choosing r.

So what this means in this case is:

100C15 x (.15)^15 x (1-.15)^(100-15)
simplified

100C15 x (.15)^15 x (.85)^85

The combinations overflow my calculator but this link provides the answer for exactly problems. http://stattrek.com/Tables/Binomial.aspx

The answer to part one is .11109, or 11.109%

Have a nice day

cdn5003

Also, for the second part, the standard deviation formula in the binomial distribution is sqrt np (1-p). in this case n=100, p=.15.

so (100)(.15)(.85) = 12.75

and sqrt of 12.75 is 3.57

so the standard deviation of the number of requests is 3.57.

Jordan fades back,swish, and that's the game!

cdn5003

here is the best link I could find on the subject, it should help when you are checking answers to problems you do on your own.

http://hyperphysics.phy-astr.gsu.edu/Hbase/math/disfcn.html

stebesplace

Looks like I myself need to go back to school too!

unc2701

cdn5003 wrote: »

This is actually an advanced problem in probability and statistics involving the binomial distribution. The difficulty is in the EXACTLY part of the question.

Might want to reel in the ego there, son. Three things here:
1)It's an extremely basic problem
2) The _exactly_ part makes it easier, not harder
3) It's generally bad form to just give out the answer- make him solve something so he'll remember it when test time comes around.

stebesplace- you were doing it the hard way, but if you worked through the expected value and did 0-100 you would have gotten the same answer- just with a LOT more work.

BaggedLancer

Damn I'm glad i'm edumikated

Mazeroth

Just curious how you guys remember this crap? Did you take stats recently, use it a lot, or just have a VERY good memory? Serious question here.

BaggedLancer

Mazeroth wrote: »

Just curious how you guys remember this crap? Did you take stats recently, use it a lot, or just have a VERY good memory? Serious question here.

Fo realz....i can't remember yesterday.

[Deleted User]

Thank you very much! This has given me a large amount of information. The statistics problem does not go into calculus level math such as integration or derivatives, so I thought I could handle this.

UNC2701, thank you very much for further explaining to me.

CDN5003, thank you very much for your help as well.

Any chance I could ask another one? You are correct, the answer to the question would not be as important as the process to come up with the answer. So, whatever you could assist me with, like the formula's and which one to use would be greatly appreciated!

Pat is a student taking a stat course. Pat is not a good student. Pat does not read the textbook, does not homework, and regularly misses class. Pat rely on luck to pass each exam. The exam consists of 10 multiple choice questions, each question has five possible answers, only of which is correct.

a) What is the probability that Pat gets no answers correct?
b) What is the expected number of correct answers that Pat will get?

part a. Pat has a 1 in 5 chance of getting the answer correct, then Pat also has a 4 in 5 chance of getting it wrong.

I do not know what to do with the information.

Halen

Mike Kozak

where am I??

Sami

I know it's a theoretical question but the fact that he goes into that class makes it practically impossible to solve with statistics. He should afterall have a little clue of what the questions are thus improving his odds.

a) Would it be as simple as 0.8^10 = 11%?
b) 20% chance of getting a right answer times 10 = 2

When it came to stats, I was like Pat, didn't like to study. That's more for accountants, not for us engineers...

unc2701

Ok, for a binomial, you have two parameters. n and theta. All the formulas for the binomial use them. n is the number of tries. theta is the probability of success on a single try. So:

Part a is just like the one yesterday. Formula is p(x)=(n C x) * (theta^x) * ((1-theta)^(n-x))

p(0)=(10 C 0) * (.2^0) * (.8)^10

Part b is easy. Expected value of a binomial is n*theta or 10*.2= 2

Now, for ANY distribution you can find the expected value by adding up x*p(x) for all values of x. so you'd do 0*p(0)+ 1*p(1) + 2*p(2).... up to 10*p(10). If you haven't learned that the expected value for a binomial is n*theta yet, you'll have to show that calculation to your teacher.

[Deleted User]

Sami,

a) That is what I thought as well. For each question, there are five possible answers, meaning .2 chance of right and .8 chance wrong.
b) Seems to make sense.

Maybe I am making this out to be harder then what it really is.

[Deleted User]

UNC, Thanks again. You are really good with this, do you gamble? lol.... well, if the odds are in your favor, I guess its not gambling!

Sami, thanks for the input. Engineer huh, ahem, Engineer Math, Engineer Physics, Linear Algebra!!!! I have been out of school many years, like 10 or so, I will never forget the endless hours of studying when I took those courses. It required novel length pages to answer single question! Unfortunately, I remember nothing, highest level of math for me is basic algebra. hahahah..

Halen

Sami

halenhoang wrote: »

UNC, Thanks again. You are really good with this, do you gamble? lol.... well, if the odds are in your favor, I guess its not gambling!

Don't know about UNC but I do gamble. The statistics they use on TV when playing Holdem aren't really statistics (or lets say they are very simplified) but if you're good at math it makes playing much easier. People without a clue about math will chase any draw and even if they manage to avoid disaster a few times the probabilities will get them in the long run. That's why I tell people who complain about "donkies" (term for inexperienced, bad player who gets lucky on a draw) to be quiet, they pay part of my bills every month...no matter how much the lucky draw cost you that time, you will get it back the next time.