Two Plus Two Older Archives  

Go Back   Two Plus Two Older Archives > General Gambling > Probability

Reply
 
Thread Tools Display Modes
  #1  
Old 10-23-2005, 01:07 PM
coopersmydog coopersmydog is offline
Member
 
Join Date: Mar 2005
Posts: 32
Default Finding a probability from a Mean and a Standard Deviation

How do I find a probability of selecting something at random when I'm given the mean and standard deviation. I'm not given a sample size or the size of the population. I'd post the problem, but I'd like to understand the concept so if someone can give me the formula or an idea of how to find it I'd appreciate it.

Thanks.
Reply With Quote
  #2  
Old 10-23-2005, 01:16 PM
AaronBrown AaronBrown is offline
Senior Member
 
Join Date: May 2005
Location: New York
Posts: 505
Default Re: Finding a probability from a Mean and a Standard Deviation

You need one more piece of information, the distribution. For example, if you are told a Poker player makes $500 in an average session, with a standard deviation of $3,000 that could mean:

(1) Half the time she makes $3,500 and the other half she loses $2,500;

(2) One time in 37 she makes $18,500 and the other 36 times she breaks even;

(3) She has a Normal bell shape of outcomes centered at $500 with 1/3 between -$2,500 and $500, 1/3 between $500 and $3,500, 1/6 below -$2,500 and 1/6 above $3,500.

(4) Or an infinite number of other possibilities.

People often assume (3), which may not be a bad idea in some situations, especially for figuring probabilities between 25% and 75%. If it's an unusual situation or you need bigger or smaller probabilities, then all you can do is fall back on some general limits.
Reply With Quote
  #3  
Old 10-23-2005, 01:31 PM
coopersmydog coopersmydog is offline
Member
 
Join Date: Mar 2005
Posts: 32
Default Re: Finding a probability from a Mean and a Standard Deviation

I wasn't given the distribution.

The question is similar to this one, not exactly the same.

A recent study found the average weight of 5 year olds is 62 pounds with a standard deviation of 8.1 pounds. If a 5 year old is selected at random, what is the probability that their weight will be less than 70 pounds.

Thanks.
Reply With Quote
  #4  
Old 10-23-2005, 01:40 PM
AaronBrown AaronBrown is offline
Senior Member
 
Join Date: May 2005
Location: New York
Posts: 505
Default Re: Finding a probability from a Mean and a Standard Deviation

I think you are meant to assume that weight at a given age is reasonably close to a Normal distribution, at least near the center. I doubt that is true, I would expect a bimodal distribution for boys and girls, and significant distortions for ethnicity, nutrition and certain medical conditions. Moreover, "five year old" probably means a group distributed in age from five to six.

However, to do the calculation if the distribution were Normal, you compute the probability of a standard Normal variate being less than 0.99 (8/8.1) standard deviations above the mean. You can do that with a table, or in Excel with Normdist(70,62,8.1,True) = 0.8383. That's probably not a bad answer for the actual data.

It's also possible this is a trick question and you are meant to answer that it cannot be determined with the information given.
Reply With Quote
  #5  
Old 10-23-2005, 01:50 PM
coopersmydog coopersmydog is offline
Member
 
Join Date: Mar 2005
Posts: 32
Default Re: Finding a probability from a Mean and a Standard Deviation

My example probably wasn't very good, because it wasn't the actual question I'm trying to answer. Anyway, the normdist function in Excel worked perfectly. Thanks for the help.

It was a multiple choice answer, so it I know it wasn't a trick question.

Thanks again.
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 05:49 PM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, vBulletin Solutions Inc.