Finding a probability from a Mean and a Standard Deviation

[coopersmydog]

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.

[AaronBrown]

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.

[coopersmydog]

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.

[AaronBrown]

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.

[coopersmydog]

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.