Looking for statistics tables...

OK....

Here goes.

If the distribution of results is truly "normal", you will have approximately 68.3% of the data points within one standard deviation of the mean (average) value. You will have approximately 95.5% of the data points within 2 standard deviations of the mean.

To calculate the standard deviation, first calculate the mean value (call it "m"), by adding up all of the individual measurements, and dividing the total by "n", the number of measuremenets.

Now, take each measurement, and calculate its difference from the mean. In otherwords, for each measurement x, calculate x-m. Don't worry about signs (positive or negative), since we'll be squaring these differences in the next step....

Take each difference from the mean, and square it. For each original measurement, you should now have one value: (x-m)^2

Take all of these squares you just calculated, and add them up to get a total. Now, divide that total by (n-1). In other words, however many original measurements you had, you'll be dividing by a number that is one less than that.

Now, take the square root of the result.

This final answer is the standard deviation, which is represented in symbols as sigma-sub-n-minus-one.

Hope this helps!

- Rick "Math geek" Dickinson

A percentile is a value for which that percentage of the measurements lies below that value.

e.g a 95th percentile value is the value where 95% of the reading of a set lie below it. The percentile value is thus dependant on the shape of the dirtribution of readings.

For a normal distribution 84.1% of the samples lie below 1 standard deviation above the mean hence (I'm reasonably sure) 1SD above the mean is the 84.1 percentile

Correspondingly 15.9 percent of readings lie below 1SD below the mean and hence 1SD below the mean is the 15.9 percentile.

The table you are talking about is "the area under the normal curve" (to the left of the ordinate Z) - see the following for a online version.

If you want a copy of a real "Z" table I can email you one.

Steve R

I remember how to compute the mean and standard deviation. have 20 year old Fortran code to do it, and a PDA that can do it.

The question was How do I convert "95th percentile" to somethiing like "1.3 standard deviations above the mean"?

Bob Kaplow NAR # 18L TRA # "Impeach the TRA BoD" >>> To reply, remove the TRABoD!

This site should have more than you want to know:

The information for your particular question is at:

Bob Kaplow wrote: > OK, it's been way too long since I took a stat course, and I can't find my > old textbook. >

Bingo! That's wht I was looking for. My search actually led me to this document, but not to the place with the table, and I didn't look around enough to trip over this.

And their Dataplot software is even available for my VMS systems!

Bob Kaplow NAR # 18L TRA # "Impeach the TRA BoD" >>> To reply, remove the TRABoD!

Bob,

If you are using experimental data and have fewer than ~120 data points, would suggest you use the cumulative t distribution rather than the normal.

Hello, so long, and regards,

-Larry (Guessing next year's R&D will be interesting :) C.

distribution?

And if you go to the HOME link, you can download the whole book, the software, or have them send you a FREE CD with the software and the book. Just the thing for folks like me who loaned their college statistics book to their little brother, only to have it vanish forever.

Bob Kaplow NAR # 18L TRA # "Impeach the TRA BoD" >>> To reply, remove the TRABoD!

Sort the data from smallest to largest. Pick the data point of your sample such that roughly 5% of your data is above that point. You will not get exactly 5%, but perhaps a little less. Then the point at which appx.

5% of your data is above this (or appx. 95% of data is below) is called a sample quantile, the 95th percentile, say x(95). You want to know c, where x(95) = xbar + c * stdev, xbar is the sample mean and stdev is the sample standard deviation. Hence, c = (x(95) - xbar)/stdev and the "95th percentile" is roughly "c standard deviations above the mean".

If the true distribution of your data is from the normal distribution, then x(95)=Z(0.95)=1.9645 appx. A way to test whether your data is from a normal distribution is to use a q-q-plot. If it's a fairly straight line, then it's normal. The q-q-plot is a plot of the ith ordered value in your data versus the (i-1/2)/n th percentile of the normal distribution Z((i-1/2)/n), where i=1,...,n and n=total number of observations. Other graphical test include looking at the histogram of your data to see if it's fairly bell shaped and not too skewed. Hopefully you have a descent samplesize n, say n>30 to work with. If n is much smaller then it's really hard to determine if the data is normally distributed or not.

THX, John

Ooops, Instead of saying x(95)=Z(0.95)=1.9645 I meant to say c=Z(0.95)=1.9645 appx. Here xbar and stdev are both sample estimates of the population mean and standard deviation.

THX, John