""He uses statistics like a drunken man uses alamppost ... for support rather than illumination." -- AndrewLang


OVERVIEW: Normal distributions are quite common inreal life settings. Any set of normally-distributed observations canbe examined efficiently by converting the data to standardizedobservations know as z-scores. A z-transformation changes a normalrandom variable with mean m and standarddeviation s into a standard normal randomvariable with mean 0 and standard deviation 1.

Any set of numbers has a mean and standard deviation. The set W ={5,10,20,65,80} has m = 36 and s = 30.56141358. If we wanted to find out howmany standard deviations a certain observation was from the mean, wewould simply calculated its distance from the mean, and then divideby the standard deviation. For instance, consider the observation 5.If we calculate (5 - 36)/30.56141358, we get z = -1.014. In otherwords, the observation 5 is slightly more than one standard deviationbelow the mean.

The standardized value (sometimes called a z-score)of an observation, x, is

z = (x - m)/s

The following table displays the standardized values forobservations from the set W.













If a variable x has a normal distribution with mean m and standard deviation s, the distribution is represented by thesymbolism N( m,s). If the x-observations are standardized, thenthe standardized distribution is also normal, and has mean = 0 andstandard deviation = 1. This distribution is represented by N(0,1).The advantage to standardizing is that it allows one to make use ofthe standard normal table that appears in statisticstextbooks. (And, which is provided on the Advanced PlacementStatistics Examination.) It is important to be able to use thistable. It's also significant to note that many useful computationscan be done on the TI-83. Here are a few examples...

Suppose a set of observations is approximately normalwith mean = 50 and standard deviation = 4.

We know that approximately 68% of the observationswill be within one standard deviation of the mean. Note that

normalcdf(46,54,50,4) = 0.6826894809
normalcdf(-1,1,0,1) = 0.6826894809

If we want to know what percent of the scores are above 56, we cannote that z56 = (56-50)/4 = 1.5.

normalcdf(56,1E99,50,4) = 0.0668072287
normalcdf(1.5,1E99,0,1) = 0.0668072287

Important to note: You canstandardize any set of numerical observations and obtain z-scores.The z-scores simply reflect how many standard deviations anobservation is from the mean. The z-scores will form a normaldistribution only if the original data set is normal. You cancorrectly use the normal distribution table to interpret z-scoresonly if the original data set is normal. Standardizing scores doesnot magically convert non-normal data into normal data.

If you have a data set in a TI-83 list, you can use the calculatorto construct a normal probability plot. This is one of theoptions when you do a stat plot on the calculator.

Extremely important for success of the Advanced PlacementStatistics Examination:
If you are given a numerical data set, always (I repeat,always) display the shape of the distribution.
Using the TI-83, this can be done very easily with a histogram or aboxplot.

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