Lesson 3
Normal Distribution & Standard Scores

Normal Distribution

Chapter 5

Sections Covered: 6.1 - 6.3; 6.5 - 6.8.

The normal (or Gaussian) distribution is the familiar unimodal, symmetric curve with thin tails that every introductory psychology textbook calls the "bell curve." (Many years ago when I was teaching at the University of Illinois, which was a leader in accommodating students with disabilities, I lectured to a class that included a blind student. I scribbled a replica of the normal curve on the chalk board and described it as looking like a bell. After class, the student politely explained to me that there are many kinds of bells--door bells, sleigh bells and the like--and would I please be a bit more specific. Touche!) The normal curve looks like a vertical cross section of the Liberty bell--with all the top attachments removed--oh, forget it.
Here is a picture of a normal distribution showing the important facts about areas under the normal curve within various standard devistion units of the center.

Certain things follow from the facts about areas in the graph for the normal curve:
• 50% of the area (and hence, half the cases in a set of data that is normally distributed) lies below the middle or mean.
• 34% of the area lies between the mean and a point one standard deviation above. Likewise, there is 34% between the mean and a standard deviation below the mean.
• It follows, then, that 16% of the data in a normal distribution lies below a point one standard deviation below the mean.

Answer each of the following questions--write down your answers. At the end, you can click on the Answers.
1. What percent of the normal distribution lies between one and two standard deviations above the mean?
2. What percent of the normal distribution lies above three standard deviations above the mean?
3. If there were 100,000 persons arrayed in a normal distribution of heights, how many would be expected to lie more than three standard deviations above the mean?

Answers to the first set of Normal Curve Questions

It is not a simple matter to calculate the area under the normal curve between to arbitrary points like 1.25 and 2.38 standard deviations above the mean.

You have four options:

1. look up values of areas under the normal curve in a printed table in a statistics textbook,
2. hope that these good people in the Netherlands have their server functioning when you need a quick reading of normal curve areas. Please note: when you enter a z-value of 1.5, say, into the calculator in the Netherlands, the area returned is the probability of being greater than 1.5 or less than -1.5; i.e., it is a two-tailed probability and must be divided by 2 to give a single tail area.
3. your best bet, if your browser will handle the Java, is to use Gary McClelland's nice Java program from the University of Colorado.
4. and, finally, just in case none of these utilities is available when you need them on the internet, you can always resort to the old-fashion way of finding normal curve areas by looking them up in a table like this one.

Please exercise either option now in answering the following questions:

1. What percent of the normal distribution lies below a point .675 standard deviations above the mean?
2. What percent of the normal distribution lies above a point that is 1.96 standard deviations above the mean?

Answers to the second set of Normal Curve Questions

Unit Normal Scores: the z-Score

All normal distributions have the same "shape" but they can have different means and different standard deviations. Once one specifies the mean and standard deviation of a normal distribution, everything alse about it is fixed (e.g., the percent of area between any two points).
For this reason, all the various normal distributions (of people's heights and weights and IQ scores) can be referred to a single table of the normal curve by standardizing a variable to a common mean and standard deviation. The simplest standardization measures the position of any point in a normal distribution in terms of its distance above or below the mean in units of the standard deviation. Thus, a standard unit normal variable has the formula
z = (X-m)/s , where m is the mean, and s is the standard deviation of the distribution of scores. Consequently, a person with a z score of +1.5 lies one and one-half standard deviations above the mean.

You are given that the distribution of adolescents IQ scores is normal in shape with a mean of 100 and a standard deviation of 15 points.
1. What is the percentile rank of a child whose IQ is 120?
2. What percent of the population of adolescents have IQ scores below 90?

Answers to the third set of Normal Curve Questions

Standard Scores

Unit normal z-scores are useful, but their properties are sometimes viewed as a disadvantage for particular applications. In these cases, one transforms them to scales that have more convenient means and standard deviations. For example, if one would multiply each z-score by 200 and then add 1000 to the product, the resulting new standard scores would have a mean of 1000 and a standard deviation of 200.
There are several particular standard score scales in such common use that it is useful to look more closely at them. In general, if a z-score is transformed via the following formula: Z = Bz + A ,then the Z-score has a mean of A and a standard deviation of B.

Some Popular Standard Scores

A Mean	B St Dev	Scale Name
500	100	SAT; GRE; LSAT; GMAT
100	15	Wechsler IQ
100	16	Stanford Binet IQ
20	5	ACT (Amer College Testing Co.)
50	10	T-scale (MMPI)

Standard scores vs. percentiles

If all one does with standard scores is convert them to percentiles, then why have both?
Percentiles and standard scores have slightly different information in them. Another way to put this is that the transformation from standard scores to their normal curve percentile equivalents is a "non-linear transformation." Very large differences between extremely large of extremely small standard scores correspond to small differences in percentiles; likewise, very small differences in standard scores near the mean correspond to large differences in percentiles.
Consider two groups of three persons each whose heights are measured both in inches and in percentiles among adult males:

 

                 Heights-inches      Heights-percentiles
                 ______________      ___________________
         Group A: 70", 72", 84"       80, 92, 99.999
         Group B: 70", 74", 76"       80, 95, 99.9
                            

                         Group A     Group B
                         _______     _______
     Mean in inches      75.333       73.333
     Mean Percentiles     90.67       91.63

Notice that Group A is taller than Group B when heights are expressed in inches, but Group B is "taller" when heights are expressed in percentiles. Is this possible? Or did I make a calculation error?

Collateral Reading

Here's more on the Normal Curve, courtesy of John Behrens.

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Assignment Three

Use this form to complete Assignment #3 and submit your work.