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What is a Bell Curve?

Updated on October 18, 2013

Characteristics

A bell curve is helpful in inferential statistics as it describes the results that a well-vetted assessment tool should produce. Some characteristics of a bell curve are

  • It has a symmetrical distribution
  • The central tendency (the axis of symmetry) represents the mean, mode, and medium of all scores concurrently
  • It contains a mean and a standard deviation
  • 68% of all scores will fall within one standard variation below the mean and one standard deviation above the mean
  • 95% of all scores will fall within two standard variation below the mean and two standard deviation above the mean
  • 99.5% of all scores will fall within three standard variation below the mean and three standard deviation above the mean

On a test such as the SAT's, the GRE's, The WAIS IQ test, or any other aptitude test the mean, or numerically averaged score (all scores added together and then divided by the total number of scores) will vary as will the standard deviation.

Example of Application

In order to use a bell curve to say anything meaningful about how an individual test result relates to many test results we must know the mean and the standard deviation of the assessment instrument being considered.

To illustrate this I'll use the WAIS (Wechsler Adult Intelligence Scale) IQ test. This test has a mean of 100 meaning that this is the average score for this test. The Standard Deviation is 15.

Knowing this we can consider the scores that will fall within one deviation below the mean and one standard deviation above the mean. This range of scores will be 85-115 (simply 15 subtracted the mean to illustrate one standard deviation below the mean and 15 added to the mean to illustrate one standard deviation above the mean)

Recall from the characteristics of a bell curve that 68% of all scores will fall in this range of one standard deviation. In other words 68% of the people tested (from a sample that is representative of the overall population will score between 85 and 115 on this test.

If we expand this to two standard deviations then we will be looking at what percentage of people scored between 70 and 130 (15x2=30, in each direction). 95% of all test takers will fall within this range.

If we look at 3 standard deviations we will be considering scores between 55 and 145. 99.5% of the testing population will receive a score in this range.

Z scores

Another convenience provided by a bell curve in working with statistical data is that we can take a score and determine just how many standard deviations it is above or below the mean.

Let's say someone scored 120 on our WAIS IQ test and we want to know where on our bell curve they fall.

The formula to use is

Z=(score-mean)/standard deviation or

Z=(120-100)/15

Z=1.33

So the score 120 is 1.33 standard deviations above the mean. We can convert this to a Normal Curve Equivalent (NCE) to find the corresponding percentile score. Percentile scores are simply where a score would be ordered for every one hundred random scores, or another way to think about it is what percentage of people scored lower and what percentage scored higher.

To convert to an NCE, or any other scale (ie.CEED score, Stanine, or Sten score) we use the formula;

new scale=(z)standard deviation'+mean', where standard deviation' is the standard deviation of the new scale and mean' is the mean of the new scale.

For NCE's, the mean is 50 and the standard deviation is 21.06, so

NCE=(1.33)21.06+50

NCE=78.0098

So the score on the WAIS IQ test of 120 falls on the 78th percentile, mean the score is higher than 78% of all score and lower than 21% of all scores.


Try It On Your Own

So now try converting a score your self;

  • Start with picking a hypothetical WAIS score at random
  • Calculate the Z using the formula Z=(score-mean)/Standard Deviation
  • Now convert to a new normed scale.
  • Remember to use the mean and standard deviation that corresponds to the scale you are using, here are some common normed scores

CEED, mean=500, S.D.=100

Stanines, mean=5, S.D.=2

Sten Scores, mean=4.5, S.D.=2

T scores, mean=50, S.D.=10

Just plug the Z score you found into the equation

New Scale Score(any of the above)=Z x (S.D. for that scale) + (mean for that scale) or,

Score'=Z(S.D.')+mean'

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