Normal Distribution

Consider some population where the numeric measurement of interest has mean m and standard deviation s. If these measurements are normally distributed then the following properties apply:

Properties:

  1. These measurements are symmetrically distributed about the mean.
  2. A histogram of these measurements will be bell shaped.
  3. These measurements satisfy the empirical rule:
    1. 68.27% of the measurements will be within one standard deviation of the mean.
    2. 95.45% of the measurements will be within two standard deviation of the mean.
    3. 99.73% of the measurements will be within three standard deviation of the mean.

Problem:

Consider the population of starting salaries for DePaul CTI'2000 graduates. Assume that these salaries are normally distributed with mean $60,000 and standard deviation $2,000.

  1. Determine the proportion of graduates that earned between $56,000 and $64,000.
  2. Determine the proportion of graduates that earned more than $64,000 .
  3. Determine the proportion of graduates that earned between $56,000 and $66,000.

Solution:

  1. Observe that $56,000 is $4,000 less than the mean and $64,000 is $4,000 more than the mean. That is we are interested in the proportion of salaries within two standard deviations of the mean and so the desired proportion is 95.45%.
  2. From the symmetry property we know that 50% of salaries are more than $60,000. Now, observe that $64,000 is $4,000 more than the mean. From the symmetry property and the empirical rule this means that (95.45/2)% of salaries will be between $60,000 and $64,000. Hence the desired proportion is 50%-(95.45/2)% which is 2.275%.
  3. Observe that $56,000 is $4,000 less than the mean and $66,000 is $6,000 more than the mean. From the symmetry property and the empirical rule this means that (95.45/2)% of salaries will be between $56,000 and $60,000. (99.73/2)% of salaries will be between $60,000 and $66,000. Hence the desired proportion is (99.73/2)%+(95.45/2)% which is 97.59%.

 

Standard Normal Distribution

Consider some population where the numeric measurement of interest is normally distributed with mean m=0 and standard deviation s=1. Such a normal distribution is referred to as a Standard Normal Distribution. Tables are available for the Standard Normal Distribution which allow us to determine the proportion of measurements between any two values to a reasonable degree of accuracy.

Table Tips:

  1. Ignore the "Height" column (hardcopy only)
  2. Scan the "z" column for the desired value (or its absolute value). The corresponding entry under the "Area" column is the proportion between -z and z.
  3. If the desired z value is not found use the closest.
  4. If the desired z value is larger than the largest entry use 100% as your proportion.

Work these problems for practice.

Transformation Rule

Given a set of measurements, known to be normally distributed, but for which m/=0 or s/=1 the proportion of measurements between two arbitrary values (say x 1 and x 2) may be determined if the x values are first transformed to z values and then the Standard Normal table used. The steps are:

  1. Convert x 1 and x 2 to z 1 and z 2 using:
    z = (x -     m)/ s
  2. Use the Standard Normal table to determine the proportion between z 1 and z 2.

Work these problems for practice.

Percentile Problems

Consider some measurement of interest, from a population, that is known to be normally distributed with mean m and standard deviation s. Let us say we would like to answer problems of the following type:
  1. Given a particular measurement, what is its percentile rank.
  2. Given a percentile rank, what is the corresponding measurement (i.e. the percentile score).

If m and s are known we may proceed as follows. Remember that for some measurement, say x, that is in the n th percentile, n% of all measurements in the population will be less than x.

  1. Determining percentile rank:
    From the definition of a percentile this is no different than any normal distribution problem that involves determining the proportion of measurements less than a particular value.

  2. Determining percentile score given rank (i.e. n%):
    We proceed in two steps (we will not discuss the case where     m=0 and s=1 since it follows naturally from the more general case).
    1. Determine z: Observe that we cannot look up n% in our table because of how our table is constructed. However we may deduce the percentage to be looked up by noticing that, due to the symmetry property, we merely double the proportion between m and the desired value. That is if n is 90 (i.e. 90 th percentile) then the proportion between m and x is 40%. Doubling gives us 80%. Similarly if n is 30 then the proportion between x and m is 20%. Doubling gives us 40%.
    2. Determine x: Apply the transformation rule. Since we know z, m, and s then, by rearranging terms, we have the expression:
      x=z s + m

Work these problems for practice.