The Abnormal Distribution I’m just misunderstood.

The Normal Distribution The normal distribution is a very common and useful distribution Describes many real variables Height Weight IQ SAT Can be used to calculate probabilities

Characteristics of the Normal Distribution Bell-shaped Symmetrical Mean, median, and mode are equal Tails extend indefinitely Area under the curve equals 1.00

Standard Scores It is often desirable to compare scores from different distributions For example, X = 100 X = 60 S = 50 S = 30 150 120 A direct comparison of those scores is not possible because each distribution has a different mean and standard deviation

Standard Scores The problem can be solved by converting the original scores to standard scores Standard scores are scores with a set mean and standard deviation Examples of standard scores IQ t scores GRE

z Scores A very useful standard score is the z score A z score states how many standard deviations the original score is above or below the mean of a distribution 150 120 X = 100 X = 60 S = 50 S = 30 1s 2s z = X - X S

Properties of z Scores The mean of any set of z scores equals to 0.0 The standard deviation of any set of z scores equals 1.00 The shape of a distribution of z scores maintains the shape of the original score distribution from which the z scores were derived

Finding Areas Under the Curve The proportion of area between two scores of a frequency distribution is equal to the proportion of cases between those two scores The relation between area and proportion permits us to determine probabilities It is necessary, then, that we be able to determine the area under the curve

The z Score Table A great deal of the work has been done for us in the z table Column 1, z Column 2, Area between mean and z Column 3, Area beyond z

Finding the Area Given a mean of 100 and a standard deviation of 20, what is the area between the mean and a score of 120? z = X - X S 120 - 100 20 = = +1.00 100 .3413 120 +1.00

Finding the Area Given a mean of 100 and a standard deviation of 20, what is the area above a score of 130? z = X - X S 130 - 100 20 = = +1.50 100 .0668 130 +1.50

Finding the Area Given a mean of 100 and a standard deviation of 20, what is the area above a score of 85? z = X - X S 85 - 100 20 = = -.75 100 .2734 .5000 85 .2734 + .5 = .7734 -.75

Finding the Area Given a mean of 100 and a standard deviation of 20, what is the area between a score of 75 and 90 ? z = X - X S 75 - 100 20 = 90 - 100 20 = = -1.25 = -.50 .1915 100 90 .3944 - .1915 = .2029 .3944 75 -1.25 -.50

Finding Scores When the Area is Known Given a mean of 100 and a standard deviation of 20, what is the cutoff score for the lower 10% ? X = (z x S) + X = (-1.28 x 20) + 100 = 74.4 100 .1000 ? -1.28

Finding Scores When the Area is Known Given a mean of 100 and a standard deviation of 20, below what score would 85% of the cases fall ? X = (z x S) + X = (1.04 x 20) + 100 = 120.8 100 .5000 .3500 ? 85% 1.04