NORMAL DISTRIBUTIONS & STANDARD (Z) SCORES

[Statistics are] the only tools by which an opening can be cut through the formidable thicket of difficulties that bars the path of those who pursue the Science of Man.

- Sir Francis Galton

The Normal Curve

¨  Theoretical only!

¨  Symmetrical; µ = midpoint = median = mode ¨  Tails of curve are infinite ¨  Aka Bell curve or Gaussian curve ¨  The “area under the curve” is determined from

standard deviations (σ) of any score from its mean (also called Z)

¨  Total area under the curve = 1.00

The Standard Normal Curve:

¨  Has a mean µ = 0 and standard deviation σ = 1 ¨  General relationships: ±1 σ = about 68.26%

±2 σ = about 95.44% ±3 σ = about 99.72%

µ -1σ +1σ +2σ +3σ -2σ -3σ

¨  (def) A unit-free standardized score that, regardless of the original units of measurement, indicates how many standard deviations a score is above or below the mean of its distribution

¨  Contains 2 parts: ¤  + or – sign (says whether it is above (+) or below (-) the mean ¤  A number (how far it is from the mean in standard deviation units)

¨  Formula:

¨  This formula changes any “raw” score into a “standardized” score (z-score!) ¤  When raw scores are transformed into z-scores, the mean of the distribution is

always 0, the standard deviation is always 1…what does this sound like?

Z scores

Only need: population mean (mu) & standard deviation (sigma)

Practice!

¨  Express each as a z-score ¤  IQ score of 135, mean is 100, standard deviation is 15 ¤ Height of 68 inches, mean is 68, standard deviation is 3 ¤ Fear of 6, mean is 3, standard deviation is 2 ¤ SAT score of 470, mean is 500, standard deviation is 100

Z scores and the Standard Normal Curve

¨  (def) normal curve for z-scores, sometimes called the z-curve ¨  Properties of the standard normal curve:

¤  Mean = 0, Standard Deviation (SD) = 1 ¤  If z = ±1, then p = .68

n  (68% of observations fall within 1 SD of the mean)

¤  If z = ±2, then p = .95 n  (95% of observations fall

within 2 SD of the mean)

¤  If z = ±3, then p = .997 n  (99% of observations fall

within 3 SD of the mean)

Example: Find the area beyond Z ¨  Table A (back of your book, p. 530) can be used to find the area

beyond the part of the curve cut off by the z-score ¤  Column C = probability associated with the area of the curve beyond z ¤  Column B = proportion of the curve between the z-score and the mean

¨  The probability of falling within ± z standard deviations of the mean is found by doubling Column B

¨  Assume the time it takes you to finish your homework is normally distributed. Mean = 30 minutes, SD = 10 minutes. How often (what percent of the time) are you finishing your homework in less than 20 minutes?

¨  Mean = 30 minutes. SD = 10 minutes. What percent of the time are you finishing your homework in less than 20 minutes? ¤  Step 1: Calculate the z-score!

¤  Step 2: Visual depiction

Example: Find the area beyond Z

1103020

10,30,20

−=−

=

===

−=

Z

andX

XZ

σµσµ

¤  Step 3: Look at Table A

¤  Answer = You finish your homework in less than 20 minutes 15.87% of the time

Practice!

¨  Assume that GRE scores are normally distributed. Mean = 500, SD = 100. Lets say you received a score of 650.

¨  Use the 3 steps to figure out how many people did better than you. That is, what percent of people got a score higher than 650?

¨  Then do it in reverse! What percent of people got a score lower than 650?

Going the other way: Finding a score

¨  Lets say a graduate school only admits students with GRE scores in the top 10% (mean = 500, SD = 100). What score do you need to be admitted? ¤ Step 1: Look up 10% or .1000 in Column C ¤ Step 2: Look to the left and see that the Z-score is 1.28 ¤ Step 3: Solve for X!

à X = 628 10050028.1

100,500,28.1−

=

===

−=

XandZ

XZ

σµσµ

Another Example (find the area beyond Z)

¨  I want to see how awesome everyone is, so I give each student an awesomeness test before the class starts. You score 130. Mean = 100, SD = 15. What percentage of the class is MORE AWESOME than you?

¨  This is answered by Column C… ¤  1:

¤  2: Visual Depiction (do on your own!) ¤  3: SeeTable A, the area beyond Z = 2 is .0228 (column C) ¤  Thus, the proportion of the class that is more awesome than you is

2.28%

215100130

15,100,130

=−

=

===

−=

Z

andX

XZ

σµσµ

Another Example (enough already!)

¨  Your friend is not quite as awesome as you are, they got a 115. Mean is still 100, SD is still 15. What percentage of the population is LESS AWESOME than him?

¨  This is answered by Column B… ¤  Step 1:

¤  Step 2: Visual Depiction (do on your own!) ¤  Step 3: Column B says that the area beyond Z = 1 is .3413 (column

B) n  But you can’t stop there. You must add .5 to this amount (for half the

population) because we are looking for people who are LESS AWESOME ¤  Thus, the proportion of the class that is less awesome than your

friend is 84.13%. Your friend is still pretty awesome

115100115

15,100,115

=−

=

===

−=

Z

andX

XZ

σµσµ

Examples Galore: Find the proportion between two scores ¨  What proportion of the class is more awesome than your

friend, but not as awesome as you? ¤ A word of caution: YOU CANNOT ADD/SUBTRACT Z-SCORES

¨  Find the difference in the proportions ¤ 1: Your score of 130 is a z-score of 2 (exceeds all but .0228

of the population) ¤ 2: His score of 115 is a z-score of 1 (exceeds all but .1587 of

the population) ¤ 3: Subtract!

n  .1587 - .0228 = .1359 … or 13.59% of the class is more awesome than your friend, but not as awesome as you.

Moving back into research…

¨  Z-scores tell us if something rare is happening in our sample

¨  α=.05 ¤ Zcritical = +1.96 ¤ Zcritical = -1.96