normal distributions & standard (z) scoresdooleykevin.com/psyc60.4.pdf · normal distributions...
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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