1. The Normal distribution and z-scores
Areas Under the Curve

2. √ Let’s Practice! x: { 3, 8, 1} (Sx)2 SS = Sx2 N x 3 8 1 m 4 (x-m) -1-3
(x-m)2 1 16 9
S(x-m)2 = SS
Find s:
S(x-m)2 = 26
S(x-m)2 N √
=√(26/3)
= 2.94 OR
SS = Sx2
(Sx)2 N __ x 3 8 1 x2 9 64 1
Sx = 12
Sx2 = 74
74 - (144/3) = 26
Then √(26/3) = 2.94

3. The Philosophy of Statistics & Standard Deviation

4. The Philosophy of Statistics & Standard Deviation
.24
.20
.16
.12
.08
.04
Proportion

5. The Philosophy of Statistics & Standard Deviation
.24
.20
.16
.12
.08
.04
Proportion

6. Standard Deviation and Distribution ShapeIQ

7. Example: IQs of Sample of Psychologists
ID IQ
1 128
2 155
3 135
4 134
5 144
6 101
7 167
8 198
9 94
10 128
11 155
12 145
x =
s =
With some simple
calculation we find:
x - x
z = s
z(144) = [ 144 – ]/ 27.91
= +0.13, “normal”
z(198) = [ 198 – ]/ 27.91
= +2.07, abnormally high
z(94) = [ 94 – ]/ 27.91
= -1.66, low side of normal

8. Forward and reverse transforms
x - m
z = s
x - x
z = s
“reverse”
x = m + z s
x = x + z s
Z- score  Raw Score
Raw score  Z-score
population
sample
Example: If population μ = 120 and σ =20
Find the raw score associated with a z-score of 2.5
x = (20)
x =
x = 170

9. Why are z-scores important?
z-scores can be used to describe how normal/abnormal scores within a distribution are
With a normal distribution, there are certain relationships between z-scores and the proportion of scores contained in the distribution that are ALWAYS true.
1. The entire distribution contains 100% of the scores
2. 68% of the scores are contained within 1 standard deviation below and above the mean
3. 95% of the scores are contained within 2 standard deviations below and above the mean

10. m= 128
s = 32
95%
68%
Z-score
What percentage of scores are contained between 96 and 160?
What percentage of scores are between 128 and 160?
If I have a total of 200 scores, how many of them are less than 128?

11. But how do we find areas associated with z-scores that are not simply
Table A in appendix D contains the areas under the normal curve indexed by Z-score.
Z-score
m= 128
s = 32
What proportion of people got a z score of 1.5 or higher?
From these tables you can determine the number
of individuals on either side of any z-score.

12. z-score
1.5

13. Examples of AREA C
2.3
-1.7

14. What percentage of people have a z-score of 0 or greater?
50%
What percentage of people have a z-score of 1 or greater?
15.87%
What percentage of people have a z-score of -2.5 or less?
.62%
What percentage of people have a z-score of 2.3 or greater?
1.07%
What percentage of people have a z-score of -1.7 or less?
4.46%

15. Examples of AREA B

16. What percentage of people have a z-score between 0 and 1?
34.13%
What percentage of people have a z-score between 0 and 2.3?
48.93%
What percentage of people have a z-score between 0 and -2.4?
49.18%
What percentage of people have a z-score between 0 and 1.27?
39.80%
What percentage of people have a z-score between 0 and 1.79?
46.33%
What percentage of people have a z-score between 0 and -3.24?
49.94%

17. Areas which require a COMBINATION of z-scores
What percentage of people have a z-score of 1 or less?
84.13%

18. m= 128
s = 32
Raw Score
What percentage of people have a z-score of -1.7 or less?
4.46%
What percentage of people have a score of 73.6 or less? 4.46%?
What z-score is required for someone to be in the bottom 4.46%?
-1.7
What score is required for someone to be in the bottom 4.46%?
128 + (-1.7)32
73.6 or below

19. What z-score is required for someone to be in the top 25%?
.68
What z-score is required for someone to be in the top 5%?
1.65
What z-score is required for someone to be in the bottom 10%?
-1.29
What z-score is required for someone to be in the bottom 70%?
.52
What z-score is required for someone to be in the top 50%?
What z-score is required for someone to be in the bottom 30%?
-.53

20. m= 128
s = 32
What percentage of scores fall between the mean and a score of 132?
Here, we must first convert this raw score to a z-score in order to be able to use what we know about the normal distribution. ( )/32 = 0.125, or rounded, 0.13.
Area B in the z-table indicates that the area contained between the mean and a z-score of .13 is .0517, which is 5.17%

21. m= 128
s = 32
What percentage of scores fall between a z-score of -1 and 1.5?
If we refer to the illustration above, it will require two separate areas added together in order to obtain the total area:
Area B for a z-score of -1:
Area B for a z-score of 1.5:
Added together, we get .7745, or 77.45%

22. m= 128
s = 32
What percentage of scores fall between a z-score of 1.2 and 2.4?
Notice that this area is not directly defined in the z-table. Again, we must use two different areas to come up with the area we need. This time, however, we will use subtraction.
Area B for a z-score of 2.4:
Area B for a z-score of 1.2:
When we subtract, we get , which is 10.69%

23. m= 128
s = 32
If my population has 200 people in it, how many people have an IQ
below a 65?
First, we must convert 65 into a z-score: (65-128)/32 = , rounded = -1.97
Since we want the proportion BELOW -1.97, we are looking for Area C of a z-score of 1.97 (remember, the distribution is symmetrical!) : = 2.44%
Last step: What is 2.44% of 200?
200(.0244) = 4.88

24. m= 128
s = 32
What IQ score would I need to have in order to make it to the top 5%?
Since we’re interested in the ‘top’ or the high end of the distribution, we want to find an Area C that is closest to .0500, then find the z-score associated with it.
The closest we can come is (always better to go under). The z-score associated with this area is 1.65.
Let’s turn this z-score into a raw score: (32) = 180.8

25. √ √ A possible type of test question: 80 - 65 9.80 SS = 2883.2
A class of 30 students takes a difficult statistics exam. The average grade turns out to be 65. Michael is a student in this class. His grade on the exam is 80. The following is known:
9.80
SS =
Assuming that these 30 students make up the population of interest, what is the approximate number of people that did better than Michael on the exam?
SS=
m = 65
N = 30 √
2883.2 30 √ SS N
s = =
= √96.11 = 9.80
z(80) = (80-65)/9.80 = 1.53
Area C for a z score of 1.53 = .0630, so about 6.3%, or 1.89 people