X X
z
S
Assume that a certain child has both a
reading and math score. How does this
child’s performance in these two areas
compare?
Reading score: 28
Math score: 14
Reading: 28
mean: 25, St. Dev.: 4
Math: 14
mean: 13 St. Dev.: 1.5
X X
z
S
28  25
Z
4
3
Z
4
Z  .75
Reading: 28
mean: 25, St. Dev.: 4
Math: 14
mean: 13 St. Dev.: 1.5
X X
Z
S
14  13
Z
1 .5
1
Z
1 .5
Z  .67
34.13 %
 = 100
= 15
13.59 %
2.28%
70
85
100
115
130
34.13 %
 = 50
13.59 %
= 7
2.28 %
36
43
50
57
64
 = 35
=3
29
32
35
38
41
What % of the population would
have scores that would fall at or
below 29?
 = 35
=3
2.28%
29
32
35
38
41

X
Z=

29
35
Z=
3
- 6
Z=
3
Z = -2
 = 35
=3
2.28%
29
32
35
38
41
What % of the population would
have scores above 36?
36
35
Z=
3
1
Z=
3
Z = .33
 = 35
=3
29
32
35
38
36
41

If Z = 1.75
Z
a column: Z score
b column: % of area between the
population mean and Z
c column: % of area between the Z
score and the tail of the
distribution.
b
c

If Z = 1.75
Z
-Z
If Z = -.80

b
c
-Z
If Z = -.80

b
c
b
c
-Z

Z
 = 35
=3
29
32
35
38
36
Z - .33
41
Prob. At z .33 = .1293 (col. B)
This means that the percentage of
times that a score would fall
between the mean and a z score of
.33 is 12.93%
Prob at .33 = .3707 (col. C)
37.07% of the population falls
beyond the z score of .33.
What % of the population will
have scores between 33 and 36?
 = 35
=3
29
32
35
33
38
36
41
Step 1: find z score for each
Step 2: use table to find
percentages.
Step 3: add percentages
33
35
Z=
3
- 2
Z=
3
Z = - .66
Prob at .66 = .2454
.2454 + .1293 = .3747
or 37.47 %