Assignment 7 Assignment 7
Name:Grace Lin
Class:碩專二甲
Number:Na2c0023
P.137 APPLICATION EXERCISES
A1.--- According to the figure above, an IQ score of 85 would represent 16 percentile
score.
A2.--- About 81.85 persentage of students scored between 70 and 115.
34.13+34.13+13.59= 81.85
A3-1--- About 5 standard deviations would Iliana be above the mean.
177-100=77 77⌯15=5.13
A3-2--- It depends on what kind of test did she take. If the test is Wechsler
Intelligence Scale Fourth Edition, it is a formal norm referenced test, it means
that Iliana has got the figure of mental giftedness. She is really intelligent
when she had a score of 177. Otherwise, we don’t know what kind of test it is
or what is the detail of the test, it is not enough to judge whether Iliana is
Intelligent or not. The only thing we can say is that Iliana had got a high score.
Her score is almost the highest.
A4-1--- Iliana’s z score wuld be 5
(177-100) ⌯15=5.13
A4-2--- Iliana’s T score wuld be 101
(10x5.13)+50=101.3
A4-3--- Iliana’s CEEB score wuld be 1013
(100x5.13)+500=1013
Student
Raw score
z score
T score
CEEB score
A
64
c
70
h
B
50
d
e
i
C
a
-1.0
F
j
D
b
-1.5
g
350
(10c)+50=70
c=2
e=(10d)+50
=(100)+50
=50
e=50
64-50=14 --- 2 (standard deviation)
c-2=d
2-2=0
d=0
f=(10-1)+50
=40
f=40
50-1(standard deviation)
=50-7
=43
g=(10-1.5)+50
=35
g=35
a=43
h=(100c)+500
50-1.5(standard deviation)
=50-10.5
=39.5
b=39.5
=(1002)+500
=700
h=700
i=(100d)+500
=(1000)+500
=500
i=500
j=(100-1)+500
=-100+500
=400
j=400
Student
Raw score
z score
T score
CEEB score
A
64
+2
70
700
B
50
0
50
500
C
43
-1.0
40
400
D
39.5
-1.5
35
350
Test
k
Raw scores
M
S
Standardized scores
M
S
A
110
60
25
500
100
B
75
60
15
50
10
C
50
11
4
0
1
k = number of items on the test
C1
a.---z scores = Test C
(-3 ≤ z ≤ 3)
b.---T scores = Test B
(20 ≤ T ≤ 80)
c.---CEEB scores = Test A
(200 ≤ CEEB ≤ 800)
C2
a.--- Test A has the largest standard deviation: 25.
b.--- Test C has the lowest mean: 11.
c.--- Test B has a negatively skewed distribution
Test A has a normal skewed distribution.
Test C has a positively skewed distribution.
C3
a.--- a raw score of 11 equals a z of 0
The raw score of 11 = mean = a z of 0
b.--- a raw score of 7 equals a T score of 40
(7-11) ⌯4=-1
T=(10-1)+50
=40
c.--- a raw score of 19 equals a CEEB score of 700
(19-11) ⌯4=2
CEEB=(1002)+500
=700
Student
Raw Score
z
T
CEEB
Shenan
77
2.07
71
707
Robert
75
1.55
65
655
Randy
72
0.77
58
577
Mitsuko
72
0.77
58
577
Millie
70
0.26
53
526
Kimi
70
0.26
53
526
Kazumoto
69
0.00
50
500
Kako
69
0.00
50
500
Joji
69
0.00
50
500
Jeanne
69
0.00
50
500
Issaku
68
-0.26
47
474
Iliana
68
-0.26
47
474
Dean
67
-0.52
45
448
Corky
64
-1.29
37
371
Bill
64
-1.29
37
371
Archie
61
-2.07
29
293
Mean
69
0
50
500
3.87
1.00
10.00
100.00
SD
Student
Mean
SD
Mode
1
2
3
4
5
6
7
8
9
Raw Score
92
93
86
83
40
70
90
24
99
10
11
12
z
T
0.43
0.48
0.11
-0.05
-2.30
-0.73
0.32
-3.14
0.79
54
55
51
50
27
43
53
19
58
CEEB
543
548
511
495
270
427
532
186
579
83
99
88
-0.05
0.79
0.22
50
58
52
495
579
522
13
14
15
16
17
18
19
80
100
90
97
97
96
90
-0.20
0.85
0.32
0.69
0.69
0.64
0.32
48
58
53
57
57
56
53
480
585
532
569
569
564
532
20
21
22
23
24
85
94
87
100
50
0.06
0.53
0.16
0.85
-1.78
51
55
52
58
32
506
553
516
585
322
83.88
19.08
90
62
-0.09
1.08
49.08
10.81
490.83
108.12
midpoint
90
median
-2.03
Skewness
3.72
Kurtosis
The skewness = -2.03
Midpoint&lt; Mean&lt; Median= Mode
Kurtosis=3.72
The data of the test in my class is not normally distributed. It is a negatively skewed
distribution. The Kurtosis is Leptokurtic.