Assessment of Learning 1

1. Haneza E. Farcasa
MEASURE OF VARIABILITY
(QUARTILE AND AVERAGE
DEVIATION)

2. QUARTILE DEVIATION
Indicates the distance we need to go above and
below the median to include the middle 50% of
the scores. It is based on the range of the middle
50% of the scores, instead of the range of the
entire set.
The formula in computing the value of the quartile
deviation is
Q= Q3 – Q1,
2
Where Q is the quartile deviation value, Q1 is
the value of the first quartile and Q3 is the value of
the third quartile.

3. STEPS IN SOLVING QUARTILE
DEVIATION
1. SOLVE FOR THEVALUE OF Q1
2. SOLVE FOR THE VALUE OF Q3
3. SOLVE FOR THE VALUE Q
USING THE FORMULA
QD= Q3 – Q1,
2

4. CLASSES f
71-74 3
75-78 10
79-82 13
83-86 18
87-90 25
91-94 19
95-98 12
n = 100

5.  We will first compute the Q1 and Q3, since only the
frequency distribution is given. Hence,
CLASSES f <cumf
71-74 3 3
75-78 10 13
79-82 13 26 1st
83-86 18 44
87-90 25 69
91-94 19 88 3rd
95-98 12 100

6. Q1 =X1b + (n/4 – cumfb ) c
fQ1
= 78.5 + ( 25 – 13) 4
13
Q1 = 82. 9

7. Q1 =X1b + (3n/4 – cumfb ) c
fQ1
= 90.5 + ( 75 - 69 ) 4
19
Q3 = 91.76

8.  The value of Q can now be obtained. Thus,
Q = Q3 – Q1
2
= 91. 76 – 82. 19
2
Q = 9.57 = 4.78
2

9. AVERAGE DEVIATION
 The average deviation refers to the arithmetic
mean of the absolute deviations of the values
from the mean of the distribution. This measure is
sometimes referred to as the mean absolute
deviation.
AD = Ʃ ǀ x – x ǀ
n
Where x represents the individual values
x is the mean of the distribution

10. EXAMPLE
X: 13, 16, 9, 6, 15, 7, 11
SOLUTION: First, we arrange the values in vertical column and
then we compute the value of the mean.
x
6
7
9
11
13
15
16
-----------
Ʃx = 77

11. X = Ʃx 77
--------- = ------- = 11
n 7
Then, we get the deviations of the individual items
from the mean. Thus,

12. X x- x
6 6-11 = -5
7 7-11= -4
9 9-11= -2
11 11-11 = 0
13 13-11= 2
15 15-11 = 4
16 16-11 = 5

13. Notice that some of the deviations from mean are
negative. Hence, we make an assumption that all
deviations are positive by introducing the absolute
value sign. Adding all these absolute deviations, we
have
x x- x ǀx-xǀ
6 -5 5
7 -4 4
9 -2 2
11 0 0
13 2 2
15 4 4
16 5 5
Ʃǀx-xǀ = 22

14. If we divide the sum of the absolute deviations by n,
then we were able to compute the value of the
average deviation. Hence,
AD = Ʃ ǀ x – x ǀ 22 = 3.14
n 7