1. Measures of Variation and
Dispersion
Statistic
s
Reporters:
Juan Miguel
Bañez
Miguel Fernando
Bilan
Lorenz Angelo
Moya

2. Range
The simplest and crudest measure of
dispersion is the range. This is defined
as the difference between the largest
and the smallest values in the
distribution. If are the values of
observations in a sample, then range
(R) of the variable X is given by:
Formula:
R = H.S. –
L.S.
R = U.B. –
L.B.

3. Examp Class Interval F <CF
le 98 – 100 2 40
* The range of the set of 95 – 97 1 38
scores : 11, 13, 12, 14, 15range92 – 94 of scores : 11,37
* The of the set 1
13, 12, 14, 15
R = U.B. – 89 – 91 6 36
L.B. R = – 88 – L.S. 6
86 H.S. 30
R = 15 – 11 83 15 –
R = – 85 11 5 24
R=4 R=4
80 – 82 9 19
77 – 79 2 10
74 – 76 3 8
74 – 73 5 5
N = 40

4. Quartile
Deviation
The inter-quartile range isspecial
A measure similar to the frequently
range (Q) to the inter-quartile range
reduced is measure of semi-
. It is the difference betweenas the
interquartile range, known the
quartile deviation) (QD), by first
third quartile (Q3 and the dividing it
quartile (Q1). by 2.
Formula:
Q3 – Q1
Q.D. =I.R. = Q –Q
3 1
or I.R.
2
2

5. Mean
Deviation
The mean deviation is an average of
absolute deviations of individual
observations from the central value of
a series. Average deviation about
mean
Formula:
M.D. = ∑ X – X
N

6. Examp
b.) a.)the entries the mean: using X and f X
Add Calculate in column X – the formula
Calculate the mean by
– le x = ∑X means we are going = add
x = ∑fx . This = 10+12+12+14 = 48 to 12
X
the entries in columnthe mean deviation of the
Find Ffx.X
X F N ∙ 4 m– X
X F Xm – X
following ungroup frequency
24 17
MD = ∑f X – 34
distribution: 2.52 42.84
3 X12 36 1.52X = ∑X 18.24
Add the column X X – X
X –X M.D. =∑ X – X
4 19 a. X76 10 12 12 14 N9.88
N 0.52
5 2810 ) 1402 4
0.48 = 47913.44
=b. X 2 3 4 5 6 7 8
141.8
6 1912 ) 1140 1.48M.D. = 428.12
F 17 12 19 28 19 9 2
7 9 12 63 0 2.48106 422.32
106
8 2 14 16 2 3.48X ≈ 4.526.96
=1
MD ≈ 1.32
N = 106 ∑fX = X – X 12.48
∑ 479 ∑f Xm – X =
=4 141.8

7. Standard
Deviation
Standard deviation is the positive
square root of the mean-square
deviations of the observations from
their arithmetic mean.
Formula:
S2 = ∑ (X – X)2
N- 1

8. Examp
a.) The mean price is :
c.) Add the column ( x – x ) and
2 X (x–x) 2
xle
= ∑fX = 90+73+78+79+83+95+77+79+74+82 = 810 =
90 81
sum up the scores.
P 81
73 64
N The price of a 250 gram-soap powder of
10
b.) The range of theleading is 95 –was=recorded from 10
10 a prices brand 73 78 P22 9
supermarkets. The prices (in pesos)
2 = ∑( x – x)2 = 427 79 4
S were:
≈ 47.44 83 4
N – 1 90, 10 78, 79, 83, 95, 77, 196 74,
73, 95 79,
–1 82
77 16
S = √47.44 ≈ 6.89
Find : 79 4
a.) The mean 74
price 49
b.) the range of the prices 0
81
∑(x – x)2 = 427
c.) the standard deviation of the
prices

9. Standard Deviation for
Group Data
The formula used in the computation of the standard deviation
which is the mean deviation method, will be very difficult to
deal with when the mean is a decimal number as shown
example in Mean Deviation. There is an easier way by which
the variance and standard deviation can be derived from the
previously used formula. This method of computing the
variance and standard deviation is called the “raw score
Formula: .
method” For Grouped Frequency
For Ungrouped Data Distribution
S2 = N∑X2 – (∑X)2
N (N – 1) S2 = N∑fXm2 – (∑fXm)2
For Ungrouped FrequencyN (N – 1)
Distribution
S2 = N∑fX2 – (∑fX)2

10. Examp
Calculate the means of the
leandoffxdata.means of the
Calculateentries in column x
Add of data.
set the the
set . 2
2 X X2
X = ∑X = 48 = 12 10 100
Calculate the standard deviation2 using the raw
X N score 4 Fx
F method. X2 12 Fx 144
X = ∑fX = 479 =
Substitute ∑fX17
a. 479 and ∑fX2 =12 14in the 144
= X 10 12 2433
2 34 4.5212 68
4
2 formula. 2 –)(∑fX)= 106(2433) – (479)2
Add N∑fX 12S2 column x2
S = the entries inX 2 3 4 9 5 6 108 196 898 –
2
3 36
b. all the scores
and square 229 441
14 7 257
= 8
N
4 get the ( N F 1) 76 12 19 28 = 304 2 = 584
and )19 –
N sum. 17 106 19 ∑X 2
16∑X 9
5 28 140 2 25 48 700
11130
Substitute ∑X = 48 and = √2.56 ≈ in the
= 2.56 ∑X = 584
6 11130 114
19 2 36 684
S2 =formula.– (∑X)S21.594(584) – (48)2 = 2336 –
N∑X2 =
7 9 2304 63 49 441
N ( N – 1)
8 2 16 64 128
4 ( 4 – 1)
N = 12 ∑fX = √2.67 ≈ ∑fx2 =
≈ 2.67 =
106 1.63
479 2433

11. End of Presentation