Measures of Variability

The measure of variability is a single number that describes how the data are scattered or how much they are bunched. It is also called the measure of dispersion or measure of spread. Closely grouped data will have relatively small values and more widely distributed data will have larger values of these measures.

Measures of Variability

Range is the least complicated measure of describing the dispersion of a set of numbers. It is the distance given by the highest observed value minus the lowest observed value in the distribution.

Range

Mean deviation (MD)- is the average distance between the mean and the scores in the distribution. This technique provides a reasonably stable estimate of variation. It is also called average deviation

Mean Deviation

Variance (σ2) – is the sum of the squares of the distribution from the mean. It is expected value of the square of the deviation from the mean.

Variance

:

Standard deviation (σ)- Is derived from the positive square root of the variance. It has been termed so because it provides a standard unit of measuring distance of various scores from the mean.

Standard Deviation

Semi-inter-quartile range (Q) - is half of the difference between P75 (or Q3) and the P25 (or Q1) in the distribution.

Semi-Inter-Quartile Range

1.Range = Highest Observed Value – Lowest Observed Value

RANGE = HOV-LOV

A. Measures of variability for ungrouped data

EXAMPLE:

GIVEN THE FOLLOWING TEST SCORES: 83, 79, 88, 82, 78, 88, AND 84. FIND THE RANGE.

SOLUTION: RANGE= 88 - 78 = 10

2. MEAN DEVIATION

MD = ∑ lx– mean l n

From the example X X lX-meanl 83 82.8571 0.1429 79 3.8571 86 3.1429 solution: 82 0.8571 78 4.8571 MD = 19.1429 88 5.1429 7 84 1.1429 MD = 2.7347 19.1429

Variance a.) σ² =∑(X-mean)² n-1 b.) σ² = n(∑X²)-(∑X)² n (n-1)

X mean lX-meanl ² X² 83 82.8571 0.0204 6,889 79 14.8772 6,241

86 9.8778 7,396 82 0.7346 6,724 78 23.5914 6,084 88 26.4494 7,744 84 1.3062 7,056 580 76.8572 48,134

Solutions: a.) σ²=76.8572b.) σ²=7(48,134)-(580)² 7-1 7(7-1) =76.8572 =336,938-336,400 6 7(6) σ² = 12.8095 = 538 42 σ² = 12.8095

4. standard deviation a.) σ=√∑ (X-mean) or σ= √n(∑X)-(∑x) n-1 n(n-1)

For the same example: a.)σ=√ 76.8572 7-1 =√76.8572 6 =√12.8095

σ =3.5790

b.)σ=√7(48,134)-(580)² 7(7-1) =√336,938-336,400 7(6) =√538 42 =√12.8095

σ = 3.5790

Semi-inter-quartile range

a.) Q= Q₃-Q₁ or Q= P₇₅-P₂₅ 2 2

For example: Q₃=P₇₅= 87 solution: Q₁=P₂₅=79 Q=87-79 2 Q= 4

Measures of variability for grouped data

range= Highest Real Limit-Lowest Real Limit

range=HLR-LLR

Range

Example: class intervals f 62 - 68 8 solution:

69 - 75 5 76 – 82 9 range= 117.5 -

61.5 83 - 89 11 90 - 96 12 range= 56 97 - 103 10 104 - 110 5 111 – 117 5 n=65

2.) MEAN DEVIATION MD= ∑f lx-meanl where:

n f= frequency x= midpoint n= sample size

class intervals f x mean lx-meanl f lx-meanl 62-68 8 65 88.6 23.6 188.8 69-75 5 72 16.6 83 76-82 9 79 9.6 86.4 83-89 11 86 2.6 28.6 90-96 12 93 4.4 52.8 97-103 10 100 11.4 114 104-110 5 107 18.4 92 111-117 5 114 25.4 127 n=65 772.6

solution:

MD= 772.6 65 MD= 11.8862

Variance a.) σ²= w²[n∑fd² -∑fd] n(n-1)

b.) σ²= n(∑x²f)-(∑fx)² n(n-1)

class intervals f d fd d² fd²

62-68 8 -4 -32 16 128 69-75 5 -3 -15 9 45 76-82 9 -2 -18 4 36 83-89 11 -1 -11 1 11 90-96 12 0 0 0 0 97-103 10 1 10 1 10 104-110 5 2 10 4 20 111-117 5 3 15 9 45 n=65 -41 295

Solution: by formula(a)

σ²=(7)²[65(295)-(-41)²] 65(65-1)

= 49[19,175-1,681] 65(64) =49 x 17,494 4160 σ²= 206.0591

class intervals f x x² x²f fx

62-68 8 65 4,225 33,800 520 69-75 5 72 5,184 25,920 360 76-82 9 79 6,241 56,169 711 83-89 11 86 7,396 81,356 946 90-96 12 93 8,649 103,788 1,116 97-103 10 100 10,000 100,000 1,000 104-110 5 107 11,449 57,245 535 111-117 5 114 12,996 64,980 570 n-65 523,258 5,758

solution

By formula (b): σ²= 65(523,258)-(5,758)² 65(65-1) = 34,011,770 - 33,254,564 65(64) = 857,206 4160 σ²= 206.0591

4.)standard deviation

a.) σ=w√n∑fd²-(∑fd)²

n(n-1)

b.) σ=√n(∑x²f)-(∑fx)²

n(n-1)

For the same example,

σ=(7)√ 19,175-1,681 σ=√65(523,258-5,758)

65(64) 65(65-1)

=(7)√17,494 =√34,011,770-33,154,564

4,160 65(64)

=(7)√4.2053 =√857,206

4,160 σ =14.3548 =√206.0591 σ=14.2548

5. semi-inter-quartile range (a) Q= Q₃-Q₁ (b) Q= P₇₅-P₂₅ 2 2

From the example, class interval f F≤ 62-68 8 8 69-75 5 13 76-82 9 22 83-89 11 33 90-96 12 45 97-103 10 55 104-110 5 60 111-117 5 65 n=65

Q₃=P₇₅= 99.125 Q₁=P₂₅=78.0278 solution: Q= 99.125-78.0278 2 = 21.0972 2 Q= 10.5486