Measures of Variations

DefinitionA measures about how far the difference between a value/data from its tendency sentral.Same center, different variation

Absolute dispersion measuresRange.

The range of a set of ungrouped data is the difference between the greatest and the least values of the set.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Range = 14 - 1 = 13 The range for grouped data are found as : R = Xi last class Xi first classR = Last class bloundary first class boundary

2. Variance Average of squared deviations of values from the meanSample variance: Population variance:

Variance for group data:

3. Standart Deviation Squared root of variationSample variance: Population variance:

Variance for group data:1-n)x(xsn1i2i=-=fi

Sample Data (Xi) : 10 12 14 15 17 18 18 24 n = 8 Mean = x = 164.242671261816)(2416)(1416)(1216)(101n)x(24)x(14)x(12)x(10s22222222==--++-+-+-=--++-+-+-=LL

Mean = 15.5 s = 3.338 11 12 13 14 15 16 17 18 19 20 21

11 12 13 14 15 16 17 18 19 20 21Data BData A

Mean = 15.5 s = .925811 12 13 14 15 16 17 18 19 20 21

Mean = 15.5 s = 4.57Data C

Relative dispersion measures1. Coefficient of VariationAlways in percentage (%)Shows variation relative to meanIs used to compare two or more sets of data measured in different units

Population Sample

2. Quartile Coefficient of VariationQ3 Q1CV=Q3 + Q1Exercise Adam and Bonnie are comparing their quiz scores in an effort to determine who is the best. Help them decide by calculating the mean, median, and mode.Adams ScoresBonnies Scores85816085 1058685857290 10080

So, who has the better quiz scores?Now, find the standard deviation for both of them.

AdamBennieMean84.584.5Median8585Mode8585

Deviations from the MeanThe deviation from the mean is the difference between a single data point and the calculated mean of the data.

Data point close to mean: small deviationData point far from mean: large deviationSum of deviations from mean is always zero.Mean of the deviations is always zero.

Deviations from Mean for Adam and Bonnie AdamBennie

Data PointDeviation from Mean8585 - 84.5 = 0.56060 84.5 = -24.5105105 84.5 = 20.58585 84.5 = 0.57272 84.5 = -12.5100100 84.5 = 15.5Sum = 0

Data PointDeviation from Mean8181 - 84.5 = -3.58585 84.5 = 0.58686 84.5 = 1.58585 84.5 = 0.59090 84.5 = 5.58080 84.5 = -4.5Sum = 0

Variance for Adam and Bennies ScoresUsing the deviations from the mean we have already calculated for Adam and Bonnie, we will find the variance for each.

Adam : s =

Bonnie: s =

s = 1417.5 5= 283.5s =65.5 5= 13.1

Standard DeviationTo find the variance, we squared the deviations from the mean, so the variance is in squared units.To return to the same units as the data, we use the square root of the variance, the standard deviation.

Standard DeviationVariance

Adam and Bonnies Standard Deviation

Adam:

Bennie:

Based on the standard deviation, Bonnies scores are better because there is less dispersion. In other words, she is more consistent than Adam.