Mathematics Measures of Dispersion PDF

MATHEMATICS AS A TOOL I: Measures of Dispersion Measures of Dispersion /Variability  Are measures of the average distance of each observation from the center of the distribution.  Describe how spread the individual values are from the mean.  Measure the homogeneity or hete...

MATHEMATICS AS A TOOL I: Measures of Dispersion Measures of Dispersion /Variability  Are measures of the average distance of each observation from the center of the distribution.  Describe how spread the individual values are from the mean.  Measure the homogeneity or heterogeneity of a particular group.  Among these measures are the range, variance, standard deviation and coefficient of variation. A small measure of variability would A BIG MEASURE OF VARIABILITY WOULD indicate that the data are: INDICATE THAT THE DATA ARE: 1. Clustered closely around the mean 1. FAR AWAY FROM THE MEAN 2. More homogeneous 2. HETEROGENEOUS 3. Less variable 3. MORE VARIABLE 4. More consistent 4. LESS CONSISTENT 5. More uniformly distributed 5. LESS UNIFORMLY DISTRIBUTED Range (R): The difference between the highest value and the lowest value in a set of data. Formula: R = HV – LV Variance (2 for population and s2 for sample): The square of standard deviation. Formula: For Ungrouped Data: ∑(𝑥−𝑥̅ )2 Population : 𝜎 2 = and 𝑁 ∑(𝑥−𝑥̅ )2 Sample : 𝑠2 = 𝑛−1 For Grouped Data: ∑ 𝑓(𝑥−𝑥̅ )2 ∑ 𝑓(𝑥−𝑥̅ )2 𝜎2 = and 𝑠2 = 𝑁 𝑛−1 Where: x is the class mark 𝑥̅ is the mean f is the frequency Standard Deviation ( for population and s for sample): The square root of the average deviation from the mean or simply the square root of the variance. Formula: For Ungrouped Data: ∑(𝑥−𝑥̅ )2 Population Standard Deviation: 𝜎 = √ and 𝑁 ∑(𝑥−𝑥̅ )2 Sample Standard Deviation: 𝑠=√ 𝑛−1 For Grouped Data: ∑ 𝑓(𝑥−𝑥̅ )2 Population: 𝜎=√ 𝑁 ∑ 𝑓(𝑥−𝑥̅ )2 Sample : 𝑠=√ 𝑛−1 Coefficient of Variation (CV): It is used to compare the variability of two or more sets of data having different units. Formula: CV = standard deviation divided by mean 𝑠 𝐶𝑉 = 𝑥̅  100% Example: The following are the sets of grades in Math of two groups consisting of five students: Male group: 100 65 75 85 95 Female Group: 84 86 85 82 83 a. Find the range, variance, standard deviation and coefficient of variation. b. Which group is more homogeneous in their Math ability? c. Which group has a greater variability of grades in Math? Solution: Male Group Female Group Range Range R = Highest Value - Lowest Value R = Highest Value - Lowest Value R = HL – LV R = HL – LV = 100 – 65 = 86 – 82 R = 35 R=4  Variance Variance x (𝑥 − 𝑥̅ ) (𝑥 − 𝑥̅ )2 x (𝑥 − 𝑥̅ ) (𝑥 − 𝑥̅ )2 65 -19 361 82 -2 4 75 -9 81 83 -1 1 85 1 1 84 0 0 95 11 121 85 1 1 100 16 256 86 2 4 ∑ 𝑥 = 420 ∑(𝑥 − 𝑥̅ )2 ∑ 𝑥 = 420 ∑(𝑥 − 𝑥̅ )2 = 820 = 10 ∑𝒙 𝟒𝟐𝟎 ∑𝒙 𝟒𝟐𝟎 ̅= Mean: 𝒙 = = 𝟖𝟒 ̅= Mean: 𝒙 = = 𝟖𝟒 𝒏 𝟓 𝒏 𝟓 2 ∑(𝑥 − 𝑥̅ )2 820 820 2 ∑(𝑥 − 𝑥̅ )2 10 10 𝑠 = = = 𝑠 = = = 𝑛−1 5−1 4 𝑛−1 5−1 4 𝒔𝟐 = 𝟐𝟎𝟓 𝒔𝒒𝒖𝒂𝒓𝒆 𝒖𝒏𝒊𝒕𝒔 𝒔𝟐 = 𝟐. 𝟓 𝒔𝒒𝒖𝒂𝒓𝒆 𝒖𝒏𝒊𝒕𝒔 Standard Deviation Standard Deviation ∑(𝑥 − 𝑥̅ )2 820 ∑(𝑥 − 𝑥̅ )2 10 𝑠=√ =√ = √205 𝑠=√ = √ = √2.5 𝑛−1 4 𝑛−1 4 𝒔 = 𝟏𝟒. 𝟑𝟐 𝒖𝒏𝒊𝒕𝒔 𝒔 = 𝟏. 𝟓𝟖 𝒖𝒏𝒊𝒕𝒔 Coefficient of Variation Coefficient of Variation 𝑠 14.32 𝑠 1.58 𝐶𝑉 =  100% = = 0.1705(100) 𝐶𝑉 =  100% = = 0.0188(100) 𝑥̅ 84 𝑥̅ 84 CV = 17.05% CV = 1.88% b. Which group is more homogeneous in their Math ability? Female Group c. Which group has a greater variability of grades in Math? Male Group Measures of Variability for GROUPED DATA Find the mean, variance and standard deviation for the ff. data. Grouped Frequency Distribution for the Entrance Examination Scores of 60 students Class intervals f 54 – 59 1 48 – 53 3 42 – 47 8 36 – 41 14 30 – 35 17 24 – 29 11 18 – 23 6 N=60 Solution: Class Intervals f x fx (𝒙 − 𝒙̅) ̅) 𝟐 (𝒙 − 𝒙 ̅)𝟐 f(𝒙 − 𝒙 54 – 59 1 56.5 56.5 22 484 484 48 – 53 3 50.5 151.5 16 256 768 42 – 47 8 44.5 356 10 100 800 36 – 41 14 38.5 539 4 16 224 30 – 35 17 32.5 552.5 -2 4 68 24 – 29 11 26.5 291.5 -8 64 704 18 – 23 6 20.5 123 -14 196 1176 N=60 ∑ 𝑓𝑋 ̅) 𝟐 ∑ 𝐟(𝒙 − 𝒙 = 2,070 = 4,224 Mean: ∑ 𝒇𝒙 𝟐, 𝟎𝟕𝟎 ̅= 𝒙 = = 𝟑𝟒. 𝟓 𝑵 𝟔𝟎 Variance: Population Variance: Sample Variance: ∑ 𝑓(𝑥−𝑥̅ )2 ∑ 𝑓(𝑥−𝑥̅ )2 𝜎2 = 𝑠2 = 𝑁 𝑛−1 4,224 4,224 4,224 = = 60−1 = 60 59 𝜎 2 = 70.4 𝑠 2 = 71.59 Standard Deviation: ∑ 𝑓(𝑥−𝜇)2 ∑ 𝑓(𝑥−𝑥̅ )2 Population: 𝜎=√ Sample : 𝑠=√ 𝑁 𝑛−1 = √70.4 = √71.59 𝝈 = 𝟖. 𝟑𝟗 𝒔 = 𝟖. 𝟒𝟔