Measure of variability
Download as PPTX, PDF7 likes937 views
This document discusses two measures of variability: quartile deviation and average deviation. Quartile deviation indicates the distance above and below the median needed to include the middle 50% of scores. It is calculated by taking the difference between the third (Q3) and first (Q1) quartiles and dividing by 2. Average deviation is the mean of the absolute deviations from the mean. It is calculated by taking the sum of the absolute value of each score's deviation from the mean and dividing by the total number of scores. The document provides examples of calculating both measures using sample data sets.
Measure of variability
- 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