Measures of Spread
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The document discusses various statistical concepts including range, mean deviation, variance, and standard deviation. It provides formulas and steps to calculate each measure. The range is the distance between the highest and lowest values. Mean deviation measures the average deviation from the mean. Variance is the average of the squared deviations from the mean and standard deviation is the square root of the variance, representing the average distance from the mean. Examples are given to demonstrate calculating each measure for both ungrouped and grouped data.
Measures of Spread
- 5. Importance of Range Tells you the distance from the smallest to largest.
- 6. Ungrouped Data R= HS-LS Whereas: R- Range HS- Highest score LS- Lowest score
- 7. Find the range of the two groups of score distribution. Group A Group B 10 (LS) 15 (LS) 12 16 15 16 17 17 25 17 26 23 28 25 30 26 35 (HS) 30 (HS)
- 8. Group A RA= HS- LS RA =35-10 RA =25 Group B RB= HS-LS RB= 30-15 RB= 15
- 9. Grouped Data R= HSUB-LSLB Whereas: R- Range HSUB – Upper boundary of highest score LSLB – Lower boundary of lowest score
- 10. Find the value of range of the scores of 50 students in Mathematics achievement test x f 25-32 3 33-40 7 41-48 5 49-56 4 57-64 12 65-72 6 73-80 8 81-88 3 89-97 2 n=50
- 11. LS= 25 LSLB= 24.5 HS= 97 HSUB= 97.5 R= HSUB-LSLB R=97.5-24.5 R= 73
- 12. end
- 15. Mean Deviation Measures the average deviation of the values from the arithmetic mean. It gives equal weight to the deviation of every score in the distribution.
- 16. A. Mean Deviation for Ungrouped Data MD = Where, MD = mean deviation value X = individual score = sample mean N = number of cases
- 17. Steps in Solving Mean Deviation for Ungrouped Data 1. Solve the mean value. 2. Subtract the mean value from each score. 3. Take the absolute value of the difference in step 2. 4. Solve the mean deviation using the formula MD =
- 18. Example 1: Find the mean deviation of the scores of 10 students in a Mathematics test. Given the scores: 35, 30, 26, 24, 20, 18, 18, 16, 15, 10
- 20. Analysis: The mean deviation of the 10 scores of students is 6.04. This means that on the average, the value deviated from the mean of 212 is 6.04.
- 22. 1. Solve for the value of the mean. 2. Subtract the mean value from each midpoint or class mark. 3. Take the absolute value of each difference. 4. Multiply the absolute value and the corresponding class frequency. 5. Find the sum of the result in step 4. 6. Solve for the mean deviation using the formula for grouped data.
- 23. Example 2: Find the mean deviation of the given below.
- 24. Analysis: The mean deviation of the 40 scores of students is 0.63. This means that on the average, the value deviated from the mean of 33.63 is 10.63.
- 26. is the square of the standard deviation. In short, having obtained the value of the standard deviation, you can already determine the value of the variance.
- 27. One of the most important measures of variation.It shows variation at the mean.
- 29. How to Calculate the Variance for Ungrouped Data 1. Find the Mean. 2. Calculate the difference between each score and the mean. 3. Square the difference between each score and the mean.
- 30. How to Calculate the Variance for Ungrouped Data 4. Add up all the squares of the difference between each score and the mean. 5. Divide the obtained sum by n – 1.
- 31. Find the Variance 35 35 35 35 35 35 210 Mean= 35 73 11 49 35 15 27 210 Mean= 35
- 32. x x-ẋ (x-ẋ)2 35 0 0 35 0 0 35 0 0 35 0 0 35 0 0 35 0 0 ∑(x-ẋ)2 0 x x-ẋ (x-ẋ)2 73 38 1444 11 -24 576 49 14 196 35 0 0 15 -20 400 27 -8 64 ∑(x-ẋ)2 2680 x x-ẋ (x-ẋ)2 35 0 0 35 0 0 35 0 0 35 0 0 35 0 0 35 0 0 ∑(x-ẋ)2 0 Mean=35
- 35. 1. Calculate the mean. 2. Get the deviations by finding the difference of each midpoint from the mean. 3. Square the deviations and find its summation. 4. Substitute in the formula.
- 36. Class Limits (1) F (2) Midpoint (3) FMp (4) _ X _ Mp - X _ (Mp-X)2 _ f( Mp-X)2 28-29 4 28.5 114.0 20.14 8.36 69.89 279.56 26-27 9 26.5 238.5 20.14 6.36 40.45 364.05 24-25 12 24.5 294.0 20.14 4.36 19.01 228.12 22-23 10 22.5 225.0 20.14 2.36 5.57 55.70 20-21 17 20.5 348.5 20.14 0.36 0.13 2.21 18-19 20 18.5 370.0 20.14 -1.64 2.69 53.80 16-17 14 16.5 231.0 20.14 -3.64 13.25 185.50 14-15 9 14.5 130.5 20.14 -5.64 31.81 286.29 12-13 5 12.5 62.5 20.14 -7.64 58.37 291.85 N= 100 ∑fMp= 2,014.0 ∑(Mp-X)2= 1,747.08
- 40. Standard Deviation •Is the most important measures of variation. •It is also known as the square root of the variance. •It is the average distance of all the scores that deviates from the mean value.
- 44. 1. Solve the mean value. 2. Subtract the mean value from each score. 3. Square the difference between the mean and each score. 4. Find the sum of step 3. 5. Solve for the population standard deviation or sample standard deviation using the formula for ungrouped data.
- 45. EXAMPLE X X - X (X – X)² 19 4.4 19.36 17 2.4 5.76 16 1.4 1.96 16 1.4 1.96 15 0.4 0.16 14 -0.6 0.36 14 -0.6 0.36 13 -1.6 2.56 12 -2.6 6.76 10 -4.6 21.16 Ʃx= 146 Ʃ(x-x)²= 60.40 x= 14.6
- 51. Steps in solving the STANDARD DEVIATION of GROUP DATA 1. Solve the mean value. 2. Subtract the mean value from each midpoint or class mark. 3. Square the difference between the mean value and midpoint or class mark. 4. Multiply the squared difference and the corresponding class frequency. 5. Find the sum of the results in step 4. 6. Solve the population standard deviation or sample standard deviation using the formula for grouped data.
- 52. EXAMPLE x f x 15-20 3 17.5 52.5 33.7 -16.2 262.44 787.32 21-26 6 23.5 141 33.7 -10.2 104.04 624.24 27-32 5 29.5 147.5 33.7 -4.2 17.64 88.2 33-38 15 35.5 532.5 33.7 1.8 3.24 48.6 39-44 8 41.5 332 33.7 7.8 60.84 486.72 45-50 3 47.5 142.5 33.7 13.8 190.44 571.32 n= 40 = 1348 Ʃf (Xm-X)²= 2606.4