measure of location or fractiles report in mmw.pptx

measure of location or fractiles- A VALUES BELOW WHICH A SPECIFIED FRACTION OR PERCENTAGE OF OBSERVATION IN A GIVEN SET MUST FALL.

1. Measures of location or fractiles

2. Measures of location or fractiles A VALUES BELOW WHICH A SPECIFIED FRACTION OR PERCENTAGE OF OBSERVATION IN A GIVEN SET MUST FALL.

3. A. PERCENTILES- V alues that divide a set of observation into 100 equal parts. P 1 (read as first percentile)- it is the value which 1% of the values fall P n – is the value below which 99% of the values fall. To compute for the nth percentile P 1 - the value of the observation of the array  

4. Example: The ff. were the scores of 8 students in a quarts . 4,3, 5, 6, 8, 6, 7, 9. Find the 70 th percentile. Solution: First, we need to arrange the data from lowest to highest. That is, 3, 4, 5, 6, 6, 7, 8, 9. Then, substitute: P 70 = = = = = 6.3   Therefore, the 70 th percentile is 7 which interpreted as 70% of the scores are below 7

5. B. Deciles- values that divide the array into 10 equal parts. D 1 -is the value below which 10% of the values fall D 9 - is the value below which 20% of the values fall

6. C. Quartiles- are values that divide the array into 4 equal parts. Q 1 -(read as first quartile) is the value below which 25% of the values fall. Q 2 - is the value below which 50% of the values fall. Q 3 - is the value below which 75% of values fall.

7. Try this to answer Calculate Q 1 , Q 2 , Q 3 , D 1 , D 4 , D 5 , D 7 , P 10 , P 25 , P 50 and P 70 for the following IQ scores: 87 90 95 96 97 98 98 99 100 100 100 100 100 101 101 102 102 102 103 104 105 107 110

8. CLASS LIMIT FREQUENCY CUMULATIVE FREQ. Cf < 46-48 1 35 43-45 1 34 40-42 2 33 37-39 3 31 34-36 3 28 31-33 4 25 28-30 7 21 25-27 5 14 22-24 3 9 19-21 2 6 4 16-18 2 4 13-15 1 2 10-12 1 1 For group data. Find Q1 , Q2 , Q3 , D1 , D4 , D5 , D7 , P10 , P25 , P50

9. Measures of dispersion

10. Measures of dispersion Central tendency measures do not reveal the variability person in the data. Dispersion is measure of variation. Dispersion is the scattered nets of the data series around it average. Dispersion is the extent in which values in a distribution differ from the average of the distribution.

11. Absolute Measure and Relative Measure Absolute Measure- Measure of the dispersion in the original unit of the data. Variability distribution can be compared provided they are given in the same unit and have the same average Relative Measure- It is the ratio of absolute measures and unit free measurement. This ratio is known as coefficient of absolute dispersion. JDP-CM-SMBT

12. 4.1.1 The Range The range of a set of measurements is the difference between the largest and the smallest values. Range (R) = maximum value - minimum value Example: The IQ scores of 5 members of the Morales' family are 108,112,127,116, and 113. Find the range. Solution: Range R=127-108=19.

13. Mean Deviation(MD) & Coefficient of MD Mean deviation It is the average of the absolute values of the deviation from the mean. Mean deviation taken from mean is given as For ungrouped data:- Mean deviation (MD) = For grouped data:- Mean deviation (MD) = where x ̅= mean f=frequency Coefficient of Mean Deviation Coe. of MD=   JDP-CM-SMBT

14. Mean deviation & Coe. of MD:- x̅ = = =82 Mean deviation (MD) = = 4.5 Coe. of MD= = = 0.0549   X 𝑥−x̅ 𝐼𝑥−x̅ 𝐼 90 8 8 82 80 -2 2 92 10 10 80 -2 2 72 -10 10 78 -4 4 82 = 656   𝛴𝐼𝑥−x̅ 𝐼 =36 X 𝑥−x̅ 𝐼𝑥−x̅ 𝐼 90 8 8 82 80 -2 2 92 10 10 80 -2 2 72 -10 10 78 -4 4 82   𝛴𝐼𝑥−x̅ 𝐼 =36 JDP-CM-SMBT

15. The Standard Deviation and the Variance Formula:

16. Example: A sample of 5 households showed the following number of household members: 3, 8, 5, 4, and 4. Find the standard deviation. Solution: ΣΧ² = 3² +8 2 +5 2 +4 2 +4 2 = 130 =(3+8+5+4+4)=(24)=576 Approximating The Standard Deviation From A Frequency Distribution where f= frequency of the ith class x 1 = class mark of the ith class = mean of the frequency distribution n= total number of observations  

17. Approximating The Variance From A Frequency Distribution or, using the computational formula where f= frequency of the ith class x=class mark of the ith class = mean of the frequency distribution n= total number of observations  

18. Example: (Refer to Scores of 110 Students in an Achievement Test) CI f cm(x) f1x1 f1 50-54 10 52 520 27040 -22 55 - 59 3 57 171 9747 -17 60 - 64 8 62 496 30752 -12 65-69 13 67 871 58357 -7 70 - 74 17 72 1224 88128 -2 75 - 79 19 77 1463 112651 3 80-84 22 82 1804 147928 8 CI f cm(x) f1x1 50-54 10 52 520 27040 -22 55 - 59 3 57 171 9747 -17 60 - 64 8 62 496 30752 -12 65-69 13 67 871 58357 -7 70 - 74 17 72 1224 88128 -2 75 - 79 19 77 1463 112651 3 80-84 22 82 1804 147928 8 85-89 13 87 1131 98397 13 90-94 4 92 368 33856 18 95-99 1 97 97 9409 23 Total 110 8145 616265

19. .99