PROBLEM SET NO. 3 Prepare a step distribution using the ungrouped scores listed below and solve for the following: Note: Show the procedure in preparing the step distribution. 1. mean X = 2. median, Md = 3. mode, Mo = 4. P%= 5. P10= 6. D =Pgo - Pio 7. P75= 8. P25 = 9. QD= 87 61 78 93 85 89 72 61 78 67 68 90 80 58 gc 59 97 4 24 54 70 86 67 60 68 56 55 86 64 53 71 73 77 92 76 72 89 58 72 72 89 58 72 62 62 48 47 79 67 SHOW YOUR SOLUTIONS 9 Ver 11212020 PROBLEM SET NO. 4 Frequency 1. mean X = 85-89 2. median, Md = 3. mode, Mo = 80-84 4. Poo= 75-79 2 5. PIO= 70-74 3 6. D = PD = 65-69 15 7. P75= 60-64 8 8. P25 = 55-59 9. QD= 6 50-54 45-49 5 40-44 35-39 SHOW YOUR SOLUTIONS\f79 68 56 74 54 64 58 66 52 25 72 55 65 74 72 38 64 72 56 76 66 67 60 55 69 40 72 38 45 15 48 58 53 48 55 29 32 48 66 55 67 49 48 47 45 54 32 MEASURES OF VARIABILITY: Spread or variation of scores: It can identify whether the data is homogeneous or heterogeneous. 1. Range: Highest Score - Lowest Score 2. Percentile Deviation (D) = Poo - Pro 3. Quartile Deviation (QD) = P75 - P25 = Q3 - Q1 2 Note: median = Pso Using the ungrouped scores above, change it into grouped scores (Step distribution) Procedure: 1. Get the Range HS - LS = 98-14 = 84 2. Divide the R by 10 or 15 (For this data, use 10) 84 =8.4~8 10 6 Ver 11212020 3. Round off to the nearest ODD number Note: Since 8 is an even number, you may use 7 or 9: 9 is closer to 8.4. . use 9 as the class interval (CI) 4. Lowest Step = Lowest score + (CI-1), I is constant Lowest step = 14 + (9-1) = 22 &: 14 22 Step Distribution Class Interval Tally Freq cf Midpoint f Md pt 95 - 103 11 2 50 99 99x2 = 198 86 - 94 w 48 90 90x3 = 270 77 - 85 9 45 81 68 - 76 36 72 59 - 67 1H - 1 33 63 50 - 58 27 54 41 -49 1H - 11 22 45 32 - 40 TH - 1I 15 37 23 - 31 28 14 -22 18 N = 50 E = 2819 Note: Cf = Cumulative Frequency (Add on the frequencies) start from the bottom f = 4 : 4+4 = 8, 8+7 = 15..... Solve for: 1. Mean (x) = f Md.pt : Mid pt = 95+103 = 99 Mean = 2819 = 56.38 N 2 50 2. Median = lower limit + J (CI): J= N - CI 2 ZN. f= frequency 2 Md: J = 50 - 22 = 25 - 22 CI = Class Interval 2 Md: 49.5 + 25 - 229 = 54.9 5 3. Mode: Mo = LL+ di CI; di =9-3=6 di + dz d2=9-3=6 = 76.5+ 6 9= 81 6+6_