![7
CLASS INTERVAL
Range HS-LS
Highest Score 71
Lowest Score 60
11
Range
]=
11 ≈ 2
√N 5
√N = √30 = 5.477225575 ≈ 5](https://image.slidesharecdn.com/standarddeviationandvariance-200313133752/85/Standard-Deviation-and-Variance-7-320.jpg)
Standard Deviation and Variance
- 2. 1) Range — Difference between max & min Formula: Highest Value – Lowest Value 2) Deviation— Difference between entry & mean 3) Variance—Sum of differences between entries and mean, divided by population or sample Formula: 4) Standard Deviation —Square root of the variance Formula: DEFINITION OF TERMS 2 Σ𝑓 𝑋 − 𝑥 ² 𝑛 − 1 Σ𝑓 𝑋 − 𝑥 ² 𝑛 − 1 Bessel’s Correction • the use of n − 1 instead of n in the formula for the sample variance and sample standard deviation, where n is the number of observations in a sample. • This method corrects the bias in the estimation of the population variance.
- 4. Data: 17, 15, 23, 9, 9, 13 𝑠2 = Σ 𝑋 − 𝑥 ² 𝑛 − 1 𝑠2 = 141.34 6 − 1 𝑠2 = 141.34 5 𝒔 𝟐 = 𝟐𝟐. 𝟓𝟔 𝑠 = 22.56 𝒔 = ±𝟒. 𝟕𝟓 UNGROUPED DATA 4 X X̅ X-X̅ (X-X̅ )² 17 14.33 2.67 7.13 15 14.33 0.67 0.45 23 14.33 8.67 75.17 9 14.33 -5.33 28.41 9 14.33 -5.33 28.41 13 14.33 -1.33 1.77 Σ 𝑋 − 𝑥 ²= 141.34
- 6. DATA 6 66 62 61 67 64 67 66 63 63 70 71 68 65 62 61 68 61 64 60 65 64 66 61 68 64 64 69 68 63 63 HEIGHT OF THE G11 STUDENTS AGES 15-19
- 7. 7 CLASS INTERVAL Range HS-LS Highest Score 71 Lowest Score 60 11 Range ]= 11 ≈ 2 √N 5 √N = √30 = 5.477225575 ≈ 5
- 8. GROUPED DATA 8 N= 30 Class Interval Frequency <Cf >Cf X f(X) X̅ X-X̅ (X-X̅ )² f(X-X̅ )² 60-61 5 5 30 60.5 302.5 64.83 -4.33 18.75 93.75 62-63 6 11 25 62.5 375 64.83 -2.33 5.43 32.58 64-65 7 18 19 64.5 451.5 64.83 -0.33 0.11 0.77 66-67 5 23 12 66.5 332.5 64.83 1.67 2.79 13.95 68-69 5 28 7 68.5 342.5 64.83 3.67 13.47 67.35 70-71 2 30 2 70.5 141 64.83 5.67 32.15 64.3 ∑f(x)= 1945 ∑f(X-X̅ )²= 272.7
- 9. Mean 9 Median 𝑥 = 𝐿𝐿 + 𝑁 2 − 𝑐𝑓 𝑓 𝑖 𝑥 = 63.5 + 30 2 − 11 7 2 𝑥 = 63.5 + 15 − 11 7 2 𝑥 = 64.64 𝑥 = Σ𝑓𝑥 𝑁 𝑥 = 1945 30 𝑥 = 64.83
- 10. Population Σ𝑓 𝑋𝑚− 𝑥 ² 𝑛 272.7 30 σ² = 9.09 VARIANCE 10 Sample Σ𝑓 𝑋𝑚− 𝑥 ² 𝑛−1 272.7 30−1 272.7 29 s² = 9.4
- 11. Population = √ σ² = √ 9.09 ≈ ±3.01 STANDARD DEVIATION 11 Sample = √ s² = √ 9.4 ≈ ±3.07
- 12. 12 SKEWNESS
- 13. 13 SKEWNESS SK = 3(Mean−median) Standard Deviation SK = 3(64.83−64.64) 3.07 SK = 0.19 Symmetric Interpreting 1. If skewness is less than −1 or greater than +1, the distribution is highly skewed. 2. If skewness is between −1 and −½ or between +½ and +1, the distribution is moderately skewed. 3. If skewness is between −½ and +½, the distribution is approximately symmetric.
- 14. 14 SKEWNESS 0 1 2 3 4 5 6 7 60-61 62-63 64-65 66-67 68-69 70-71 Frequency Height
- 15. -3 -2 -1 0 1 2 3 Example: x = 65 μ = 64.83 σ = 3.07 Z SCORE 15 Solution: = 65 − 64.83 3.07 = 0.55438 ≈ 0.55
- 16. 16 FINDING THE PERCENTAGE UNDER THE NORMAL DISTRIBUTION Z~N(0,1) P(Z) Q(Z) R(Z) -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3
- 17. 17 P(Z<0.55) = 0.70884 = 70.84% FINDING THE PERCENTAGE UNDER THE NORMAL DISTRIBUTION -3 -2 -1 0 1 2 3
- 18. 18 ACTIVITY Class Interval Frequency <Cf Xm fXm X̅ Xm-X̅ (Xm-X̅ )² f(Xm-X̅ )² 37-45 1 50 41 41 82.4 -41.4 1713.96 1713.96 46-54 0 49 50 0 82.4 -32.4 1049.76 0 55-63 2 49 59 118 82.4 -23.4 547.56 1095.12 64-72 6 47 68 408 82.4 -14.4 207.36 1244.16 73-81 14 41 77 1078 82.4 -5.4 29.16 408.24 82-90 13 27 86 1118 82.4 3.6 12.96 168.48 91-99 11 14 95 1045 82.4 12.6 158.76 1746.36 100-108 3 3 104 312 82.4 21.6 466.56 1399.68 Find the skewness N= 50 ∑fxm= 4120 ∑f(Xm-X̅ )²= 7776 Clue: Find the mean, median and S.D. first i= 9
- 19. Siberian Dogs Mean: 48.5 lbs. S.D.: 2.1 lbs. Problem: 1.) Sketch the normal curve 2.) Find the percentage - between 46.4 lbs. and 50.6 lbs. Tip: Find the Z-scores first. Then, if you have a calculator, use Q( and add both of them together. 19 ACTIVITY -3 -2 -1 0 1 2 3
- 20. THANK YOU 20