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Chapter 4 The Normal Distribution EPS 625 Statistical Methods Applied to Education I

Chapter 4 The Normal Distribution EPS 625 Statistical Methods Applied to Education I. Figure 4.3, page 85. Beginning on Page 634. What is the proportion of scores in a normal distribution between the mean and z = +0.52?. Answer: 19.85%. What is the proportion of scores

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Chapter 4 The Normal Distribution EPS 625 Statistical Methods Applied to Education I

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  1. Chapter 4 The Normal Distribution EPS 625 Statistical Methods Applied to Education I

  2. Figure 4.3, page 85

  3. Beginning on Page 634

  4. What is the proportion of scores in a normal distribution between the mean and z = +0.52? Answer: 19.85%

  5. What is the proportion of scores in a normal distribution between the mean and z = -1.89? Answer: 47.06%

  6. What is the proportion of scores in a normal distribution between z = -0.19 and z = +3.02? Area A = 7.53% (and) Area B = 49.87% Therefore, the Area of Interest = 7.53 + 49.87 = 57.40%

  7. What is the proportion of scores in a normal distribution between z = +0.19 and z = +1.12? Area A = 7.53% (and) Area B (total) = 36.86% Therefore, the Area of Interest* = 36.86 - 7.53 = 29.33%

  8. What is the proportion of scores in a normal distribution between z = -1.09 and z = -3.02? Area A = 36.27% (and) Area B (total) = 49.87% Therefore, the Area of Interest* = 49.87 – 36.27 = 13.60%

  9. What is the proportion of scores in a normal distribution above z = +0.87? Area A (total beyond z = 0.00) = 50.00% (and) Area B = 30.78% Therefore, the Area of Interest = 50.00 – 30.78 = 19.22% Or – use “Area Beyond z” Column Locate z = 0.87 and you will find 19.22%

  10. What is the proportion of scores in a normal distribution below z = +1.28? We know that Area A (total beyond z = 0.00) = 50.00% We find Area B = 39.97% Therefore, the Area of Interest = 50.00 + 39.97 = 89.97%

  11. What is the proportion of scores in a normal distribution above z = -2.00? We know that Area B (total beyond z = 0.00) = 50.00% We find Area A = 47.72% Therefore, the Area of Interest = 50.00 + 47.72 = 97.72%

  12. What is the proportion of scores in a normal distribution below z = -0.52? Area A = 19.85% (and) Area B (total beyond z = 0.00) = 50.00% Therefore, the Area of Interest = 50.00 – 19.85 = 30.15% Or – use “Area Beyond z” Column Locate z = 0.52 and you will find 30.15%

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