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Chapter 5 The Normal Curve and Standard Scores EPS 525 Introduction to Statistics

Chapter 5 The Normal Curve and Standard Scores EPS 525 Introduction to Statistics. 0.13. 0.13. Raw Score (IQ): 52 68 84 100 116 132 148. Beginning on Page 529. What is the proportion of scores in a normal distribution

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Chapter 5 The Normal Curve and Standard Scores EPS 525 Introduction to Statistics

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  1. Chapter 5 The Normal Curve and Standard Scores EPS 525 Introduction to Statistics

  2. 0.13 0.13

  3. Raw Score (IQ): 52 68 84 100 116 132 148

  4. Beginning on Page 529

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

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

  7. 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%

  8. 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%

  9. What is the proportion of scores in a normal distribution between z = -1.09 and z = -3.02? Area A = 36.21% (and) Area B (total) = 49.87% Therefore, the Area of Interest* = 49.87 – 36.21 = 13.66%

  10. 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%

  11. 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%

  12. 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%

  13. 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%

  14. -.51 .51 Knowing that the mean is 50 and s = 3, what are the raw scores that bound the middle 39% of the normal distribution? 48.47 51.53 Using Table A – we find that z = + .51 will bound the middle 39% of the scores. Area Between Mean and z = .1950 (39/2 = .1950) Solve for the lower X: Solve for the upper X: Therefore, the two raw scores that bound the middle 39% are 48.47 and 51.53

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