Worked solution P31-Easy-1. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{54.5-71}{11}=-1.50,P(X\le 54.5)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-2. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{51.0-61}{10}=-1.00,P(X\le 51.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-3. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{78.0-75}{6}=0.50,P(X\le 78.0)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-4. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{80.0-70}{8}=1.25,P(X\le 80.0)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{84.0-64}{10}=2.00,P(X\le 84.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-6. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{45.0-60}{10}=-1.50,P(X\le 45.0)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-7. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{62.0-71}{9}=-1.00,P(X\le 62.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-8. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{63.0-60}{6}=0.50,P(X\le 63.0)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-9. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{90.2-74}{13}=1.25,P(X\le 90.2)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-10. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{75.0-61}{7}=2.00,P(X\le 75.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-11. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{55.0-70}{10}=-1.50,P(X\le 55.0)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-12. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{64.0-77}{13}=-1.00,P(X\le 64.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-13. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{69.5-63}{13}=0.50,P(X\le 69.5)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-14. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{79.8-71}{7}=1.25,P(X\le 79.8)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-15. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{93.0-73}{10}=2.00,P(X\le 93.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-16. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{56.0-74}{12}=-1.50,P(X\le 56.0)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-17. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{69.0-80}{11}=-1.00,P(X\le 69.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-18. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{70.0-64}{12}=0.50,P(X\le 70.0)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-19. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{75.0-60}{12}=1.25,P(X\le 75.0)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-20. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{77.0-61}{8}=2.00,P(X\le 77.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Easy-21. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{65.0-74}{6}=-1.50,P(X\le 65.0)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-1. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{58.0-76}{12}=-1.50,P(X\le 58.0)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-2. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{53.0-62}{9}=-1.00,P(X\le 53.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-3. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{66.5-61}{11}=0.50,P(X\le 66.5)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-4. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{81.8-73}{7}=1.25,P(X\le 81.8)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{79.0-63}{8}=2.00,P(X\le 79.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-6. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{44.5-61}{11}=-1.50,P(X\le 44.5)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-7. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{57.0-66}{9}=-1.00,P(X\le 57.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-8. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{83.5-79}{9}=0.50,P(X\le 83.5)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-9. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{90.5-78}{10}=1.25,P(X\le 90.5)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-10. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{83.0-67}{8}=2.00,P(X\le 83.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-11. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{61.0-70}{6}=-1.50,P(X\le 61.0)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-12. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{66.0-79}{13}=-1.00,P(X\le 66.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-13. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{79.0-73}{12}=0.50,P(X\le 79.0)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-14. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{79.2-68}{9}=1.25,P(X\le 79.2)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-15. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{80.0-60}{10}=2.00,P(X\le 80.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-16. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{47.5-61}{9}=-1.50,P(X\le 47.5)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-17. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{57.0-68}{11}=-1.00,P(X\le 57.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-18. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{82.5-76}{13}=0.50,P(X\le 82.5)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-19. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{82.5-70}{10}=1.25,P(X\le 82.5)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-20. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{87.0-63}{12}=2.00,P(X\le 87.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Tough-21. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{63.0-78}{10}=-1.50,P(X\le 63.0)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-1. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{45.0-63}{12}=-1.50,P(X\le 45.0)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-2. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{61.0-70}{9}=-1.00,P(X\le 61.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-3. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{75.0-70}{10}=0.50,P(X\le 75.0)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-4. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{86.0-76}{8}=1.25,P(X\le 86.0)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{86.0-64}{11}=2.00,P(X\le 86.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-6. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{51.5-71}{13}=-1.50,P(X\le 51.5)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-7. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{52.0-61}{9}=-1.00,P(X\le 52.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-8. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{69.5-66}{7}=0.50,P(X\le 69.5)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-9. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{76.8-63}{11}=1.25,P(X\le 76.8)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-10. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{84.0-60}{12}=2.00,P(X\le 84.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-11. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{55.5-69}{9}=-1.50,P(X\le 55.5)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-12. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{73.0-80}{7}=-1.00,P(X\le 73.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-13. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{78.5-72}{13}=0.50,P(X\le 78.5)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-14. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{90.0-75}{12}=1.25,P(X\le 90.0)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-15. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{100.0-80}{10}=2.00,P(X\le 100.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-16. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{53.0-62}{6}=-1.50,P(X\le 53.0)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-17. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{58.0-67}{9}=-1.00,P(X\le 58.0)=\Phi (-1.00)\approx 0.1587.$$ Interpretation: The value is 1.00 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-18. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{67.5-63}{9}=0.50,P(X\le 67.5)=\Phi (0.50)\approx 0.6915.$$ Interpretation: The value is 0.50 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-19. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{88.8-80}{7}=1.25,P(X\le 88.8)=\Phi (1.25)\approx 0.8944.$$ Interpretation: The value is 1.25 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-20. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{78.0-66}{6}=2.00,P(X\le 78.0)=\Phi (2.00)\approx 0.9772.$$ Interpretation: The value is 2.00 standard deviations above the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.

Worked solution P31-Toughest-21. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{45.0-63}{12}=-1.50,P(X\le 45.0)=\Phi (-1.50)\approx 0.0668.$$ Interpretation: The value is 1.50 standard deviations below the model mean. Validity: A normal probability statement requires a reasonable normal model; standardization alone does not establish shape. Error to reject: A z-score is a signed standard-deviation distance, whereas a percentile is a cumulative proportion below a value.