Worked solution P30-Easy-1. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{47.5-67}{13}=-1.50,P(X\le 47.5)=\Phi (-1.50)\approx 0.0668.$$ 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 P30-Easy-2. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{65.0-73}{8}=-1.00,P(X\le 65.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Easy-3. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{81.5-76}{11}=0.50,P(X\le 81.5)=\Phi (0.50)\approx 0.6915.$$ 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 P30-Easy-4. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{80.2-64}{13}=1.25,P(X\le 80.2)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Easy-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{82.0-66}{8}=2.00,P(X\le 82.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-Easy-6. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{60.0-72}{8}=-1.50,P(X\le 60.0)=\Phi (-1.50)\approx 0.0668.$$ 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 P30-Easy-7. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{62.0-73}{11}=-1.00,P(X\le 62.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Easy-8. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{83.5-78}{11}=0.50,P(X\le 83.5)=\Phi (0.50)\approx 0.6915.$$ 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 P30-Easy-9. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{85.2-74}{9}=1.25,P(X\le 85.2)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Easy-10. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{98.0-76}{11}=2.00,P(X\le 98.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-Easy-11. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{59.0-71}{8}=-1.50,P(X\le 59.0)=\Phi (-1.50)\approx 0.0668.$$ 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 P30-Easy-12. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{68.0-80}{12}=-1.00,P(X\le 68.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Easy-13. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{67.0-64}{6}=0.50,P(X\le 67.0)=\Phi (0.50)\approx 0.6915.$$ 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 P30-Easy-14. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{85.0-75}{8}=1.25,P(X\le 85.0)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Easy-15. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{84.0-62}{11}=2.00,P(X\le 84.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-Easy-16. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{64.5-75}{7}=-1.50,P(X\le 64.5)=\Phi (-1.50)\approx 0.0668.$$ 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 P30-Easy-17. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{66.0-73}{7}=-1.00,P(X\le 66.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Easy-18. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{76.0-70}{12}=0.50,P(X\le 76.0)=\Phi (0.50)\approx 0.6915.$$ 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 P30-Easy-19. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{76.5-64}{10}=1.25,P(X\le 76.5)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Easy-20. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{91.0-77}{7}=2.00,P(X\le 91.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-Tough-1. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{53.0-65}{8}=-1.50,P(X\le 53.0)=\Phi (-1.50)\approx 0.0668.$$ 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 P30-Tough-2. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{69.0-77}{8}=-1.00,P(X\le 69.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Tough-3. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{78.0-73}{10}=0.50,P(X\le 78.0)=\Phi (0.50)\approx 0.6915.$$ 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 P30-Tough-4. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{80.8-67}{11}=1.25,P(X\le 80.8)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Tough-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{81.0-63}{9}=2.00,P(X\le 81.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-Tough-6. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{49.0-67}{12}=-1.50,P(X\le 49.0)=\Phi (-1.50)\approx 0.0668.$$ 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 P30-Tough-7. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{71.0-79}{8}=-1.00,P(X\le 71.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Tough-8. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{81.0-78}{6}=0.50,P(X\le 81.0)=\Phi (0.50)\approx 0.6915.$$ 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 P30-Tough-9. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{75.2-64}{9}=1.25,P(X\le 75.2)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Tough-10. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{99.0-73}{13}=2.00,P(X\le 99.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-Tough-11. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{56.5-70}{9}=-1.50,P(X\le 56.5)=\Phi (-1.50)\approx 0.0668.$$ 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 P30-Tough-12. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{55.0-63}{8}=-1.00,P(X\le 55.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Tough-13. 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 P30-Tough-14. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{88.2-72}{13}=1.25,P(X\le 88.2)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Tough-15. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{82.0-66}{8}=2.00,P(X\le 82.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-Tough-16. 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 P30-Tough-17. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{67.0-74}{7}=-1.00,P(X\le 67.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Tough-18. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{83.5-77}{13}=0.50,P(X\le 83.5)=\Phi (0.50)\approx 0.6915.$$ 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 P30-Tough-19. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{78.0-68}{8}=1.25,P(X\le 78.0)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Tough-20. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{92.0-66}{13}=2.00,P(X\le 92.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-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 P30-Toughest-2. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{51.0-63}{12}=-1.00,P(X\le 51.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Toughest-3. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{65.5-62}{7}=0.50,P(X\le 65.5)=\Phi (0.50)\approx 0.6915.$$ 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 P30-Toughest-4. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{88.2-72}{13}=1.25,P(X\le 88.2)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Toughest-5. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{86.0-72}{7}=2.00,P(X\le 86.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-Toughest-6. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{61.0-79}{12}=-1.50,P(X\le 61.0)=\Phi (-1.50)\approx 0.0668.$$ 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 P30-Toughest-7. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{63.0-74}{11}=-1.00,P(X\le 63.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Toughest-8. 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 P30-Toughest-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 P30-Toughest-10. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{82.0-66}{8}=2.00,P(X\le 82.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-Toughest-11. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{60.0-72}{8}=-1.50,P(X\le 60.0)=\Phi (-1.50)\approx 0.0668.$$ 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 P30-Toughest-12. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{65.0-72}{7}=-1.00,P(X\le 65.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Toughest-13. The standardized score is z=0.50; the modeled proportion at or below x is approximately 0.6915. $$z=\frac{80.5-74}{13}=0.50,P(X\le 80.5)=\Phi (0.50)\approx 0.6915.$$ 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 P30-Toughest-14. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{86.8-78}{7}=1.25,P(X\le 86.8)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Toughest-15. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{95.0-73}{11}=2.00,P(X\le 95.0)=\Phi (2.00)\approx 0.9772.$$ 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 P30-Toughest-16. The standardized score is z=-1.50; the modeled proportion at or below x is approximately 0.0668. $$z=\frac{41.5-61}{13}=-1.50,P(X\le 41.5)=\Phi (-1.50)\approx 0.0668.$$ 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 P30-Toughest-17. The standardized score is z=-1.00; the modeled proportion at or below x is approximately 0.1587. $$z=\frac{60.0-72}{12}=-1.00,P(X\le 60.0)=\Phi (-1.00)\approx 0.1587.$$ 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 P30-Toughest-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 P30-Toughest-19. The standardized score is z=1.25; the modeled proportion at or below x is approximately 0.8944. $$z=\frac{73.0-63}{8}=1.25,P(X\le 73.0)=\Phi (1.25)\approx 0.8944.$$ 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 P30-Toughest-20. The standardized score is z=2.00; the modeled proportion at or below x is approximately 0.9772. $$z=\frac{97.0-71}{13}=2.00,P(X\le 97.0)=\Phi (2.00)\approx 0.9772.$$ 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.