Z-Scores and Standardization

Which is more impressive?

• Scoring 720 on the SAT Math section
• Scoring 28 on the ACT Math section

These tests have different scales, different means, and different standard deviations. Raw scores can’t be compared directly!

Exam scores have mean $\mu = 75$ and standard deviation $\sigma = 8$.

What is the z-score for a student who scored 91?

$z = \frac{91 - 75}{8} = \frac{16}{8} = 2.0$

Interpretation: This student scored 2 standard deviations above the mean—an excellent performance!

SAT Math: Mean = 500, SD = 100 ACT Math: Mean = 20, SD = 5

Student A: SAT Math = 720 $z = \frac{720 - 500}{100} = 2.2$

Student B: ACT Math = 28 $z = \frac{28 - 20}{5} = 1.6$

Conclusion: Student A performed relatively better (2.2 standard deviations above mean vs. 1.6).

Original data: 60, 70, 80, 90, 100 (mean = 80, SD = 14.14)

Z-scores:

• 60: $z = (60-80)/14.14 = -1.41$
• 70: $z = (70-80)/14.14 = -0.71$
• 80: $z = (80-80)/14.14 = 0$
• 90: $z = (90-80)/14.14 = +0.71$
• 100: $z = (100-80)/14.14 = +1.41$

Verify: Mean of z-scores = 0, SD of z-scores = 1 ✓

Adult male heights: Mean = 70 inches, SD = 3 inches

Within 1 SD (z between -1 and 1):

• Range: 70 ± 3 = 67 to 73 inches
• Contains: 68% of men

Within 2 SD (z between -2 and 2):

• Range: 70 ± 6 = 64 to 76 inches
• Contains: 95% of men

Within 3 SD (z between -3 and 3):

• Range: 70 ± 9 = 61 to 79 inches
• Contains: 99.7% of men

Only 0.3% of men are shorter than 61” or taller than 79”!

What percentage of values fall below z = 1.5?

From z-table: P(Z < 1.5) = 0.9332 or 93.32%

Interpretation: 93.32% of values in a standard normal distribution are below z = 1.5.

Given a normal distribution with $\mu = 100$ and $\sigma = 15$:

Type 1: P(X < value) — “less than” P(X < 115)?

• $z = (115-100)/15 = 1.0$
• P(Z < 1.0) = 0.8413 (from table)

Type 2: P(X > value) — “greater than” P(X > 115)?

• Use complement: P(X > 115) = 1 - P(X < 115)
• = 1 - 0.8413 = 0.1587

Type 3: P(a < X < b) — “between” P(90 < X < 115)?

• $z_1 = (90-100)/15 = -0.67$ → P(Z < -0.67) = 0.2514
• $z_2 = (115-100)/15 = 1.0$ → P(Z < 1.0) = 0.8413
• P(90 < X < 115) = 0.8413 - 0.2514 = 0.5899

IQ scores: $\mu = 100$, $\sigma = 15$

What IQ score is at the 90th percentile?

Step 1: Find z-score for 90th percentile

• From z-table: z = 1.28 (area = 0.90)

Step 2: Convert z back to x $x = \mu + z \cdot \sigma = 100 + 1.28(15) = 100 + 19.2 = 119.2$

An IQ of 119 is at the 90th percentile.

Class exam: Mean = 78, SD = 10

Student scores: 95, 82, 71, 45, 88

Z-scores:

• 95: z = (95-78)/10 = 1.7 (typical)
• 82: z = (82-78)/10 = 0.4 (typical)
• 71: z = (71-78)/10 = -0.7 (typical)
• 45: z = (45-78)/10 = -3.3 (outlier!)
• 88: z = (88-78)/10 = 1.0 (typical)

The score of 45 is more than 3 standard deviations below the mean—investigate this student’s situation.

A patient’s cholesterol is 245 mg/dL. Population: Mean = 200 mg/dL, SD = 25 mg/dL

$z = \frac{245 - 200}{25} = 1.8$

The patient is 1.8 standard deviations above average—elevated but not extremely unusual. About 3.6% of the population has higher cholesterol (P(Z > 1.8) ≈ 0.036).