Standard Normal and Z-Scores

Overview

Z-scores standardize values from any normal distribution to the standard normal distribution, enabling probability calculations using a single reference table.

Z-Score Formula

$$Z = \frac{X - \mu}{\sigma}$$

Where:

• $X$ = raw score
• $\mu$ = population mean
• $\sigma$ = population standard deviation

Interpretation

Z-Score	Meaning
$Z = 0$	At the mean
$Z = 1$	One SD above mean
$Z = -1$	One SD below mean
$Z = 2$	Two SD above mean

The z-score tells you how many standard deviations a value is from the mean.

Standard Normal Distribution

$$Z \sim N(0, 1)$$

Properties:

• Mean = 0
• Standard deviation = 1
• Total area = 1
• Symmetric about 0

Using Z-Tables

Z-tables give $P(Z < z)$, the area to the left of $z$.

Common Z-Values and Probabilities

$Z$	$P(Z < z)$	$P(Z > z)$
-2.58	0.0049	0.9951
-1.96	0.0250	0.9750
-1.645	0.0500	0.9500
0	0.5000	0.5000
1.645	0.9500	0.0500
1.96	0.9750	0.0250
2.58	0.9951	0.0049

Key Percentiles

Percentile	Z-Score
1st	-2.326
5th	-1.645
10th	-1.282
25th	-0.674
50th	0
75th	0.674
90th	1.282
95th	1.645
99th	2.326

Probability Calculations

Left Tail: $P(Z < z)$

Read directly from z-table

Right Tail: $P(Z > z)$

$$P(Z > z) = 1 - P(Z < z)$$

Between Two Values: $P(z_1 < Z < z_2)$

$$P(z_1 < Z < z_2) = P(Z < z_2) - P(Z < z_1)$$

Two-Tailed: $P(\lvert Z \rvert > z)$

$$P(\lvert Z \rvert > z) = 2 \times P(Z > z)$$

Examples

Example 1: Finding Z-Score

IQ scores: $\mu = 100$, $\sigma = 15$. Convert IQ of 130 to z-score.

$$Z = \frac{130 - 100}{15} = 2.0$$

Interpretation: An IQ of 130 is 2 standard deviations above the mean.

Example 2: Finding Probability

SAT scores: $\mu = 500$, $\sigma = 100$. Find $P(X < 620)$.

$$Z = \frac{620 - 500}{100} = 1.2$$ $$P(Z < 1.2) = 0.8849$$

Approximately 88.5% score below 620.

Example 3: Finding a Value from Z

Given $P(X < k) = 0.90$, find $k$.

$$\text{From table: } Z = 1.282$$ $$k = \mu + Z \times \sigma = 500 + 1.282 \times 100 = 628.2$$

Example 4: Comparing Values from Different Distributions

Test A: Score 85, $\mu = 75$, $\sigma = 5$ → $Z = 2.0$

Test B: Score 90, $\mu = 80$, $\sigma = 10$ → $Z = 1.0$

The score of 85 on Test A is relatively better (higher z-score).

Example 5: Probability Between Values

Heights: $\mu = 170$ cm, $\sigma = 10$ cm. Find $P(160 < X < 180)$.

$$Z_1 = \frac{160 - 170}{10} = -1.0$$ $$Z_2 = \frac{180 - 170}{10} = 1.0$$ $$P(-1 < Z < 1) = P(Z < 1) - P(Z < -1) = 0.8413 - 0.1587 = 0.6826$$

Applications

Application	What Z-Scores Tell You
Class grades	Position relative to class
Quality control	How unusual a measurement is
Research	Identifying outliers ($\lvert Z \rvert > 3$)
Standardized tests	Comparison across different forms

Converting Back

From z-score to raw score:

$$X = \mu + Z \times \sigma$$