Normal Distribution Probability: Z-Score to Area Worked Examples (Step-by-Step)
Every intro statistics class revolves around the normal distribution and the z-score. But the worked-example side — given a real-world problem, produce the probability — is where students get stuck. This guide walks through the full translation from real-world question to z-score to area to probability, with examples for left tail, right tail, between-value, and outside ranges.
What You'll Learn
- ✓Convert real-world values to z-scores using μ and σ
- ✓Look up probability areas from a standard normal table or calculator
- ✓Apply the right approach for left tail, right tail, between, and outside probabilities
- ✓Use the empirical rule (68-95-99.7) for quick sanity checks
- ✓Recognize when data is not normal and adjust the approach
1. Direct Answer: The Z-Score to Probability Workflow
The normal distribution is defined by two parameters: mean (μ) and standard deviation (σ). To find the probability that a value falls in some range, you translate the range into a z-score range, then look up the area under the normal curve corresponding to that range. The workflow has three steps: 1. Convert x-values to z-scores: z = (x - μ) / σ. This tells you how many standard deviations above or below the mean a value is. 2. Determine the type of probability you need: left tail (P(X < x)), right tail (P(X > x)), between two values (P(a < X < b)), or outside a range (P(X < a or X > b)). 3. Look up the area(s) under the standard normal curve and compute. Standard normal tables (or calculators) give cumulative probabilities P(Z < z) — the area to the LEFT of z. Key conversions: - Left tail P(X < x): look up P(Z < z) directly - Right tail P(X > x): use 1 - P(Z < z) - Between P(a < X < b): P(Z < z_b) - P(Z < z_a) - Outside P(X < a or X > b): P(Z < z_a) + (1 - P(Z < z_b)), or equivalently 1 - P(a < X < b) Sanity check using the empirical rule: - About 68% of data within ±1σ of the mean - About 95% of data within ±2σ of the mean - About 99.7% of data within ±3σ of the mean A z-score beyond ±3 represents less than 0.3% of the distribution — unusual. This content is for educational purposes only and supports statistics student learning.
Key Points
- •z = (x - μ) / σ — converts any normal value to standard normal
- •Standard normal tables give P(Z < z), the cumulative left-tail probability
- •Right-tail: 1 - left-tail; Between: subtract left-tails
- •Empirical rule: 68-95-99.7 for sanity checks at ±1, ±2, ±3σ
- •Most problems reduce to careful z-score calculation + correct table lookup
2. Worked Example 1: Left-Tail Probability
Problem: SAT scores are normally distributed with mean μ = 1050 and standard deviation σ = 200. Find the probability that a randomly selected student scores less than 1200. Step 1 — Convert to z-score: z = (1200 - 1050) / 200 = 150 / 200 = 0.75 Step 2 — Type of probability: left tail, P(X < 1200) = P(Z < 0.75) Step 3 — Look up P(Z < 0.75) in the standard normal table: P(Z < 0.75) = 0.7734 Interpretation: about 77.34% of SAT scores are below 1200. Equivalently, about 22.66% are above 1200 (1 - 0.7734). Sanity check: 1200 is 0.75 standard deviations above the mean. Since about 50% of the distribution is below the mean, and we add part of the distribution between 0 and 0.75 SD above, we expect the answer to be between 50% and 84% (84% would be at +1 SD). 77% is in this range — reasonable. Common mistake: looking up P(Z < -0.75) = 0.2266 by accident. Always double-check which direction z is and which tail you want.
Key Points
- •Left tail problems: look up z-score directly in the standard normal table
- •Always identify the direction (less than vs greater than) before the table lookup
- •Sanity check against the empirical rule
- •Double-check sign of z — positive z is above the mean, negative is below
- •Most students do the arithmetic right; errors come from misreading the table
3. Worked Example 2: Right-Tail Probability
Problem: Adult male heights are normally distributed with mean 70 inches and standard deviation 3 inches. Find the probability that a randomly selected man is taller than 76 inches. Step 1 — Convert to z-score: z = (76 - 70) / 3 = 6 / 3 = 2.00 Step 2 — Type of probability: right tail, P(X > 76) = P(Z > 2.00) = 1 - P(Z < 2.00) Step 3 — Look up P(Z < 2.00): P(Z < 2.00) = 0.9772 Step 4 — Compute right tail: P(Z > 2.00) = 1 - 0.9772 = 0.0228 Interpretation: about 2.28% of adult men are taller than 76 inches (about 6'4"). This matches the empirical rule (about 2.5% are more than 2σ above the mean). Sanity check: 76 inches is 2 standard deviations above the mean. The empirical rule says about 2.5% of the distribution is beyond 2σ in either direction (5% total, split evenly). 2.28% is the precise value. Alternative method (modern): use software or a graphing calculator directly. In TI-84: normalcdf(76, 1E99, 70, 3) = 0.02275. In Excel: 1 - NORM.DIST(76, 70, 3, TRUE) = 0.02275. Both match the table lookup.
Key Points
- •Right tail: compute 1 - left-tail probability
- •normalcdf(lower, upper, μ, σ) on calculator skips the z-score conversion
- •Table-based approach is essential for understanding the concept
- •Calculator approach is faster for exam problems where time matters
- •Always verify with empirical rule that the answer is reasonable
4. Worked Example 3: Between-Two-Values Probability
Problem: IQ scores are normally distributed with mean 100 and standard deviation 15. Find the probability that a randomly selected person has IQ between 85 and 120. Step 1 — Convert both x-values to z-scores: For x = 85: z₁ = (85 - 100) / 15 = -15 / 15 = -1.00 For x = 120: z₂ = (120 - 100) / 15 = 20 / 15 = 1.33 Step 2 — Type of probability: between two values P(85 < X < 120) = P(-1.00 < Z < 1.33) = P(Z < 1.33) - P(Z < -1.00) Step 3 — Look up both values: P(Z < 1.33) = 0.9082 P(Z < -1.00) = 0.1587 Step 4 — Subtract: P(85 < X < 120) = 0.9082 - 0.1587 = 0.7495 Interpretation: about 74.95% of people have IQ between 85 and 120. This range is roughly -1 to +1.33 standard deviations. Sanity check: The empirical rule says about 68% fall within ±1σ. We're asking about a range that's -1σ to about +1.33σ — slightly wider on the upper side. So the probability should be slightly more than 68%. 74.95% is consistent. Common mistake: subtracting in the wrong order. Always subtract the smaller z-score's area from the larger one. You want the LARGER z minus the SMALLER z.
Key Points
- •Between two values: P(a < X < b) = P(Z < z_b) - P(Z < z_a)
- •Convert both x-values to z-scores first
- •Subtract the smaller-z area from the larger-z area
- •Common mistake: subtracting in wrong order
- •Calculator: normalcdf(85, 120, 100, 15) = 0.7495
5. Worked Example 4: Outside a Range (Complement)
Problem: A factory produces widgets with weights normally distributed, μ = 500 grams, σ = 10 grams. Quality control rejects widgets outside the range 480 to 520 grams. What percentage of widgets are rejected? Step 1 — Convert cutoffs to z-scores: For 480: z₁ = (480 - 500) / 10 = -20 / 10 = -2.00 For 520: z₂ = (520 - 500) / 10 = 20 / 10 = 2.00 Step 2 — Type of probability: outside the range P(X < 480 or X > 520) = P(Z < -2.00) + P(Z > 2.00) OR use complement: = 1 - P(-2.00 < Z < 2.00) = 1 - [P(Z < 2.00) - P(Z < -2.00)] Step 3 — Look up values: P(Z < -2.00) = 0.0228 P(Z < 2.00) = 0.9772 Step 4 — Compute (using complement approach): P(-2 < Z < 2) = 0.9772 - 0.0228 = 0.9544 P(outside) = 1 - 0.9544 = 0.0456 Or using the direct approach: P(outside) = P(Z < -2) + P(Z > 2) = 0.0228 + (1 - 0.9772) = 0.0228 + 0.0228 = 0.0456 Both methods give the same answer. Interpretation: about 4.56% of widgets are rejected. This is the empirical rule: about 5% of the distribution falls beyond ±2σ from the mean. Practical implication: if the factory produces 10,000 widgets per day, roughly 456 will be rejected. A factory making capital investment decisions based on yield needs this calculation to be accurate.
Key Points
- •Outside a range: complement is often easiest
- •P(outside) = 1 - P(inside range)
- •Alternative: add left-tail below lower cutoff + right-tail above upper cutoff
- •Both methods give the same answer — pick whichever is clearer
- •±2σ captures about 95% — so ~5% outside matches empirical rule
6. Reading the Standard Normal Table
Standard normal tables typically look up cumulative probabilities P(Z < z) for z-values from -3.49 to 3.49. The table is structured with: - Row labels: first two digits of z (e.g., 1.2, 1.5) - Column labels: second decimal place (0.00, 0.01, 0.02, ..., 0.09) - Cell values: P(Z < z) for that z combination Example: finding P(Z < 1.23) - Find row 1.2 - Find column 0.03 - Intersection gives 0.8907 For negative z-values, use the symmetry of the normal distribution: P(Z < -z) = 1 - P(Z < z) Example: P(Z < -1.23) = 1 - 0.8907 = 0.1093 Some tables provide separate negative-z sections; others only cover positive z. If your table is positive-only, the symmetry formula is essential. Alternative to tables: - Graphing calculator: TI-84 normalcdf(lower, upper) for standard normal, or normalcdf(lower, upper, μ, σ) for any normal - Excel: NORM.S.DIST(z, TRUE) for standard normal cumulative - Python: scipy.stats.norm.cdf(z) for cumulative at z - R: pnorm(z) for cumulative at z Practice tip: do 5-10 table lookups manually to build intuition, then rely on calculator/software for speed. Understanding the table structure is valuable even when you use technology.
Key Points
- •Standard normal table gives P(Z < z) — cumulative left-tail probability
- •Row = first two digits of z; Column = second decimal
- •For negative z: P(Z < -z) = 1 - P(Z < z)
- •Modern alternatives: TI-84 normalcdf, Excel NORM.S.DIST, Python scipy, R pnorm
- •Master table reading first; use calculator for speed
7. When Data Is NOT Normal and the Approach Changes
Normal distribution methods only apply when data is approximately normal. Common situations where it doesn't apply: 1. Highly skewed distributions. Income, house prices, and failure times are often right-skewed (long tail on the high end). The mean and median differ significantly, and the empirical rule doesn't hold. 2. Bimodal distributions. Some populations have two modes (e.g., heights of mixed-gender groups have two peaks). Normal methods treat this as a single normal, which is misleading. 3. Heavy-tailed distributions. Stock returns have heavier tails than normal — extreme moves occur more often than a normal model predicts. The empirical rule's 99.7% within ±3σ underestimates extreme events. 4. Count data. Counts (number of defects, number of accidents) are Poisson or discrete, not continuous. Normal approximation works for large counts (rule of thumb: if the Poisson mean > 20) but not small ones. 5. Bounded data. Percentages (0-100), ratings (1-5 stars), and probabilities are bounded. Pure normal distribution can predict values outside the valid range, which is nonsensical. 6. Small samples. The Central Limit Theorem eventually gives sample means a normal distribution, but small samples (n < 30) from skewed parent distributions may not follow normal shape. How to check if data is approximately normal: - Histogram: should be bell-shaped, symmetric - Q-Q plot (quantile-quantile): should be approximately a straight line - Skewness: close to 0 (not > 1 or < -1) - Kurtosis: close to 3 (excess kurtosis close to 0) - Shapiro-Wilk test: p-value > 0.05 suggests normal is plausible (though this test is sensitive to sample size) If data is not normal: use non-parametric tests (Wilcoxon, Mann-Whitney, Kruskal-Wallis) instead of normal-theory tests. Or apply transformations (log, square root) to make skewed data more normal-like. Or use appropriate non-normal distributions (lognormal for positive skewed, beta for bounded data, Poisson for counts).
Key Points
- •Normal distribution requires approximately symmetric, bell-shaped data
- •Skewed, bimodal, bounded, or count data may need different approaches
- •Check normality with histogram, Q-Q plot, skewness/kurtosis
- •If not normal: use non-parametric tests or transformations
- •Empirical rule only applies to (approximately) normal distributions
Key Takeaways
- ★z = (x - μ) / σ converts any normal to standard normal
- ★Standard normal table gives P(Z < z), the left-tail cumulative probability
- ★Right tail: 1 - P(Z < z)
- ★Between: P(Z < z_b) - P(Z < z_a)
- ★Outside: 1 - P(inside) or sum of both tails
- ★Empirical rule: 68% within ±1σ, 95% within ±2σ, 99.7% within ±3σ
- ★P(Z < -z) = 1 - P(Z < z) for symmetric tables
- ★TI-84 normalcdf(lower, upper, μ, σ) gives probabilities without z-score step
- ★Always sanity-check answer with empirical rule
- ★Normal methods require approximately normal data (check first)
Practice Questions
1. A test has scores that are normally distributed with μ = 75 and σ = 10. What is the probability that a randomly selected score is above 90?
2. Heights of women are normally distributed with μ = 64 inches and σ = 2.5 inches. What is the probability that a randomly selected woman is between 60 and 66 inches tall?
3. A factory's product has length normally distributed with μ = 100 mm and σ = 2 mm. Specifications are 97 to 103 mm. What percent of product is within spec?
4. SAT scores are normally distributed with μ = 1050 and σ = 200. A college accepts only students scoring above the 85th percentile. What score is required?
5. Is the normal distribution appropriate for modeling income in a country?
FAQs
Common questions about this topic
Because only ONE table is needed for all normal distributions. Without standardization, every (μ, σ) combination would need its own table — impractical. Standardizing via z = (x - μ) / σ reduces every normal problem to the standard normal (μ = 0, σ = 1), for which a single table suffices. Modern calculators can compute probabilities directly without standardization, but z-scores remain the conceptual foundation.
Nothing. For continuous distributions like the normal, P(X = any specific value) = 0 because there are infinitely many possible values. So P(X < 50) = P(X ≤ 50). This is different from discrete distributions (binomial, Poisson) where exact values have positive probability. In continuous distributions, the inclusive vs exclusive distinction doesn't matter.
Very accurate for bell-shaped normal distributions. The exact percentages are 68.27% within ±1σ, 95.45% within ±2σ, and 99.73% within ±3σ. Rounding to 68-95-99.7 loses precision but maintains utility for quick checks. For non-normal data, these percentages can be significantly off.
Sometimes. The normal approximation to the binomial requires np > 5 and n(1-p) > 5. Normal approximation to the Poisson requires the mean > 20. When approximating discrete with continuous, apply continuity correction (e.g., P(X = 5) becomes P(4.5 < X < 5.5)). For small samples, exact binomial or Poisson calculations are preferred.
Use a calculator with normalcdf or statistical software. Tables are limited in precision (typically 2 decimal places for z). For z = 1.2345 (beyond table precision), calculators give the exact value. Modern practice relies on software; tables remain useful for concept building and quick reference.
Yes. Snap a photo of any normal distribution problem and StatsIQ converts x-values to z-scores, looks up areas under the standard normal curve, handles left-tail/right-tail/between/outside probability questions, and shows the empirical rule check. It also identifies when data is NOT approximately normal and suggests appropriate alternatives (non-parametric tests or different distributions). This content is for educational purposes only and does not constitute statistical advice.