Z-Score Calculator

Normal Distribution Probability Calculator

Probability under any normal distribution N(μ, σ²) via standardization to z = (x − μ)/σ.

Question

Probability that X is at most x — the area to the left.

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Solution

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Worked Examples

Left Tail · IQ Scores

P(IQ ≤ 120) when μ = 100 and σ = 15

Adult IQ scores are normally distributed with mean 100 and standard deviation 15. What fraction of adults score at most 120?

  1. Identify the distribution: N(μ = 100, σ = 15).
  2. Standardize: z = (120 − 100)/15 = 1.3333.
  3. Read the left-tail area: Φ(1.3333) ≈ 0.9088.
  4. P(X ≤ 120) ≈ 0.9088 — about 90.88% of adults.

A z-score of 1.33 corresponds to the ~91st percentile. Anyone with an IQ ≤ 120 is in the bottom 91% of the population.

Right Tail · Adult Heights

P(Height ≥ 75 in) when μ = 70 and σ = 3

U.S. adult male heights are roughly normal with mean 70 inches and standard deviation 3 inches. What fraction is at least 75 inches (6'3'')?

  1. Identify the distribution: N(μ = 70, σ = 3).
  2. Standardize: z = (75 − 70)/3 = 1.6667.
  3. Read the right-tail area: 1 − Φ(1.6667) ≈ 0.0478.
  4. P(X ≥ 75) ≈ 0.0478 — about 4.78% of adult men.

z = 1.67 sits very near the standard one-tailed 5% cutoff (z ≈ 1.6449), so 75 inches falls at roughly the 95th percentile of male height.

Between · Empirical Rule

P(85 ≤ X ≤ 115) when μ = 100 and σ = 15

On the IQ scale, what is the probability of a score within one standard deviation of the mean?

  1. Identify the distribution: N(μ = 100, σ = 15).
  2. Standardize the bounds: z₁ = (85 − 100)/15 = −1, z₂ = (115 − 100)/15 = 1.
  3. Compute the middle area: Φ(1) − Φ(−1) ≈ 0.8413 − 0.1587 = 0.6827.
  4. P(85 ≤ X ≤ 115) ≈ 0.6827 — the textbook ~68% from the empirical rule.

Within ±1σ of the mean: ~68%. Within ±2σ: ~95%. Within ±3σ: ~99.7%. This is the empirical rule.

Outside · Quality Control

P(X ≤ 9.95 or X ≥ 10.05) when μ = 10 and σ = 0.02

A factory produces bolts with diameter normally distributed with μ = 10 mm and σ = 0.02 mm. The spec range is 9.95 to 10.05 mm. What is the defect rate?

  1. Identify the distribution: N(μ = 10, σ = 0.02).
  2. Standardize: z₁ = (9.95 − 10)/0.02 = −2.5, z₂ = (10.05 − 10)/0.02 = 2.5.
  3. Combine the two tails: Φ(−2.5) + (1 − Φ(2.5)) ≈ 0.0062 + 0.0062 = 0.0124.
  4. Defect rate ≈ 0.0124 — about 1.24% of bolts are out of spec.

The two-tail combined area at ±2.5σ is roughly 1.24%, often cited as the failure rate at a 2.5-sigma process. Six Sigma quality (±6σ) puts the rate at parts per billion.

Standardize, then read the area

Probabilities under any normal distribution N(μ, σ²) reduce to a standard-normal lookup. Convert the raw value x to a z-score by subtracting the mean and dividing by the standard deviation, then read Φ(z) — the cumulative area to the left — from a z-table or the normal CDF. The same transformation handles right-tail, between, and outside-interval probabilities.

z = (x − μ) / σ, then P(X ≤ x) = Φ(z)

How It Works

A normal distribution is fully described by two numbers: its mean μ (where the bell curve is centered) and its standard deviation σ (how wide it spreads). To find the probability that a normal random variable falls in some interval, you don't need a separate table for every (μ, σ) pair — you standardize. Subtracting the mean and dividing by σ converts any normal value x into a z-score, and the z-score's tail area Φ(z) is the same for every distribution. This calculator applies that two-step recipe automatically: enter μ, σ, and the cutoff x (or x₁ and x₂), pick which area you want — left, right, between, or outside — and the calculator returns the probability along with the underlying z-scores so you can verify the work against a printed z-table.

Example Problem

Adult IQ scores are normally distributed with μ = 100 and σ = 15. What is the probability that a randomly chosen adult has an IQ of at most 120?

  1. Identify the distribution: N(μ = 100, σ = 15).
  2. Compute the z-score for x = 120: z = (120 − 100)/15 = 20/15 ≈ 1.3333.
  3. Read the left-tail area from the standard normal: Φ(1.3333) ≈ 0.9088.
  4. Therefore P(X ≤ 120) ≈ 0.9088, or about 90.88%.
  5. Interpret: roughly 91 out of every 100 adults score at or below an IQ of 120.

The standard normal CDF Φ(z) is the same lookup regardless of μ and σ — that is the whole point of standardization. Once z is known, any normal probability becomes a z-table problem.

Key Concepts

The four common questions for a normal random variable map to four standard-normal lookups: P(X ≤ x) is the left-tail area Φ(z), P(X ≥ x) is the right-tail area 1 − Φ(z), P(x₁ ≤ X ≤ x₂) is Φ(z₂) − Φ(z₁), and P(X ≤ x₁ or X ≥ x₂) is Φ(z₁) + (1 − Φ(z₂)). The empirical rule is the most-cited shortcut: about 68% of values fall within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ. Those numbers come straight from Φ(1) − Φ(−1) ≈ 0.6827, Φ(2) − Φ(−2) ≈ 0.9545, and Φ(3) − Φ(−3) ≈ 0.9973.

Applications

  • Test scores: SAT, ACT, IQ, and standardized assessments are designed around normal distributions
  • Quality control: process capability, Six Sigma defect rates, and acceptance sampling
  • Finance: log-returns are often modeled as normal for risk and option pricing
  • Biostatistics: heights, blood pressure, and laboratory reference ranges
  • Reliability engineering: lifetime distributions and fatigue analysis
  • Psychometrics: percentile ranks for clinical and educational measurements

Common Mistakes

  • Forgetting to standardize — looking up x directly in a z-table when you should be looking up z = (x − μ)/σ
  • Using variance σ² in place of σ when computing the z-score
  • Mixing up P(X ≤ x) with P(X ≥ x) — they sum to 1, not equal to each other
  • Mistaking the empirical rule's 68/95/99.7% for exact values — they are approximations of Φ(1), Φ(2), Φ(3)
  • Applying the normal model to data that is heavily skewed or bounded (e.g., income, reaction times)
  • Using N(μ, σ) when a textbook expects N(μ, σ²) — the second parameter is variance in formal notation, but most calculators take σ

Frequently Asked Questions

What is a normal distribution?

A normal distribution is a symmetric bell-shaped probability distribution defined by two parameters: the mean μ (the center) and the standard deviation σ (the spread). About 68% of values fall within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ.

How do I find the probability under a normal distribution?

Standardize: convert your raw value x to a z-score using z = (x − μ)/σ, then look up the corresponding tail area Φ(z) from a z-table or the standard normal CDF. The calculator above does both steps automatically.

What is the difference between a normal distribution and the standard normal?

The standard normal is the special normal distribution with μ = 0 and σ = 1. Every other normal distribution can be reduced to the standard normal by the z-score transformation, which is why z-tables are universally useful.

What is P(X ≤ 120) when μ = 100 and σ = 15?

z = (120 − 100)/15 = 1.3333, and Φ(1.3333) ≈ 0.9088. So P(X ≤ 120) ≈ 0.9088, or about 90.88%.

What is the 68-95-99.7 rule?

It is the empirical rule for the normal distribution: approximately 68% of values lie within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ. The exact values are Φ(1) − Φ(−1) ≈ 0.6827, Φ(2) − Φ(−2) ≈ 0.9545, and Φ(3) − Φ(−3) ≈ 0.9973.

How do I compute P(a ≤ X ≤ b)?

Compute z₁ = (a − μ)/σ and z₂ = (b − μ)/σ, then take Φ(z₂) − Φ(z₁). The answer is the area under the bell curve between the two z-scores.

What does P(X ≥ x) equal?

P(X ≥ x) = 1 − P(X ≤ x) = 1 − Φ(z) where z = (x − μ)/σ. This is the right-tail area — the complement of the left-tail area.

When should I not use a normal distribution?

When the underlying data is heavily skewed, has hard bounds (e.g., must be positive), or shows obvious multimodality. In those cases the normal model can give misleading probabilities; alternatives include lognormal, exponential, or empirical distributions.

Reference: Standardization to a z-score is presented in every introductory statistics text; this calculator uses Abramowitz & Stegun's rational approximation for Φ(z) (|error| < 7.5 × 10⁻⁸).

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