Normal Distribution Calculator

Enter the mean and standard deviation, pick the question, and get the probability (or the value) with its z-score and the one-, two- and three-sigma bands.

Used for the below and above questions.
Used for the between question.
Used for the between question.
Used for the inverse question: the percentage of the distribution below the x you are looking for.

Result

97.725%

z-score

2

Explanation

130 is 2 standard deviations from the mean. P(X < 130) = Φ(2) = 97.725%.

How it works

The normal (Gaussian) distribution is the bell curve. It is fixed by two numbers: the mean μ, where the curve is centred, and the standard deviation σ, which sets how wide it is. Heights, measurement errors, test scores and averages of many small effects are often close to normal, and many statistical methods assume it.

Every normal question is answered by converting to a z-score, z = (x − μ) ÷ σ, which says how many standard deviations x is from the mean, and then reading the standard normal cumulative distribution Φ(z): the probability that a standard normal variable is below z. The probability above x is 1 − Φ(z), and the probability between a and b is Φ(z_b) − Φ(z_a). This replaces the printed z-table with a direct computation.

The inverse question runs the other way: given a cumulative probability p, find z with Φ(z) = p and convert back with x = μ + zσ. This is how you find the cut-off for the top 2.5%, the 90th percentile of a score distribution, or a critical value for a confidence interval.

The 68–95–99.7 rule is the same calculation at z = ±1, ±2 and ±3: about 68.27% of a normal distribution lies within one standard deviation of the mean, 95.45% within two and 99.73% within three.

Formula

z = (x − μ) ÷ σ
P(X < x) = Φ(z)              P(X > x) = 1 − Φ(z) = Φ(−z)
P(a < X < b) = Φ(z_b) − Φ(z_a)
x at cumulative probability p:  x = μ + Φ⁻¹(p) × σ
Φ(z) = ½ + φ(z) × (z + z³/3 + z⁵/15 + z⁷/105 + …),  φ(z) = e^(−z²/2) ÷ √(2π)

Example

IQ scores are modelled as normal with mean 100 and standard deviation 15. For x = 130 the z-score is (130 − 100) ÷ 15 = 2, and Φ(2) = 0.97725, so 97.72% of people score below 130 and 2.28% score above it.

Between 85 and 115 (z = −1 to 1) lies Φ(1) − Φ(−1) = 0.84134 − 0.15866 = 68.27% of the distribution. The score with 97.5% of the distribution below it is 100 + 1.959964 × 15 = 129.40.

Assumptions and limitations

  • The variable is exactly normally distributed with the mean and standard deviation you enter. Real data is often only roughly normal, especially in the tails.
  • σ is the population standard deviation. Using a sample standard deviation from a small sample makes the probabilities approximate.
  • Φ(z) is computed by Marsaglia's series and is accurate to about 1e-15 for |z| ≤ 8; tail probabilities below that level are shown as 0.
  • The inverse uses Acklam's rational approximation refined with one Halley step, accurate to double precision; the cumulative probability must be strictly between 0% and 100%.
  • Probabilities are shown as percentages rounded to four decimal places.

Frequently asked questions

What is the difference between P(X < x) and P(X ≤ x)?

Nothing for a normal distribution. It is continuous, so the probability of landing exactly on x is zero and the two are equal. The same goes for the between question: it does not matter whether the ends are included.

How do I get a z-table critical value such as 1.96?

Use the inverse question with mean 0 and standard deviation 1. A cumulative probability of 97.5% gives z = 1.959964, the two-sided 95% critical value; 95% gives 1.644854 and 99.5% gives 2.575829.