Normal distribution calculator

Set the mean and standard deviation, pick the region, and see exactly which slice of the bell curve the probability is.

Find
P(X < 115)0.8413
As a percentage84.13%
557085100115130145115
  1. z = (115 − 100) / 15 = 1
  2. P(Z < 1) = 0.8413

What the normal distribution is

The normal distribution is the symmetric, bell-shaped curve fixed entirely by two numbers: the mean μ, which sets its centre, and the standard deviation σ, which sets its width. Its density is f(x) = e^(−(x − μ)² / 2σ²) / (σ√(2π)). Probability is area under that curve, and the total area is exactly 1.

How this calculator gets its answer

  1. Standardise. Convert each x to z = (x − μ) / σ, the number of standard deviations from the mean.
  2. Find the cumulative area. Φ(z), the area left of z, is the value in a z table. Here it is computed numerically to about 15 digits.
  3. Combine areas. Right tail = 1 − Φ(z); between = Φ(z₂) − Φ(z₁); outside = 1 − between.

Worked example: IQ scores

IQ tests are scaled to μ = 100 and σ = 15. The share of people above 130 is P(Z > 2) = 1 − 0.9772 = 0.0228, about 2.3%. The score that marks the top 10% is x = 100 + 1.2816 × 15 ≈ 119.2, which is the inverse normal in action.

Not sure your data are normal? Plot them first with the histogram maker and turn on the normal overlay.

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An AI tutor that sees the numbers you entered above and explains them. It can make mistakes, so check anything important against the worked steps.

Questions students ask

How do you calculate a normal distribution probability?

Convert x to a z-score with z = (x − μ) / σ, then find the cumulative area to the left of z (from a z table or the normal CDF). For "greater than", subtract from 1; for "between", subtract the two cumulative areas.

What is the inverse normal (invNorm)?

Inverse normal goes the other way: you give a probability, such as 0.90, and it returns the x-value with that much area to its left. On a TI-84 this is invNorm(area, μ, σ).

Why is P(X = a) zero for a normal distribution?

A continuous distribution spreads probability over an interval, and a single point has no width, so its area is zero. That is also why "<" and "≤" give the same answer.

When can I use the normal distribution?

When the data are roughly symmetric and bell-shaped (heights, measurement errors, many test scores), or for sample means of large samples thanks to the central limit theorem.

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