How do I use the standard normal distribution to compute probabilities?

For X ~ N(μ, σ²), standardize z = (x − μ)/σ, so P(X x₀) = 1 − Φ(z₀).

Why are the normal probability integrals computed numerically?

The normal PDF has no elementary antiderivative, so the CDF Φ(z) cannot be expressed in closed form using only the elementary functions. Instead, it is computed numerically — typically via a rational approximation like the Abramowitz & Stegun formula, accurate to about 7.5 × 10⁻⁸.

What is the relationship between the three normal probabilities?

They satisfy two identities: P(X x₀) = 1 (the left and right tails partition the total probability), and P(x₀ < X < x₁) = P(X < x₁) − P(X < x₀) (the interval probability is the difference of two cumulative probabilities).

Normal Probability Calculator — Cumulative, Between & Above

Q: What is the formula for the normal probability?

The PDF is f(x) = (1/(σ√(2π))) · exp(−(x − μ)²/(2σ²)). The probability of X falling in an interval is the integral of this density over the interval. Because the integral has no elementary closed form, it is computed numerically via the standard normal CDF.

Overview

For a normally distributed random variable \( X \sim N(\mu, \sigma^2) \), the probability of falling in an interval is the area under the density curve. This page provides three calculators: the probability that \( X \) lies in an interval \( (x_0, x_1) \), the probability that \( X < x_0 \), and the probability that \( X > x_0 \).

Probability Density and Integrals

The probability density function of normally distributed \( X \) with mean \( \mu \) and standard deviation \( \sigma \) is:

\[ f_X(x;\mu,\sigma) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \quad x \in \mathbb{R} \]

Probability between \( x_0 \) and \( x_1 \)

\[ P(x_0 < X < x_1) = \int_{x_0}^{x_1} \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x-\mu)^2}{2\sigma^2}}\, dx \]

Normal density curve with the area between x₀ and x₁ shaded — Figure 1 — \( P(x_0 < X < x_1) \) as the shaded area between \( x_0 \) and \( x_1 \).

Probability below \( x_0 \)

\[ P(X < x_0) = \int_{-\infty}^{x_0} \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x-\mu)^2}{2\sigma^2}}\, dx \]

Normal density curve with the left-tail area below x₀ shaded — Figure 2 — \( P(X < x_0) \) as the shaded left-tail area.

Probability above \( x_0 \)

\[ P(X > x_0) = \int_{x_0}^{\infty} \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x-\mu)^2}{2\sigma^2}}\, dx \]

Normal density curve with the right-tail area above x₀ shaded — Figure 3 — \( P(X > x_0) \) as the shaded right-tail area.

Numerical computation. These integrals have no elementary closed form, so they are computed numerically via the standard normal CDF \( \Phi(z) \). The standard-normal transformation is \( z = (x - \mu)/\sigma \), giving \( P(X < x_0) = \Phi(z_0) \), \( P(X > x_0) = 1 - \Phi(z_0) \), and \( P(x_0 < X < x_1) = \Phi(z_1) - \Phi(z_0) \).

Distribution Parameters

Enter the mean \( \mu \) and standard deviation \( \sigma > 0 \) used by all three calculators below.

Mean \( \mu \): Standard deviation \( \sigma \): Decimal places:

1. Probability that \( X \) lies between two values

2. Probability that \( X < x_0 \)

3. Probability that \( X > x_0 \)

Cross-checks. The results satisfy \( P(X < x_0) + P(X > x_0) = 1 \) and \( P(x_0 < X < x_1) = P(X < x_1) - P(X < x_0) \). Use these identities to verify any answer.

Worked Example — Standard Normal

With \( \mu = 0 \), \( \sigma = 1 \), and default calculator values \( (x_0, x_1) = (0, 2) \):

View solution

Compute the z-scores:

\[ z_0 = \frac{0 - 0}{1} = 0, \qquad z_1 = \frac{2 - 0}{1} = 2 \]

Between:

\[ P(0 < X < 2) = \Phi(2) - \Phi(0) \approx 0.9772 - 0.5000 = 0.4772 \]

Below:

\[ P(X < 2) = \Phi(2) \approx 0.9772 \]

Above:

\[ P(X > 2) = 1 - \Phi(2) \approx 0.0228 \]

Cross-check: \( P(X < 2) + P(X > 2) = 0.9772 + 0.0228 = 1.0000 \) ✓

Practice Problems

Practice 1 — General normal

For \( X \sim N(50, 8^2) \), find \( P(X < 42) \).

Answer

\( z = (42 - 50)/8 = -1 \), so \( P(X < 42) = \Phi(-1) \approx 0.1587 \).

Practice 2 — Interval probability

For \( X \sim N(100, 15^2) \), find \( P(85 < X < 115) \).

Answer

\( z_0 = (85 - 100)/15 = -1 \), \( z_1 = (115 - 100)/15 = 1 \).

\[ P(-1 < Z < 1) = \Phi(1) - \Phi(-1) \approx 0.6827 \]

(This is the "68%" empirical rule for \( \mu \pm 1\sigma \).)

Practice 3 — Right tail

For \( X \sim N(70, 5^2) \), find \( P(X > 80) \).

Answer

\( z = (80 - 70)/5 = 2 \), so \( P(X > 80) = 1 - \Phi(2) \approx 0.0228 \).

Frequently Asked Questions

What is the formula for the normal probability density?

\( f(x) = \dfrac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x-\mu)^2}{2\sigma^2}} \) for \(x \in \mathbb{R}\). Probabilities of intervals are obtained by integrating this density over the interval.

How do I compute normal probabilities using the standard normal?

Standardize \(z = (x - \mu)/\sigma\), then use \(P(X < x_0) = \Phi(z_0)\), \(P(X > x_0) = 1 - \Phi(z_0)\), and \(P(x_0 < X < x_1) = \Phi(z_1) - \Phi(z_0)\).

Why can't the normal integral be computed in closed form?

The function \(e^{-x^2}\) has no elementary antiderivative. The standard normal CDF \( \Phi \) must be computed numerically — typically via a rational approximation (Abramowitz & Stegun 26.2.17) accurate to about \(7.5 \times 10^{-8}\).

What identities link the three normal probabilities?

\(P(X < x_0) + P(X > x_0) = 1\), and \(P(x_0 < X < x_1) = P(X < x_1) - P(X < x_0)\). Both follow directly from the area interpretations of the integrals.

More References & Links

Last updated: October 20, 2026