Why are boundary probabilities P₀ = 0 and P₀ = 1 not valid?

For P₀ = 0 the threshold is −∞; for P₀ = 1 it is +∞. The standard normal quantile Φ⁻¹(0) = −∞ and Φ⁻¹(1) = +∞, so these boundary values don't produce finite x₀. The calculator rejects P₀ = 0 and P₀ = 1 with a clear message.

Inverse Normal Probability Calculator — Find x from a Probability

Q: What is the inverse normal problem?

The inverse normal problem asks: given a target probability P₀ and a normal distribution N(μ, σ²), find the value x₀ such that P(X x₀) = P₀, or P(x₀ < X < x₁) = P₀ for a centered interval. The answer is x₀ = μ + σ·Φ⁻¹(P₀), where Φ⁻¹ is the standard normal quantile function.

Q: How do I find the inverse normal value?

For a left-tail probability P₀, compute z = Φ⁻¹(P₀) and then x₀ = μ + σ·z. For a right-tail probability P₀, compute z = Φ⁻¹(1 − P₀) and x₀ = μ + σ·z. For a centered interval probability P₀, compute z = Φ⁻¹((1 + P₀)/2) and the bounds are x₀ = μ − σ·z and x₁ = μ + σ·z.

Q: What is the connection to critical values?

Critical values in hypothesis testing and confidence intervals are inverse normal outputs. For a two-sided 95% confidence interval, the critical value z = Φ⁻¹(0.975) ≈ 1.96. For a left-tail test at α = 0.05, the critical value is z = Φ⁻¹(0.05) ≈ −1.645.

Overview

This calculator solves the inverse problem for the normal distribution: given a target probability \( P_0 \), find the corresponding value \( x_0 \) — or the two bounds \( [x_0, x_1] \) for a centered interval. Three modes are supported:

Left-tail: find \( x_0 \) such that \( P(X < x_0) = P_0 \)
Right-tail: find \( x_0 \) such that \( P(X > x_0) = P_0 \)
Centered interval: find \( [x_0, x_1] \) symmetric about \( \mu \) such that \( P(x_0 < X < x_1) = P_0 \)

Normal Distribution Formulas

The probability density function of \( X \sim N(\mu, \sigma^2) \) is:

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

Probabilities of a normal random variable
Event	Probability
Below \( x_0 \)	\( P(X < x_0) = \displaystyle\int_{-\infty}^{x_0} f_X(x)\,dx \)
Above \( x_0 \)	\( P(X > x_0) = \displaystyle\int_{x_0}^{\infty} f_X(x)\,dx \)
Between \( x_0 \) and \( x_1 \)	\( P(x_0 < X < x_1) = \displaystyle\int_{x_0}^{x_1} f_X(x)\,dx \)

Standard-normal transformation. Every calculation reduces to the standard normal \( Z = (X - \mu)/\sigma \) via the quantile function \( z_0 = \Phi^{-1}(P_0) \), then inverts to \( x_0 = \mu + \sigma z_0 \).

Distribution Parameters

Set the mean, standard deviation, and decimal places before using any of the three calculators below.

Mean \( \mu \): Std. dev. \( \sigma \): Decimal places:

1. Left-tail inverse: find x₀ such that P(X < x₀) = P₀

Enter \( P_0 \in (0, 1) \) to find the threshold \( x_0 \) with left-tail area \( P_0 \).

Left-tail result

2. Right-tail inverse: find x₀ such that P(X > x₀) = P₀

Enter \( P_0 \in (0, 1) \) to find the threshold \( x_0 \) with right-tail area \( P_0 \).

Right-tail result

3. Centered interval: find [x₀, x₁] with P(x₀ < X < x₁) = P₀

Enter \( P_0 \in (0, 1) \) to find the bounds \( [x_0, x_1] \) symmetric about \( \mu \) that contain probability \( P_0 \).

Centered interval bounds

Boundary cases. For \( P_0 = 0 \) and \( P_0 = 1 \) the threshold is \( -\infty \) or \( +\infty \) respectively — the calculator rejects these values with a message.

Worked Example — Standard Normal

With \( \mu = 0 \) and \( \sigma = 1 \) (standard normal), and \( P_0 = 0.2 \):

View solution

Left-tail. Solve \( \Phi(x_0) = 0.2 \):

\[ x_0 = \Phi^{-1}(0.2) \approx -0.8416 \]

Right-tail. Solve \( P(X > x_0) = 0.2 \), i.e. \( \Phi(x_0) = 0.8 \):

\[ x_0 = \Phi^{-1}(0.8) \approx 0.8416 \]

Centered interval (P₀ = 0.6). Solve \( \Phi(z) - \Phi(-z) = 0.6 \), i.e. \( \Phi(z) = 0.8 \):

\[ z = \Phi^{-1}(0.8) \approx 0.8416 \] \[ [x_0, x_1] = [-\sigma z, +\sigma z] \approx [-0.8416, 0.8416] \]

Observation. For the standard normal, the left-tail and right-tail thresholds for the same \( P_0 \) are negatives of each other — a consequence of symmetry about 0.

Practice Problems

Practice 1 — Left-tail critical value

For \( X \sim N(100, 15^2) \), find \( x_0 \) such that \( P(X < x_0) = 0.05 \).

Answer

\( z = \Phi^{-1}(0.05) \approx -1.6449 \).

\[ x_0 = 100 + 15(-1.6449) \approx 75.33 \]

Practice 2 — Right-tail critical value

For \( X \sim N(100, 15^2) \), find \( x_0 \) such that \( P(X > x_0) = 0.10 \).

Answer

\( z = \Phi^{-1}(0.90) \approx 1.2816 \).

\[ x_0 = 100 + 15(1.2816) \approx 119.22 \]

Practice 3 — 95% centered interval

For \( X \sim N(0, 1) \), find the centered 95% interval \( [x_0, x_1] \).

Answer

Solve \( \Phi(z) = 0.975 \):

\[ z = \Phi^{-1}(0.975) \approx 1.9600 \] \[ [x_0, x_1] \approx [-1.9600, 1.9600] \]

(This is the source of the "±1.96" rule of thumb for 95% confidence intervals.)

Frequently Asked Questions

What is the inverse normal problem?

Given a target probability \(P_0\) and a normal distribution \(N(\mu, \sigma^2)\), find \(x_0\) such that \(P(X < x_0) = P_0\), or \(P(X > x_0) = P_0\), or \(P(x_0 < X < x_1) = P_0\) for a centered interval. Answer: \(x_0 = \mu + \sigma\,\Phi^{-1}(P_0)\).

How do I find the inverse normal value?

Left-tail: \(z = \Phi^{-1}(P_0)\), then \(x_0 = \mu + \sigma z\). Right-tail: \(z = \Phi^{-1}(1 - P_0)\), then \(x_0 = \mu + \sigma z\). Centered: \(z = \Phi^{-1}((1+P_0)/2)\), then \([x_0, x_1] = [\mu - \sigma z, \mu + \sigma z]\).

Why are \(P_0 = 0\) and \(P_0 = 1\) not valid?

\(\Phi^{-1}(0) = -\infty\) and \(\Phi^{-1}(1) = +\infty\) — the threshold is infinite, not a finite \(x_0\). The calculator rejects these boundary values with a clear message.

What is the connection to critical values?

Critical values in hypothesis testing and confidence intervals are inverse normal outputs. Two-sided 95% CI: \(z = \Phi^{-1}(0.975) \approx 1.96\). Left-tail test at \(\alpha = 0.05\): \(z = \Phi^{-1}(0.05) \approx -1.645\).

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Last updated: October 20, 2026