acalculator

InvNorm calculator: x from an area

Type an area (a probability such as 0.95), the mean and the standard deviation, and say where the area lies. The inverse normal calculator finds x, like invNorm on a graphing calculator, and shades the area on the curve.

Your numbers

Area is
x
1.644853627

An area of 0.95 under this normal curve lies below 1.644853627.

z-score
1.644853627
The area lies
below 1.644853627

x: 1.644853627. An area of 0.95 under this normal curve lies below 1.644853627.

Where is the area?

How to calculate

Finds the value x with a given area under a normal curve to its left, to its right, or in the center (invNorm), with the z-score and a chart of the shaded area.

Example with the default inputs (Area (probability) 0.95, Area is Left of x, Mean (μ) 0, Standard deviation (σ) 1): An area of 0.95 under this normal curve lies below 1.644853627.

Method: x = μ + σz. Left: Φ(z) = area. Right: 1 − Φ(z) = area. Center: Φ(z) − Φ(−z) = area, with ends μ ± σz. Φ is the standard normal CDF.

  • The values follow a normal distribution with the mean and standard deviation you type.
  • The area is a probability, more than 0 and less than 1 (at least 10⁻³⁰⁰).
  • Left and right match invNorm(area, μ, σ, LEFT) and RIGHT; center splits the rest of the area evenly between the two tails.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Area (probability) 0.95, Area is Left of x, Mean (μ) 0, Standard deviation (σ) 1 gives x 1.644854, z-score 1.644854.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.7.1 Cumulative Distribution Function of the Standard Normal Distribution (table of the area from 0 to z: 0.44950 at z = 1.64, 0.45053 at 1.65, 0.47500 at 1.96). https://www.itl.nist.gov/div898/handbook/eda/section3/eda3671.htm (1.6449 lies between 1.64 and 1.65); NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.1 Normal Distribution (the percent point function, computed numerically). https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm
  2. Area (probability) 0.95, Area is Center, Mean (μ) 0, Standard deviation (σ) 1 gives Lower end -1.959964, Upper end 1.959964, The area lies between −1.959963985 and 1.959963985.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.7.1 Cumulative Distribution Function of the Standard Normal Distribution (table of the area from 0 to z: 0.44950 at z = 1.64, 0.45053 at 1.65, 0.47500 at 1.96). https://www.itl.nist.gov/div898/handbook/eda/section3/eda3671.htm (the area from 0 to 1.96 is 0.475, so the middle 95% runs from −1.96 to 1.96)
  3. Area (probability) 0.1, Area is Right of x, Mean (μ) 100, Standard deviation (σ) 15 gives x 119.223273, z-score 1.281552.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.1 Normal Distribution (the percent point function, computed numerically). https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm
  4. Area (probability) 0.5, Area is Left of x, Mean (μ) 7, Standard deviation (σ) 2 gives x 7, z-score 0.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.1 Normal Distribution (the percent point function, computed numerically). https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm

How it works

Let Φ be the standard normal cumulative distribution function (the area to the left of z under the curve with mean 0 and SD 1). For an area A, a mean μ and a standard deviation σ:

  • Left of x: find z with Φ(z) = A. Then x = μ + σz.
  • Right of x: find z with 1 − Φ(z) = A, which is z = −(the left z for A). Then x = μ + σz.
  • Center: find z ≥ 0 with Φ(z) − Φ(−z) = A. The ends are μ − σz and μ + σz; x shows the upper end.

z is the percent point function of the standard normal (Φ⁻¹), found numerically to full double precision. To stay precise, the page solves on the side with the most digits: for a left area below 0.25 it solves Φ(z) = A; above 0.75 it solves Φ(−z) = 1 − A; in between it uses the center form with 2A − 1. The center form P(|Z| < z) = A is erf(z/√2), summed as its series, which stays precise when z is near 0 (an area of 10⁻³⁰⁰ gives z = 1.2533 × 10⁻³⁰⁰). A center area above ½ solves Φ(−z) = (1 − A) ÷ 2.

Rules

  • The area is at least 10⁻³⁰⁰ and less than 1.
  • The mean is from −10¹² to 10¹². The standard deviation is from 10⁻¹² to 10¹².
  • An area of exactly 0.5 to the left (or right) gives x = μ and z = 0.
  • The text "The area lies" reads "below x", "above x", or "between lower and upper", with each number to 10 significant digits and a true minus sign (−).

Output format. x, z and the ends are double-precision numbers shown to 10 significant digits.

Worked examples by hand

invNorm(0.95, 0, 1), left. The table gives the area from 0 to z as 0.44950 at 1.64 and 0.45053 at 1.65, so Φ(1.64) = 0.94950 and Φ(1.65) = 0.95053, and z is between 1.64 and 1.65. Numerically z = 1.644854, and x = 0 + 1 × z = 1.644854.

Center 0.95, mean 0, SD 1. Each tail holds (1 − 0.95) ÷ 2 = 0.025, so the upper end has Φ(z) = 0.975 and z = 1.959964. The middle 95% runs from −1.959964 to 1.959964.

Right 0.1, mean 100, SD 15. The top 10% starts where Φ(z) = 0.9: z = 1.281552. x = 100 + 15 × 1.281552 = 119.2233.

Left 0.5, mean 7, SD 2. Half the area lies below the mean, so z = 0 and x = 7.

Other questions people ask

What does invNorm do?

It works backwards from an area to a value. invNorm(0.95, 0, 1) is the z with 95% of the standard normal curve to its left: about 1.6449.

How do I find x for a mean and standard deviation that are not 0 and 1?

Find z for the area on the standard normal curve, then x = μ + zσ. The top 10% of a scale with mean 100 and SD 15 starts at 100 + 1.28155 × 15 ≈ 119.22.

What is the difference between left, right and center?

Left finds x with the area below it. Right finds x with the area above it, which is the same as left with 1 − area. Center finds the two ends, μ − zσ and μ + zσ, with the area between them and the rest split evenly between the two tails.

Why is the middle 95% from −1.96 to 1.96?

The 5% left over splits into 2.5% in each tail. The standard normal table gives an area of 0.475 from 0 to 1.96, so 0.5 + 0.475 = 97.5% lies below 1.96 and 2.5% above it.

Can the area be 0 or 1?

No. An area of 0 or 1 would need x at minus or plus infinity. The page takes areas more than 0 (at least 10⁻³⁰⁰) and less than 1.

Is the inverse normal the same as the percentile?

For a left area, yes: invNorm(0.9, μ, σ) is the 90th percentile of the normal distribution with mean μ and SD σ.