What is the normal distribution area?
Enter the mean, the standard deviation, and one or two bounds to get the area under the normal curve, shaded on the bell curve.
- P(a < X < b)
- 0.682689
For a normal distribution with mean 0 and standard deviation 1, P(a < X < b) = 0.682689.
- Outside the bounds
- 0.317311
- P(X < a)
- 0.158655
- P(X > b)
- 0.158655
- z-score of a
- -1
- z-score of b
- 1
P(a < X < b): 0.682689. For a normal distribution with mean 0 and standard deviation 1, P(a < X < b) = 0.682689.
Where is the area under the curve?
How to calculate
Computes the probability that a normal random variable falls between two bounds, below one, or above one, with the tails outside and the z-scores.
Example with the default inputs (Mean (μ) 0, Standard deviation (σ) 1, Lower bound (a) -1, Upper bound (b) 1): For a normal distribution with mean 0 and standard deviation 1, P(a < X < b) = 0.682689.
Method: z = (x − μ) ÷ σ; P(a < X < b) = Φ(z(b)) − Φ(z(a)), where Φ is the standard normal cumulative distribution function; an empty bound is −∞ or +∞.
- X follows a normal distribution with the mean and standard deviation you enter.
- Each area is worked out from the tail it lies in, not by subtracting from 1, so tiny probabilities keep their precision.
Worked examples
Each example is checked against the calculator on every build.
- Mean (μ) 0, Standard deviation (σ) 1, Lower bound (a) -1, Upper bound (b) 1 gives P(a < X < b) 0.682689, Outside the bounds 0.317311, P(X < a) 0.158655.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.1 Normal Distribution. https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm
- Mean (μ) 0, Standard deviation (σ) 1, Lower bound (a) -1.96, Upper bound (b) 1.96 gives P(a < X < b) 0.950004, Outside the bounds 0.049996.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.1 Normal Distribution. https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm (table: Φ(1.96) = 0.97500)
- Mean (μ) 100, Standard deviation (σ) 15, Upper bound (b) 130 gives P(a < X < b) 0.97725, P(X > b) 0.02275, z-score of b 2.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.1 Normal Distribution. https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm
- Mean (μ) 5, Standard deviation (σ) 2, Lower bound (a) 7 gives P(a < X < b) 0.158655, z-score of a 1.Source: OpenStax, Introductory Statistics 2e, §6.2 Using the Normal Distribution. https://openstax.org/books/introductory-statistics-2e/pages/6-2-using-the-normal-distribution
- Mean (μ) 0, Standard deviation (σ) 1, Lower bound (a) 10 gives P(a < X < b) 0.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.1 Normal Distribution. https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm
- Mean (μ) 0, Standard deviation (σ) 1, Lower bound (a) 8, Upper bound (b) 9 gives P(a < X < b) 0.Source: NIST/SEMATECH e-Handbook of Statistical Methods, §1.3.6.6.1 Normal Distribution. https://www.itl.nist.gov/div898/handbook/eda/section3/eda3661.htm
How it works
X follows a normal distribution with mean μ and standard deviation σ. Enter a lower bound a, an upper bound b, or both. An empty lower bound means −∞ and an empty upper bound means +∞.
- The z-scores: z(a) = (a − μ) ÷ σ and z(b) = (b − μ) ÷ σ, worked out exactly from the decimals typed, then rounded once to the nearest 64-bit float. Every probability below uses these rounded z-scores.
- Φ(z) is the standard normal cumulative distribution function, Φ(z) = ½ × erfc(−z ÷ √2), with Φ(−∞) = 0 and Φ(+∞) = 1.
- The results:
- P(a < X < b) = Φ(z(b)) − Φ(z(a)).
- P(X < a) = Φ(z(a)), shown when there is a lower bound.
- P(X > b) = Φ(−z(b)), shown when there is an upper bound.
- Outside the bounds = P(X < a) + P(X > b) = 1 − P(a < X < b).
- z-score of a and z-score of b, each shown when its bound is given.
To keep small areas precise, the area between the bounds is worked out from the tail it lies in: Φ(z(b)) − Φ(z(a)) when both z-scores are 0 or below, Φ(−z(a)) − Φ(−z(b)) when both are 0 or above, and 1 − Φ(z(a)) − Φ(−z(b)) when the bounds are on either side of the mean.
For a narrow range, two nearly equal tails would cancel, so the area is the integral of the normal density instead. With both bounds given, let w = (b − a) ÷ σ, worked out exactly from the decimals typed. When w × max(1, the smaller of |z(a)| and |z(b)|) ≤ 1, P(a < X < b) = ∫ φ(z) dz from z(a) to z(a) + w, where φ(z) = e^(−z²/2) ÷ √(2π), by 10-point Gauss–Legendre quadrature (Abramowitz and Stegun, table 25.4). This keeps about 12 significant digits even for a range such as 1 to 1.0000001.
Rules
- The mean is from −10¹² to 10¹². The standard deviation is from 10⁻¹² to 10¹². Each bound is from −10¹⁵ to 10¹⁵.
- Enter at least one bound; with neither there is no answer.
- With both bounds, the lower bound must be below the upper bound; otherwise there is no answer.
- Φ is good to at least 12 significant digits, including far into the tails (checked against Python's math.erfc).
- Results show up to 6 decimal places (6 significant digits below 0.0001), with halves rounded up.
Worked examples by hand
Within one standard deviation (μ = 0, σ = 1, a = −1, b = 1). Φ(1) − Φ(−1) = erf(1 ÷ √2) = 0.682689. Outside: 0.317311, split evenly: P(X < −1) = 0.158655.
The middle 95% (a = −1.96, b = 1.96). NIST's table gives Φ(1.96) = 0.97500, so the area is about 1 − 2 × 0.025 = 0.95; to more digits 0.950004, and outside 0.049996.
Below 130 when μ = 100, σ = 15 (no lower bound). z(b) = (130 − 100) ÷ 15 = 2. P(X < 130) = Φ(2) = 0.97725; P(X > 130) = 0.02275.
Above 7 when μ = 5, σ = 2 (no upper bound). z(a) = (7 − 5) ÷ 2 = 1. P(X > 7) = Φ(−1) = 0.158655.
A far tail (μ = 0, σ = 1, a = 10, no upper bound). P(X > 10) = ½ × erfc(10 ÷ √2) = 7.61985 × 10⁻²⁴.
Between 8 and 9 standard deviations (a = 8, b = 9). Φ(−8) − Φ(−9) = 6.22096 × 10⁻¹⁶ − 1.12859 × 10⁻¹⁹ = 6.21983 × 10⁻¹⁶.
Other questions people ask
How do I find a probability from a normal distribution?
Turn each bound into a z-score, z = (x − μ) ÷ σ, then read the area from the standard normal cumulative distribution Φ. The probability between a and b is Φ(z(b)) − Φ(z(a)). With mean 100 and standard deviation 15, the chance of a value below 130 is Φ(2) = 0.97725.
How do I get P(X < b) or P(X > a)?
Leave the other bound empty. With only an upper bound b, the calculator gives P(X < b); with only a lower bound a, it gives P(X > a). The area outside the bounds is shown too.
What is the empirical rule?
For any normal distribution, about 68.27% of values lie within 1 standard deviation of the mean, 95.45% within 2, and 99.73% within 3. Enter bounds of μ − σ and μ + σ to see the first: 0.682689.
Is P(X < b) the same as P(X ≤ b)?
Yes. A normal distribution is continuous, so any single value has probability 0, and it makes no difference whether the bounds are included.
Why does the calculator show such tiny probabilities?
Far from the mean the tails are very thin but not zero: P(Z > 10) is about 7.6 × 10⁻²⁴. The calculator works out each tail directly instead of subtracting from 1, so these small areas keep their precision.
How do I find the value for a given probability?
That is the inverse question. The z-score calculator finds the z for a percentile, and the value is then x = μ + z × σ. The critical value calculator gives the z or t that cuts off a tail of a given size.