Calculating the expected range of normal samples

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The previous post looked at the expected IQ range in a jury of 12. This post will look more generally at computing the expected range of n samples from a N(0, 1) random variable. This will give the expected range in units of σ, i.e. multiply the results by σ if your σ isn’t 1.

As mentioned in the previous post, the expected range is given by

where φ and Φ are the PDF and CDF of a standard normal. The integral can be calculated in closed form for n ≤ 5, but in general it requires numerical integration [1].

The following Python code can compute dn.

from scipy.stats import norm<br>from scipy.integrate import quad<br>import numpy as np

def d(n):<br>integrand = lambda x: x*norm.pdf(x)*norm.cdf(x)**(n-1)<br>res, info = quad(integrand, -np.inf, np.inf)<br>return 2*n*res

For large n we have the asymptotic approximation

which we could implement in Python by

def approx(n):<br>return 2*norm.ppf((n - 0.375)/(n + 0.25))

For very large n the asymptotic expression may be more accurate than the integral due to numerical integration error.

Here are a few example values.

|-----+-------|<br>| n | d_n |<br>|-----+-------|<br>| 2 | 1.128 |<br>| 3 | 1.693 |<br>| 5 | 2.326 |<br>| 10 | 3.078 |<br>| 12 | 3.258 |<br>| 23 | 3.858 |<br>| 50 | 4.498 |<br>| 100 | 5.015 |<br>|-----+-------|

[1] Order Statistics by H. A. David. John Wiley & Sons. 1970.

One thought on &ldquo;Calculating the expected range of normal samples&rdquo;

Blaise F Egan

26 May 2026 at 15:38

I just asked Microsoft Copilot to create the code for Mathematica (which I have, but I am a complete newbie at). I was amazed at how easy it was.

Create Mathematica code to numerically evaluate the integral, for n=12, of $2*n* \int_{-infty}^ {infty} \Phi(x)^{n-1} *x*\phi(x) dx$

where $\Phi(x)$ is the CDF of the standard normal and $\phi(x)$ is the PDF.

It gave me

n = 12;

Φ[x_] := CDF[NormalDistribution[], x];

φ[x_] := PDF[NormalDistribution[], x];

N[ 2 n NIntegrate[ Φ[x]^(n – 1) * x * φ[x], {x, -∞, ∞},

Method -> "GlobalAdaptive", WorkingPrecision -> 30 ],

20

It ran for a fraction of a second and gave me:

3.2584552797438259806

with comments:

Notes on the implementation

WorkingPrecision -> 30 ensures numerical stability because the integrand involves a product of a CDF power and a PDF.

Method -> "GlobalAdaptive" helps with tails and smoothness.

The integral converges rapidly, so {-∞, ∞} is fine; Mathematica handles the truncation automatically.

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