Normal Models For Continuous Measurements

Introduction

A continuous measurement is often modeled by describing where its distribution is centered and how widely it varies around that center. The normal model expresses these features with two parameters: a location, which determines the center, and a positive scale, which determines the spread. Once those parameters are specified, the model gives a precise way to calculate the probability that a measurement falls in a chosen interval.

This tutorial defines the normal model for a continuous response, identifies the values it can take, and develops the standardization calculation used for interval probabilities. The model is an assumption about a response distribution; whether it is appropriate for a particular application depends on the measurement and the scientific context.

Defining a Normal Response Model

Definition: A random variable \(X\) has a normal distribution with location parameter \(\mu\) and scale parameter \(\sigma\), written \(X\sim N(\mu,\sigma^2)\), when \(\mu\in\mathbb{R}\), \(\sigma>0\), and its density is

f(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left\{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right\},\qquad x\in\mathbb{R}.

The parameter \(\mu\) is the location, and \(\sigma\) is the scale (standard deviation). The variance is \(\sigma^2\). The support is all of \(\mathbb{R}\).

The density is positive at every real number, and the model assigns probability zero outside its support. Because the support is all real numbers, a normal model does not impose a finite lower or upper bound on possible measurements. In an application where the response is physically bounded, that feature may be an approximation that must be considered rather than ignored.

The notation \(N(\mu,\sigma^2)\) places the variance, not the standard deviation, in the second position. For example, \(N(100,15^2)\) has location 100, standard deviation 15, and variance 225. The condition \(\sigma>0\) is required for the displayed density; setting \(\sigma=0\) does not give a continuous normal density.

Standardization and Interval Probabilities

The normal family has a common shape. Changing \(\mu\) shifts that shape along the real line, while changing \(\sigma\) stretches or compresses it. This makes it possible to convert a measurement \(X\) into a standardized value \(Z=(X-\mu)/\sigma\). The resulting random variable has the standard normal distribution, \(N(0,1)\).

Definition: The standard normal density is \(\phi(z)=(2\pi)^{-1/2}\exp(-z^2/2)\), for \(z\in\mathbb{R}\). Its cumulative distribution function is

\Phi(z)=\int_{-\infty}^{z}\phi(t)\,dt.

Thus \(\Phi(z)\) is the probability that a standard normal random variable is at most \(z\).

Theorem: If \(X\sim N(\mu,\sigma^2)\), with \(\mu\in\mathbb{R}\) and \(\sigma>0\), then \(Z=(X-\mu)/\sigma\sim N(0,1)\). For any real numbers \(a\le b\),

P(a\le X\le b)=\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right).

Proof. Under the change of variable \(x=\mu+\sigma z\), the density of \(Z=(X-\mu)/\sigma\) is the density of \(X\) evaluated at \(\mu+\sigma z\), multiplied by the absolute Jacobian \(\sigma\). Therefore,

f_Z(z)=\sigma f_X(\mu+\sigma z) =\sigma\frac{1}{\sigma\sqrt{2\pi}}\exp\left\{-\frac{1}{2}\left(\frac{\mu+\sigma z-\mu}{\sigma}\right)^2\right\} =\phi(z).

This proves the standardization claim. For the interval, the same substitution maps \(a\le X\le b\) to \((a-\mu)/\sigma\le Z\le(b-\mu)/\sigma\), since \(\sigma>0\). Integrating the standard normal density over these limits gives

P(a\le X\le b) =\int_{(a-\mu)/\sigma}^{(b-\mu)/\sigma}\phi(z)\,dz =\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right).

A continuous distribution assigns probability zero to an individual point: the integral of its density over a singleton is zero. Thus using strict or non-strict inequalities at either endpoint does not change this interval probability. \(\square\)

Worked Example: A Symmetric Interval Around the Location

Suppose a laboratory measurement is modeled as \(X\sim N(100,15^2)\). Find the probability that a measurement is between 85 and 115.

First record the model parameters: \(\mu=100\) and \(\sigma=15\). Standardize each endpoint:

\frac{85-100}{15}=-1,\qquad \frac{115-100}{15}=1.

Therefore,

P(85\le X\le115)=\Phi(1)-\Phi(-1)=2\Phi(1)-1.

Using \(\Phi(1)\approx0.841344746\), the probability is approximately \(0.682689492\), or \(0.6827\) to four decimal places. The interval is one standard deviation below to one standard deviation above the location.

Worked Example: An Interval Above the Location

Let \(X\sim N(50,4^2)\). Find the probability that \(X\) lies between 54 and 58.

Here \(\mu=50\) and \(\sigma=4\), so the standardized endpoints are

\frac{54-50}{4}=1,\qquad \frac{58-50}{4}=2.

Applying the interval formula,

P(54\le X\le58)=\Phi(2)-\Phi(1).

Using \(\Phi(2)\approx0.977249868\) and \(\Phi(1)\approx0.841344746\), the difference is approximately \(0.135905122\). Thus the requested probability is \(0.1359\) to four decimal places. The endpoints are one and two standard deviations above the location, so this interval is not symmetric around the center.

Worked Example: A One-Sided Measurement Threshold

A device’s measurement is modeled as \(X\sim N(72,8^2)\). Find the probability that a measurement exceeds 80.

The threshold standardized is

\frac{80-72}{8}=1.

The probability above 80 is the upper tail beyond one standard deviation above the location. Using the complement of the CDF,

P(X>80)=1-\Phi(1)\approx1-0.841344746=0.158655254\approx0.1587.

The same result holds for \(P(X\ge80)\), because a continuous normal distribution assigns probability zero to the single value 80.

Why the Location and Scale Matter

The location and scale determine how a measurement interval sits relative to the distribution. Standardizing converts its endpoints into distances measured in standard deviations from the location. An endpoint above \(\mu\) has a positive standardized value; an endpoint below \(\mu\) has a negative one. The interval probability is then the difference between the CDF values at the upper and lower standardized endpoints.

The parameters also describe the distribution’s center and spread in the usual moment sense. The following derivation verifies this interpretation directly from the density. It also confirms why \(\sigma\), rather than \(\sigma^2\), is the scale used to standardize values.

Result: For \(X\sim N(\mu,\sigma^2)\), \(E[X]=\mu\) and \(\operatorname{Var}(X)=\sigma^2\).

Derivation. First consider \(Z\sim N(0,1)\). Its density is symmetric about zero, so the integral of the odd function \(z\phi(z)\) over the real line is zero. This integral exists absolutely because \(|z|\exp(-z^2/2)\) is integrable. Hence \(E[Z]=0\). Also, \(\phi'(z)=-z\phi(z)\). Integration by parts, with the boundary term \(z\phi(z)\) tending to zero at both ends, gives

E[Z^2]=\int_{-\infty}^{\infty}z^2\phi(z)\,dz =\int_{-\infty}^{\infty}z\{-\phi'(z)\}\,dz =\left[-z\phi(z)\right]_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\phi(z)\,dz =1.

The standardization result gives \(X=\mu+\sigma Z\) in distribution. Linearity of expectation and the variance rule for adding a constant and multiplying by a constant therefore yield

E[X]=\mu+\sigma E[Z]=\mu,\qquad \operatorname{Var}(X)=\sigma^2\operatorname{Var}(Z)=\sigma^2\{E[Z^2]-(E[Z])^2\}=\sigma^2.

Thus \(\mu\) is the mean, and \(\sigma\) is the standard deviation. \(\square\)

Common Mistakes and Checks

Normal interval calculations are short, but a few errors can change the answer substantially. Before reporting a probability, check the model parameters, the signs and order of the standardized endpoints, and the precision of any numerical subtraction.

Using the variance as the standard deviation. In \(N(\mu,\sigma^2)\), divide endpoint differences by \(\sigma\), not by \(\sigma^2\). For \(N(50,4^2)\), the standard deviation is 4, not 16.
Reversing the interval endpoints. The upper standardized endpoint goes into the first CDF term. For \(a\le b\), the probability is \(\Phi((b-\mu)/\sigma)-\Phi((a-\mu)/\sigma)\), a nonnegative difference.
Reporting more precision than the calculation supports. Rounded CDF values can produce a subtraction that differs in the last displayed digit from the difference of unrounded values. Keep sufficient precision during calculation, then round the final probability.
Treating the normal support as physically bounded. The model assigns positive probability throughout the real line. A normal model for a quantity that cannot be negative may still be used as an approximation, but its implications should be checked in context.

A useful reasonableness check is to compare an interval with the location and scale. A symmetric interval around \(\mu\) must have equal tail probabilities on both sides. An interval far into one tail should have a smaller probability than a comparable interval near the center. These checks do not replace the calculation, but they can reveal an incorrect standardization or subtraction.

Check Your Understanding

Use the normal model definition and standardization formula to answer the following.

For \(X\sim N(30,6^2)\), state the location, scale, variance, and support.
For \(X\sim N(100,15^2)\), write the standardized endpoints for the interval from 70 to 130.
If \(X\sim N(50,4^2)\), express \(P(46\le X\le54)\) using \(\Phi\), and explain why the two endpoints are symmetric relative to the location.
Why does a normal model with \(\sigma=0\) not satisfy the continuous-density definition given here?
Explain why \(P(X>c)=P(X\ge c)\) for a normal random variable and any real number \(c\).