Normal Location-Scale Parameterization

Location and Scale in the Normal Model

A normal model is often used when measurements fluctuate around a typical value and are described on a particular measurement scale. Its two parameters answer different questions: where is the distribution centered? and how spread out is it? Keeping those roles distinct makes it possible to change units, compare observations to a standard reference, and calculate probabilities without repeatedly integrating a new density.

In this tutorial, $\mu$ is the location and $\sigma$ is the scale. The scale is positive, and the variance is its square. This distinction is important because normal notation conventionally places the variance—not the standard deviation—in the second argument.

Definition: For $\mu\in\mathbb{R}$ and $\sigma>0$, the normal distribution with location $\mu$ and scale $\sigma$, written $X\sim N(\mu,\sigma^2)$, has density $$ f_X(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left\{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right\},\qquad -\infty<x<\infty. $$ The parameter $\mu$ is the location, $\sigma$ is the scale (the standard deviation), and the variance is $\sigma^2$. The standard normal distribution is $N(0,1)$, with density $\phi(z)=(2\pi)^{-1/2}\exp(-z^2/2)$ and CDF $\Phi(z)=\int_{-\infty}^{z}\phi(t)\,dt$.

The density formula is a shifted and rescaled version of the standard normal density: $f_X(x)=\sigma^{-1}\phi((x-\mu)/\sigma)$. The factor $1/\sigma$ adjusts the density height when the horizontal scale changes; it ensures the density still integrates to one. The standardization $z=(x-\mu)/\sigma$ expresses a value as its signed distance from $\mu$, measured in units of $\sigma$.

Standardizing a Normal Variable

The standard normal variable gives one common scale for all normal models. The following result justifies standardization directly from the density.

Theorem: If $X\sim N(\mu,\sigma^2)$, where $\sigma>0$, then $$ Z=\frac{X-\mu}{\sigma}\sim N(0,1). $$ Consequently, for every real $x$, $$ P(X\le x)=\Phi\left(\frac{x-\mu}{\sigma}\right). $$

Proof. The map $x\mapsto z=(x-\mu)/\sigma$ is strictly increasing, with inverse $x=\mu+\sigma z$ and derivative $dx/dz=\sigma$. Applying the change-of-variables formula to the density of $X$,

f_Z(z)=f_X(\mu+\sigma z)\sigma =\frac{1}{\sigma}\phi(z)\sigma =\phi(z).

Thus $Z$ has the standard normal density. Since the transformation is increasing, the event $X\le x$ is exactly the event $Z\le (x-\mu)/\sigma$, giving the stated CDF formula. $\square$

The parameters also have their usual moment interpretation. The standard normal density is symmetric about zero, so its mean is zero. For the variance, integration by parts gives

E[Z^2]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^2e^{-z^2/2}\,dz =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-z^2/2}\,dz =1.

In the integration by parts, use $z^2e^{-z^2/2}=z(ze^{-z^2/2})$; the boundary term $-ze^{-z^2/2}$ tends to zero at both limits, and the remaining integral is the normalized Gaussian integral. Since $X=\mu+\sigma Z$, linearity of expectation and the variance rule for multiplication by a constant yield $E[X]=\mu$ and $\operatorname{Var}(X)=\sigma^2$. This confirms why $\mu$ is called location and $\sigma$ scale.

Worked Example: Standardizing a Measurement

Suppose an adult’s resting heart rate $X$, in beats per minute, is modeled as $N(70,8^2)$. Find the standardized value for $X=82$, and express the probability that the heart rate is at most 82.

The location is $\mu=70$, and the scale is $\sigma=8$ beats per minute. Therefore,

z=\frac{82-70}{8}=\frac{12}{8}=1.5.

The observation is 1.5 standard deviations above the model’s location. By the standardization result, $P(X\le82)=\Phi(1.5)$. Using the supplied standard normal value $\Phi(1.5)\approx0.9332$, the probability is approximately $0.9332$. The standardization determines which standard normal probability is needed; evaluating $\Phi$ numerically is a separate table or software step.

Affine Changes of Location and Scale

A measurement may be shifted or recorded in different units. If $Y=a+bX$, the constant $a$ shifts every value and $b$ multiplies every value. For a normal variable, the transformed distribution remains normal. The absolute value of $b$ matters for spread: a negative multiplier reverses direction but cannot produce a negative standard deviation.

Theorem: If $X\sim N(\mu,\sigma^2)$, with $\sigma>0$, and $a,b\in\mathbb{R}$ with $b\ne0$, then $$ Y=a+bX\sim N(a+b\mu,\;b^2\sigma^2). $$ Its location is $a+b\mu$, and its scale is $|b|\sigma$.

Proof. The inverse transformation is $x=(y-a)/b$, and its absolute derivative is $1/|b|$. The change-of-variables formula gives

f_Y(y)=\frac{1}{|b|}f_X\left(\frac{y-a}{b}\right) =\frac{1}{|b|\sigma}\phi\left(\frac{y-a-b\mu}{b\sigma}\right).

Because $\phi(u)=\phi(-u)$, the last density can be written with denominator $|b|\sigma$ inside its squared argument:

f_Y(y)=\frac{1}{|b|\sigma}\phi\left(\frac{y-(a+b\mu)}{|b|\sigma}\right).

This is the normal density with location $a+b\mu$ and scale $|b|\sigma$, so its variance is $b^2\sigma^2$. $\square$

For probability calculations, standardize the endpoints. If $l<u$, continuity of the normal distribution means endpoint inclusion does not change the probability, and

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

The upper standardized endpoint appears first because the CDF gives probability to the left of its argument. For an affine transformation with a negative multiplier, it is usually safest to transform the event carefully or use the transformed normal model and order the resulting endpoints from low to high.

Worked Example: Probability Within an Interval

Let $X\sim N(70,8^2)$. Calculate $P(62\le X\le82)$.

Standardize each endpoint using the same location and scale:

\frac{62-70}{8}=-1,\qquad \frac{82-70}{8}=1.5.

Thus the event becomes $-1\le Z\le1.5$, and subtracting the lower-tail probabilities gives

P(62\le X\le82)=\Phi(1.5)-\Phi(-1) \approx0.9332-0.1587 =0.7745.

The interval contains approximately $77.45\%$ of the modeled probability. Both endpoints must be standardized: using only the upper endpoint would give the probability below 82, not the probability between 62 and 82.

Worked Example: A Negative Multiplier and a Transformed Interval

Suppose $X\sim N(4,1.5^2)$ and a reported score is $Y=20-2X$. Find $P(10<Y\le16)$.

First transform the model. Here $a=20$ and $b=-2$, so $Y$ has location $20-2(4)=12$, scale $|-2|(1.5)=3$, and variance $3^2=9$. Hence $Y\sim N(12,9)$. Directly standardizing the ordered score endpoints gives

\frac{10-12}{3}=-\frac{2}{3},\qquad \frac{16-12}{3}=\frac{4}{3}.

Therefore,

P(10<Y\le16) =\Phi\left(\frac{4}{3}\right)-\Phi\left(-\frac{2}{3}\right) \approx0.9088-0.2525 =0.6563.

As a check on the negative sign, transform the event back to $X$. The inequality $10<20-2X\le16$ gives $2\le X<5$: the upper bound on $Y$ produces the lower bound on $X$, while the lower bound on $Y$ produces the upper bound on $X$. Standardizing these $X$-endpoints gives $(2-4)/1.5=-4/3$ and $(5-4)/1.5=2/3$. Thus the probability is $\Phi(2/3)-\Phi(-4/3)$, which equals the previous expression by the symmetry identity $\Phi(-z)=1-\Phi(z)$.

Common Mistakes and Why the Parameterization Matters

The location-scale form is useful because it separates a model’s shape from its units. Standardization lets a single reference distribution handle calculations for many choices of $\mu$ and $\sigma$. Affine transformations then account for recoding, unit conversion, or a score that reverses direction. These operations are common in measurement and modeling, but they are also frequent sources of avoidable errors.

Using the variance as the scale. In $N(\mu,\sigma^2)$, the standardizing denominator is $\sigma$, not $\sigma^2$. If the model is $N(70,64)$, its scale is $8$, not $64$.
Forgetting the absolute value for a negative multiplier. If $Y=a+bX$, its standard deviation is $|b|\sigma$, even when $b<0$. Its variance is $b^2\sigma^2$.
Leaving an inequality unreversed after dividing by a negative number. When transforming an event through $Y=a+bX$ with $b<0$, the order of the bounds reverses. Check the transformed bounds by testing a value or use the model for $Y$ directly.
Subtracting CDF values in the wrong order. For an interval, subtract the CDF at the lower endpoint from the CDF at the upper endpoint.

Key takeaway: For $X\sim N(\mu,\sigma^2)$, standardize with $(X-\mu)/\sigma$. Under $Y=a+bX$, the transformed location is $a+b\mu$, the scale is $|b|\sigma$, and the variance is $b^2\sigma^2$.

Check Your Understanding

For each question, keep the scale distinct from the variance and show how the endpoints or parameters are transformed.

If $X\sim N(12,25)$, what are its location and scale, and what is the standardized value of $X=17$?
For $X\sim N(50,10^2)$, express $P(40\le X\le65)$ using $\Phi$, without evaluating the CDFs.
If $X\sim N(3,2^2)$ and $Y=5+4X$, state the location, scale, and variance of $Y$.
If $X\sim N(0,1)$ and $Y=8-3X$, what is the location and scale of $Y$, and why is its scale positive?
Explain why standardizing both endpoints is necessary when calculating the probability that a normal variable falls within an interval.

Normal Location-Scale Parameterization

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