Normal Location-Scale Parameters

Why Location and Scale Matter

A normal model describes a continuous measurement using two parameters. One places the density along the measurement axis; the other determines how broadly the density is spread. These interpretations are useful when specifying a model, checking whether its parameters match a stated mean and standard deviation, and explaining what a fitted model implies.

The notation used here is \(X\sim N(\mu,\sigma^2)\). The second argument is the variance, not the standard deviation. In this model, \(\mu\) is the location parameter and \(\sigma\) is the scale parameter. Keeping \(\sigma\) and \(\sigma^2\) distinct is essential in both formulas and interpretation.

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

f_X(x)=\frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right), \qquad -\infty<x<\infty.

The density is defined on the whole real line. It assigns positive probability to every interval of positive length on the real line, including intervals below any physical lower bound. It assigns probability zero to any single point; probabilities for intervals come from areas under the density. If negative values or other physically impossible values make the model unsuitable, a different model may be needed.

Interpreting the Two Parameters

The location parameter \(\mu\) determines where the density is centered. Its highest point occurs at \(x=\mu\), and changing \(\mu\) shifts the density without changing its shape or spread. The scale parameter \(\sigma\) controls spread: a larger \(\sigma\) produces a wider, flatter density, while a smaller \(\sigma\) produces a narrower, taller one.

The parameter \(\sigma\) is the standard deviation, and the variance is \(\sigma^2\). Thus \(\mu\) has the same units as \(X\), \(\sigma\) also has the same units as \(X\), and \(\sigma^2\) has squared units. For example, if \(X\) is measured in hours, \(\mu\) and \(\sigma\) are measured in hours, while the variance is in hours squared.

Theorem: If \(X\sim N(\mu,\sigma^2)\) with \(\sigma>0\), then

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

This is why \(\mu\) and \(\sigma\) are also called the mean and standard deviation of the normal distribution. They are not merely visual descriptors: they equal the distribution’s first two central summaries.

Why the Parameters Equal the Mean and Variance

A brief derivation makes the connection precise. In the expectation integral, substitute \(z=(x-\mu)/\sigma\), so \(x=\mu+\sigma z\). The normal density times \(dx\) becomes the standard normal density \(\phi(z)\) times \(dz\). The standard normal density integrates to 1, and \(z\phi(z)\) is odd, so its integral over the real line is 0. Therefore:

E[X] =\int_{-\infty}^{\infty}(\mu+\sigma z)\phi(z)\,dz =\mu\int_{-\infty}^{\infty}\phi(z)\,dz +\sigma\int_{-\infty}^{\infty}z\phi(z)\,dz =\mu.

For the variance, subtract the mean \(\mu\). The squared deviation is \(\sigma^2z^2\). Since \(\phi'(z)=-z\phi(z)\), integration by parts gives \(\int_{-\infty}^{\infty}z^2\phi(z)\,dz=1\): the boundary term \(-z\phi(z)\) vanishes at both ends, and the remaining integral is \(\int\phi(z)\,dz=1\). Hence:

\operatorname{Var}(X) =E[(X-\mu)^2] =\sigma^2\int_{-\infty}^{\infty}z^2\phi(z)\,dz =\sigma^2.

This argument uses the standard normal density’s total integral and symmetry. It establishes the parameter interpretation needed here without requiring probability calculations for particular intervals.

The Density’s Position, Spread, and Peak

The exponent in the density depends on the squared distance from \(\mu\), scaled by \(2\sigma^2\). At a fixed \(\sigma\), the density is largest where \((x-\mu)^2\) is smallest, namely at \(x=\mu\). Substituting \(x=\mu\) gives the maximum density:

f_X(\mu)=\frac{1}{\sigma\sqrt{2\pi}}.

Thus, doubling \(\sigma\) halves the peak height and increases the variance by a factor of four. This inverse relationship between scale and peak height preserves the total area under the density at 1. Density values have inverse units: if \(X\) is measured in hours, a density value has units of per hour. A density is not itself a probability.

The following table summarizes what changes when one parameter changes while the other is held fixed.

Parameter change	Effect on the density	Effect on mean and variance
Increase \(\mu\), hold \(\sigma\) fixed	Shifts the entire density to the right	Mean increases; variance is unchanged
Decrease \(\mu\), hold \(\sigma\) fixed	Shifts the entire density to the left	Mean decreases; variance is unchanged
Increase \(\sigma\), hold \(\mu\) fixed	Widens and lowers the density	Mean is unchanged; variance increases to \(\sigma^2\)
Decrease \(\sigma\), hold \(\mu\) fixed	Narrows and raises the density	Mean is unchanged; variance decreases to \(\sigma^2\)

Worked Example: Interpreting a Model for Processing Times

Hypothetical scenario. A firm models the processing time \(X\), in hours, for a particular type of document as \(X\sim N(3.5,0.8^2)\). No interest rate or mortality assumption is involved. Identify the distribution’s parameters, calculate the density at 4.3 hours, and state how the model’s center and spread should be interpreted. Round the density to four decimal places.

Method. Read \(\mu\) and \(\sigma\) from the notation, then substitute them into the density formula. The model gives \(\mu=3.5\) hours and \(\sigma=0.8\) hours, so its variance is \(\sigma^2=0.8^2=0.64\) hours squared. Its mean is 3.5 hours.

At \(x=4.3\), the difference from the location is \(4.3-3.5=0.8\) hours. The exponent is therefore \(-0.8^2/(2(0.8^2))=-0.5\). Substitution gives:

\begin{aligned} f_X(4.3) &=\frac{1}{0.8\sqrt{2\pi}}e^{-0.5}\\ &=\frac{0.6065307}{0.8(2.5066283)}\\ &=\frac{0.6065307}{2.0053026}\\ &\approx 0.3025\text{ per hour}. \end{aligned}

The density’s center is 3.5 hours, and its scale is 0.8 hours. The calculated value is a density, not the probability that processing takes exactly 4.3 hours; that probability is zero for a continuous model. Common mistake avoided: interpreting \(0.8^2\) as the standard deviation. The standard deviation is 0.8 hours; 0.64 is the variance.

Check. The location difference at 4.3 hours equals one standard deviation, so the exponent is \(-1/2\), as used. The density is positive and has units of per hour.

Worked Example: Comparing Two Sensor Models

Hypothetical scenario. Two calibrated sensors are modeled separately. Sensor A’s reading \(X_A\), in units, has distribution \(N(72,3^2)\); sensor B’s reading \(X_B\) has distribution \(N(75,6^2)\). No interest rate or mortality assumption is involved. For each model, give the mean, variance, standard deviation, and peak density, then compare the centers and spreads. Round peak densities to four decimal places.

Method. Use the first parameter as \(\mu\), and take the square root of the second parameter to obtain \(\sigma\). For sensor A, \(\mu_A=72\), \(\sigma_A=3\), and \(\operatorname{Var}(X_A)=3^2=9\). Its peak density is:

f_{X_A}(72)=\frac{1}{3\sqrt{2\pi}} =\frac{1}{3(2.5066283)} =\frac{1}{7.5198849} \approx 0.1330\text{ per unit}.

For sensor B, \(\mu_B=75\), \(\sigma_B=6\), and \(\operatorname{Var}(X_B)=6^2=36\). Its peak density is:

f_{X_B}(75)=\frac{1}{6\sqrt{2\pi}} =\frac{1}{6(2.5066283)} =\frac{1}{15.0397698} \approx 0.0665\text{ per unit}.

Sensor B’s center is 3 units higher, since \(75-72=3\). Its standard deviation is twice sensor A’s, and its variance is four times as large, since \(36/9=4\). Its peak density is half as high because \(1/(6\sqrt{2\pi})\) is half of \(1/(3\sqrt{2\pi})\). Thus the second model is centered farther to the right and is more spread out. Common mistake avoided: comparing variances as if they were standard deviations. The standard deviations are 3 and 6 units, not 9 and 36 units.

Check. The peak ratio is \(0.0665/0.1330\approx0.5\), consistent with sensor B having twice the scale. Its larger variance agrees with its wider, lower density.

Worked Example: Recovering Scale From a Peak Density

Hypothetical scenario. A measurement \(T\), in minutes, is modeled as normal with mean 40 minutes. A report gives the model’s peak density as 0.0798 per minute, rounded to four decimal places. No interest rate or mortality assumption is involved. Estimate the standard deviation and variance, write the approximate model, and calculate its density at 45 minutes. Round the density to four decimal places.

Method. The peak of a normal density is \(1/(\sigma\sqrt{2\pi})\). Because the reported peak is rounded, the scale recovered from it is an estimate. Solving for \(\sigma\) gives:

\sigma=\frac{1}{0.0798\sqrt{2\pi}} =\frac{1}{0.0798(2.5066283)} =\frac{1}{0.2000289} \approx 4.9993\text{ minutes}.

To a sensible precision, the estimated standard deviation is 5.00 minutes and the estimated variance is \(5.00^2=25.0\) minutes squared. The approximate model is \(T\sim N(40,25)\). At 45 minutes, the difference from the mean is \(45-40=5\) minutes, equal to the estimated standard deviation. Using \(\sigma=5\) in the density calculation:

\begin{aligned} f_T(45) &=\frac{1}{5\sqrt{2\pi}} \exp\left(-\frac{(45-40)^2}{2(5^2)}\right)\\ &=\frac{1}{5(2.5066283)}e^{-0.5}\\ &=\frac{0.6065307}{12.5331415}\\ &\approx 0.0484\text{ per minute}. \end{aligned}

The model is centered at 40 minutes, with an estimated scale of 5 minutes. Common mistake avoided: treating the reported peak as a probability or solving for variance directly. The peak formula determines \(\sigma\); variance is then its square. Check. A density at a point one estimated standard deviation from the mean is the peak multiplied by \(e^{-0.5}\), giving approximately \(0.0798(0.6065)=0.0484\) per minute.

A Common Parameterization Trap

The notation \(N(\mu,\sigma^2)\) places the variance, not the standard deviation, in the second position. A model written \(X\sim N(50,16)\) has \(\mu=50\), variance 16, and standard deviation 4. Substituting 16 as the scale into the density formula is incorrect.

For example, the incorrect peak calculation would be \(1/(16\sqrt{2\pi})\approx0.0249\). The correct calculation uses \(\sigma=\sqrt{16}=4\):

f_X(50)=\frac{1}{4\sqrt{2\pi}} =\frac{1}{10.0265} \approx 0.0997.

The incorrect answer is too small by a factor of four. Before using a normal density, translate the notation into \(\mu\) and \(\sigma\) explicitly. Also check units: the density’s units must be the inverse of the measurement unit.

When the Model Is Appropriate

A normal model is a mathematical approximation, not a guarantee that every real measurement can take any real value. Its symmetric density and unbounded support may be reasonable for some measurements and unsuitable for quantities with strict limits, strong skewness, or a hard lower bound. In a hypothetical insurance application, for instance, a normal model for a nonnegative claim amount could assign positive probability to negative amounts. Whether that is acceptable depends on the purpose and accuracy required.

For a stated normal model, the key interpretation is direct: \(\mu\) determines the center and equals the mean; \(\sigma\) determines the scale and equals the standard deviation; and the variance is \(\sigma^2\). These facts let you translate between a verbal description, a distribution notation, and the shape of its density. This parameter interpretation is aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, identify the location and scale carefully before calculating. Densities are rounded to four decimal places where needed.

A hypothetical inspection time \(X\), in hours, has distribution \(N(6,1.5^2)\). Give its mean, standard deviation, and variance, including units. (Answer: mean \(6\) hours; standard deviation \(1.5\) hours; variance \(2.25\) hours squared.)
A hypothetical measurement has distribution \(N(-2,3^2)\). Calculate its density at its location. (Answer: \(1/(3\sqrt{2\pi})\approx0.1330\) per measurement unit.)
A normal density has peak height \(1/(4\sqrt{2\pi})\) per unit and mean 18 units. Give its scale, variance, and distribution notation. (Answer: \(\sigma=4\) units; variance \(16\) units squared; \(X\sim N(18,16)\).)
Two hypothetical models are \(Y\sim N(10,4)\) and \(Z\sim N(10,16)\). Compare their standard deviations, variances, and peak heights. (Answer: standard deviations 2 and 4; variances 4 and 16; \(Y\)'s peak is twice \(Z\)'s.)
A hypothetical duration \(W\), in hours, has distribution \(N(6,1.5^2)\). Calculate its density at 7.5 hours. (Answer: \(e^{-0.5}/(1.5\sqrt{2\pi})\approx0.1613\) per hour.)

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