Normal Moments and Symmetry

Why Moments and Symmetry Matter

A normal model is summarized by its location parameter and its spread, but probability calculations depend on how those parameters relate to the density. Deriving the moments from the density makes that relationship precise: the location parameter is the mean, and the square of the scale parameter is the variance. The same density also has a useful symmetry, so deviations equally far above and below the mean have equal probabilities.

These facts are useful when interpreting a model and checking calculations. Symmetry can simplify a central interval or a two-tail probability, but it does not mean that every interval has probability determined by its width alone. For an interval not centered at the mean, standardize both endpoints and use the normal cumulative distribution function.

Definition: A continuous random variable $X$ has a normal distribution with mean $\mu$ and variance $\sigma^2$, written $X\sim N(\mu,\sigma^2)$, when $\sigma>0$ and its density is $$ f_X(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right),\qquad -\infty<x<\infty. $$ Here $\mu$ is a real number and $\sigma$ is the positive standard deviation. The second parameter in $N(\mu,\sigma^2)$ is the variance, not the standard deviation.

Deriving the Standard Normal Moments

Begin with the standard normal variable $Z\sim N(0,1)$, whose density is $$ \phi(z)=\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right). $$ The density is even: $\phi(-z)=\phi(z)$. In particular, $z\phi(z)$ is an odd function. The mean is defined by integrating $z$ against the density over its full support. The positive and negative contributions cancel:

E[Z]=\int_{-\infty}^{\infty}z\phi(z)\,dz=0.

To find the variance, first calculate the second moment. Differentiating the standard normal density gives $\phi'(z)=-z\phi(z)$. Integration by parts then yields

\begin{aligned} 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\\ &=0+1=1. \end{aligned}

The boundary term is zero because $z\phi(z)$ tends to zero at both ends of the real line, and the density integrates to one. Since $E[Z]=0$, the variance is $E[Z^2]-(E[Z])^2=1$. Thus the standard normal has mean zero and variance one.

Moments of a General Normal Variable

For $X\sim N(\mu,\sigma^2)$, set $z=(x-\mu)/\sigma$ in the density integral. The density and the standard normal density satisfy $f_X(x)\,dx=\phi(z)\,dz$. Therefore the moment integrals can be expressed using the standard normal moments just derived:

\begin{aligned} E[X] &=\int_{-\infty}^{\infty}x f_X(x)\,dx\\ &=\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. \end{aligned}

For the variance, use $X-\mu=\sigma z$ in the defining integral. This gives

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

This establishes from the density that the parameters in $N(\mu,\sigma^2)$ are exactly the mean and variance. The derivation assumes $\sigma>0$; if $\sigma=0$, the model is degenerate at $\mu$ and does not have the stated continuous density.

Theorem: If $X\sim N(\mu,\sigma^2)$, then $E[X]=\mu$, $\operatorname{Var}(X)=\sigma^2$, and the standard deviation is $\sigma$. The density is symmetric about $\mu$: for every $t$, $f_X(\mu+t)=f_X(\mu-t)$. Consequently, for $t>0$, $$ P(X<\mu-t)=P(X>\mu+t), $$ and, because a continuous variable assigns probability zero to any single point, $P(X<\mu)=P(X>\mu)=0.5$.

What Symmetry Does and Does Not Tell You

Symmetry follows directly from the squared term in the density: replacing $x-\mu$ by $-(x-\mu)$ leaves the density unchanged. Integrating equal density values over matching intervals on the two sides of the mean gives equal probabilities. For example, the intervals $(\mu-2\sigma,\mu-\sigma)$ and $(\mu+\sigma,\mu+2\sigma)$ have the same probability.

Symmetry does not imply that two intervals of equal length have equal probabilities if they are not mirror images about the mean. The density is higher near $\mu$ than far out in either tail. For an arbitrary interval $(a,b)$, standardize both endpoints:

P(a<X<b) = P\left(\frac{a-\mu}{\sigma}<Z<\frac{b-\mu}{\sigma}\right) = \Phi\left(\frac{b-\mu}{\sigma}\right) -\Phi\left(\frac{a-\mu}{\sigma}\right),

where $\Phi(z)=P(Z\le z)$ is the standard normal cumulative distribution function. This formula uses continuity, so it does not matter whether either endpoint is included. It applies to any finite interval in a normal model; symmetry is a shortcut only when the standardized endpoints are opposites or when the requested tails are mirror images.

Worked Example: A Central Measurement Range

A hypothetical instrument’s repeated calibration error $X$, measured in units, is modeled as $N(250,20^2)$. Find its mean and variance, then calculate the probability that a measurement lies between 230 and 270 units. Use standard normal values $\Phi(1)=0.84134$ and $\Phi(-1)=0.15866$. Method: read the moments from the normal density parameters, then standardize the interval. The calculation avoids treating 20 as the variance.

The model parameters are $\mu=250$ and $\sigma=20$, so $\sigma^2=20^2=400$. From the moment result, $E[X]=250$ and $\operatorname{Var}(X)=400$.

For the probability, standardize each endpoint:

\frac{230-250}{20}=-1,\qquad \frac{270-250}{20}=1.

The standardized interval is symmetric about zero. Therefore,

\begin{aligned} P(230<X<270) &=P(-1<Z<1)\\ &=\Phi(1)-\Phi(-1)\\ &=0.84134-0.15866\\ &=0.68268. \end{aligned}

Rounded to four decimal places, the probability is $0.6827$. As a check, symmetry gives equal probabilities below 230 and above 270; together those tails have probability $1-0.68268=0.31732$, or about $0.15866$ each.

Worked Example: Equal Tails Outside a Central Range

For a hypothetical claims-handling process, resolution time $T$, in days, is modeled as $N(42,6^2)$. Find the probability that resolution takes fewer than 36 days or more than 48 days. Use $\Phi(1)=0.84134$. Method: standardize the two cutoffs and use symmetry to combine equal tails. The calculation avoids adding the probability between the cutoffs instead of outside them.

The mean is 42 days and the standard deviation is 6 days. The two cutoffs are one standard deviation below and above the mean:

\frac{36-42}{6}=-1,\qquad \frac{48-42}{6}=1.

Thus the requested events are $Z<-1$ and $Z>1$. By symmetry, their probabilities are equal. The lower-tail probability is

P(Z<-1)=\Phi(-1)=1-\Phi(1)=1-0.84134=0.15866.

The upper tail also has probability $0.15866$, so the disjoint events have combined probability

P(T<36\text{ or }T>48)=0.15866+0.15866=0.31732.

Rounded to four decimal places, the probability is $0.3173$. The central probability is its complement, $1-0.31732=0.68268$, which agrees with the probability of being within one standard deviation of the mean in the preceding example.

Worked Example: An Interval That Is Not Centered at the Mean

A hypothetical sensor output $Y$ is modeled as $N(75,10^2)$ units. Find the probability that the output is greater than 65 and at most 87 units. Use $\Phi(1.2)=0.88493$ and $\Phi(-1)=0.15866$. Method: standardize both endpoints and subtract cumulative probabilities. The calculation avoids incorrectly doubling a tail probability when the interval is not centered at the mean.

The lower endpoint is one standard deviation below the mean, and the upper endpoint is 1.2 standard deviations above it:

\frac{65-75}{10}=-1,\qquad \frac{87-75}{10}=1.2.

The interval is not symmetric about the mean, so the equal-tail shortcut does not apply. Instead,

\begin{aligned} P(65<Y\le87) &=P(-1<Z\le1.2)\\ &=\Phi(1.2)-\Phi(-1)\\ &=0.88493-0.15866\\ &=0.72627. \end{aligned}

Rounded to four decimal places, the probability is $0.7263$. A check on the direction is that the interval contains the mean and extends farther above it than below it, so its probability should be greater than the probability between $-1$ and $1$, approximately $0.6827$. The computed value satisfies this check.

Common Mistakes and Exam Checks

The density provides a reliable way to check both parameter interpretation and probability setup. Before using a normal table or calculator, identify the model’s mean and standard deviation, then express each cutoff in standard deviation units. Use symmetry only after confirming that the relevant standardized cutoffs are mirror images.

Confusing variance and standard deviation. In $N(\mu,\sigma^2)$, the second parameter is the variance. If a model is $N(250,400)$, its standard deviation is $\sqrt{400}=20$, not 400.
Assuming a mean is a median for every distribution. Here the mean bisects probability because the normal density is symmetric. This reasoning does not automatically apply to an arbitrary distribution.
Applying symmetry to a noncentral interval. Equal interval widths do not guarantee equal probabilities. The intervals must be reflections around $\mu$ for the symmetry argument to work.
Mixing up central and tail probabilities. If a question asks for values outside a symmetric range, calculate both tails and add them. The probability inside the range is the complement.
Reversing a cumulative probability subtraction. For $a<b$, use $\Phi((b-\mu)/\sigma)-\Phi((a-\mu)/\sigma)$; the result should be nonnegative.

The density-based derivation explains why the normal parameters have their usual interpretation, while symmetry provides a principled shortcut rather than a guess. These checks support accurate probability work aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, identify the normal parameters before calculating, and use symmetry only when the interval or tails are symmetric about the mean.

A hypothetical sensor reading $R$ is modeled as $N(80,10^2)$. What are its mean and variance, and what is $P(R>80)$? (Answer: mean $80$, variance $100$, and probability $0.5000$)
A hypothetical inspection duration $D$ has distribution $N(30,4^2)$. Given $\Phi(1)=0.84134$ and $\Phi(-1)=0.15866$, find $P(26<D<34)$. (Answer: $0.6827$)
A hypothetical measurement error $W$ has distribution $N(100,12^2)$. Given $\Phi(2)=0.97725$, find the probability that $W<76$ or $W>124$. (Answer: $0.0455$)
A hypothetical account balance $R$, in dollars, is modeled as $N(500,40^2)$. Given $\Phi(-1)=0.15866$ and $\Phi(1.2)=0.88493$, find $P(460<R\le548)$. (Answer: $0.7263$)
A hypothetical standardized quality score $Q$ has distribution $N(0,1)$. Given $\Phi(1.5)=0.93319$, find $P(|Q|<1.5)$, and explain which property allows the calculation from one table value. (Answer: $0.8664$, using symmetry about zero)

What You'll Learn