The Normal Distribution $N(\mu,\sigma^2)$

The standard normal distribution $Z\sim N(0,1)$ is the bell curve every other normal is measured against. Its density is $\varphi(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2}$ — symmetric about $0$, peaked at the origin, and decaying rapidly in both tails. Proposition 13.6 records its two defining moments: $\mathbb{E}[Z]=0,\qquad \mathrm{Var}(Z)=\mathbb{E}[Z^2]=1.$ The mean is $0$ because $z\varphi(z)$ is an odd integrand on a symmetric domain, so its integral vanishes; the variance is $1$ because $\int_{-\infty}^{\infty}z^2\varphi(z)\,dz=1$ (a standard integration by parts), and $\mathrm{Var}(Z)=\mathbb{E}[Z^2]-(\mathbb{E}[Z])^2=1-0=1$. There is no elementary antiderivative for $\varphi$, so probabilities are not computed by hand but read from the standard-normal c.d.f. $\Phi(z)=\mathbb{P}(Z\le z)=\int_{-\infty}^{z}\frac{1}{\sqrt{2\pi}}e^{-t^2/2}\,dt,$ tabulated or built into software. Symmetry gives the frequently used identity $\Phi(-z)=1-\Phi(z)$, so a table of positive $z$-values suffices for all of them.

The whole normal family is produced from $Z$ by affine transformations. Fix a mean $\mu\in\mathbb{R}$ and a standard deviation $\sigma>0$ and set $X=\sigma Z+\mu.$ Scaling by $\sigma$ widens or narrows the bell and shifting by $\mu$ slides its centre, giving $\mathbb{E}[X]=\sigma\,\mathbb{E}[Z]+\mu=\mu$ and $\mathrm{Var}(X)=\sigma^2\,\mathrm{Var}(Z)=\sigma^2$. Transforming the density accordingly, Definition 13.5 says $X$ has the normal distribution with mean $\mu$ and variance $\sigma^2$, written $X\sim N(\mu,\sigma^2)$, if $f(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right).$ Read the notation carefully: the first slot is the mean $\mu$ and the second slot is the variance $\sigma^2$ — not the standard deviation. So $N(-3,4)$ has mean $-3$, variance $4$, and standard deviation $\sigma=\sqrt{4}=2$. The graph is a symmetric bell centred at $\mu$; a larger $\sigma$ makes it shorter and wider, a smaller $\sigma$ taller and narrower (compare $N(0,1)$ with $N(2,2.25)$, where $\sigma=1.5$).

Normality is preserved by any affine map. Proposition 13.7: if $X\sim N(\mu,\sigma^2)$ and $a\neq0$, $b\in\mathbb{R}$, then $aX+b\sim N\!\big(a\mu+b,\;a^2\sigma^2\big).$ The new mean is $a\mu+b$ (apply $\mathbb{E}[aX+b]=a\mathbb{E}[X]+b$) and the new variance is $a^2\sigma^2$ (apply $\mathrm{Var}(aX+b)=a^2\mathrm{Var}(X)$); crucially, the shape stays a bell, so the result is again exactly normal. The most important special case runs the construction $X=\sigma Z+\mu$ in reverse. Taking $a=\tfrac1\sigma$ and $b=-\tfrac{\mu}{\sigma}$ gives the standardized variable $Z=\frac{X-\mu}{\sigma}\sim N(0,1),$ which re-expresses the value $x$ as a $z$-score — how many standard deviations $x$ lies above (or below) the mean. Subtracting $\mu$ recentres to $0$; dividing by $\sigma$ rescales the variance to $1$. Standardization is the bridge from any normal to the single tabulated distribution $N(0,1)$.

Standardization turns every normal probability into a lookup in $\Phi$. Since $X\le x$ exactly when $Z=\frac{X-\mu}{\sigma}\le\frac{x-\mu}{\sigma}$, $\mathbb{P}(X\le x)=\Phi\!\left(\frac{x-\mu}{\sigma}\right),$ and a two-sided probability is a difference, $\mathbb{P}(a\le X\le b)=\Phi\!\big(\tfrac{b-\mu}{\sigma}\big)-\Phi\!\big(\tfrac{a-\mu}{\sigma}\big)$. For $X\sim N(-3,4)$ (so $\sigma=2$), Example 13.14 gives $\mathbb{P}(X\le-1.7)=\Phi\!\big(\tfrac{-1.7+3}{2}\big)=\Phi(0.65)\approx0.742$. Applying this to the symmetric band $|X-\mu|\le k\sigma$ gives the 68–95–99.7 (empirical) rule: because $|X-\mu|\le k\sigma$ is the same event as $|Z|\le k$, $\mathbb{P}(|X-\mu|\le k\sigma)=2\Phi(k)-1,$ which equals about $0.683$ for $k=1$, $0.954$ for $k=2$, and $0.997$ for $k=3$. Equivalently, the chance of straying more than $2\sigma$ from the mean is $2\big(1-\Phi(2)\big)\approx0.046$ (Example 13.15) — so a normal variable sits within two standard deviations of its mean over $95\%$ of the time.