Normal Distribution
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 3.7: “Probability” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/3-7-probability |
| Textbook used in class | Stewart, Calculus, Section 8.5: “Probability” |
Opening Scenario
Measurement errors, heights of adults, IQ scores, exam grades -- many naturally occurring quantities cluster around a central value and taper off symmetrically. The mathematical model that captures this is the normal distribution (the “bell curve”). Its pdf involves $e^{-x^2}$, whose integral over the whole real line cannot be expressed in elementary closed form, yet equals $\sqrt{\pi}$ -- a remarkable fact that ties together $e$, $\pi$, and infinity.
In practice, probabilities under the normal curve are found using a standard table (or calculator), not by antidifferentiating directly. The calculus connection is that those tabulated values are definite integrals.
Quick Reference
Normal distribution with mean $\mu$ and standard deviation $\sigma$: $$f(x) = \frac{1}{\sigma\sqrt{2\pi}}\,e^{-(x-\mu)^2/(2\sigma^2)}, \quad -\infty < x < \infty.$$
Standard normal $Z \sim N(0,1)$ ($\mu = 0$, $\sigma = 1$): $$\phi(z) = \frac{1}{\sqrt{2\pi}}\,e^{-z^2/2}.$$
Standardization. If $X \sim N(\mu, \sigma^2)$, set $Z = \dfrac{X-\mu}{\sigma}$. Then $Z \sim N(0,1)$.
Probability: $$P(a \leq X \leq b) = P\!\left(\frac{a-\mu}{\sigma} \leq Z \leq \frac{b-\mu}{\sigma}\right) = \Phi\!\left(\frac{b-\mu}{\sigma}\right) - \Phi\!\left(\frac{a-\mu}{\sigma}\right),$$ where $\Phi(z) = \displaystyle\int_{-\infty}^z \phi(t)\,dt$ is the standard normal CDF.
Key Concepts
1. Shape of the Bell Curve
The normal pdf $f(x)$ is:
- Symmetric about $x = \mu$ (mean = median = mode).
- Bell-shaped: rises from near zero, reaches its maximum at $x = \mu$, then falls back to near zero.
- The maximum value is $f(\mu) = \dfrac{1}{\sigma\sqrt{2\pi}}$.
- Inflection points at $x = \mu \pm \sigma$.
- As $|x-\mu|$ increases, $f(x) \to 0$ exponentially fast.
The width (spread) is controlled by $\sigma$: larger $\sigma$ gives a wider, flatter bell; smaller $\sigma$ gives a taller, narrower bell. In both cases the total area equals 1.
2. Why $e^{-x^2}$ Is Not Integrable in Closed Form
The antiderivative of $e^{-x^2}$ is not an elementary function. The improper integral $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$ is proved using a change to polar coordinates (covered in multivariable calculus). In MATH162, take this normalization as given: $$\int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)}\,dx = 1.$$
3. Standardization and the $z$-Score
Any normal random variable $X \sim N(\mu, \sigma^2)$ can be converted to a standard normal $Z \sim N(0,1)$ by the transformation $Z = (X-\mu)/\sigma$. A $z$-score measures how many standard deviations a value is from the mean.
Probabilities for $X$ are found by:
- Converting the bounds to $z$-scores: $z_1 = (a-\mu)/\sigma$, $z_2 = (b-\mu)/\sigma$.
- Looking up $P(z_1 \leq Z \leq z_2)$ in a standard normal table.
4. The 68-95-99.7 Rule
For a normal distribution:
- $P(\mu - \sigma \leq X \leq \mu + \sigma) \approx 0.6827$ (68.27%).
- $P(\mu - 2\sigma \leq X \leq \mu + 2\sigma) \approx 0.9545$ (95.45%).
- $P(\mu - 3\sigma \leq X \leq \mu + 3\sigma) \approx 0.9973$ (99.73%).
These correspond to $P(-1 \leq Z \leq 1)$, $P(-2 \leq Z \leq 2)$, $P(-3 \leq Z \leq 3)$ for the standard normal.
Worked Example
Suppose $X \sim N(100, 225)$ (mean 100, variance 225, so $\sigma = 15$). Using the 68-95-99.7 rule:
(a) What is $P(85 \leq X \leq 115)$?
$85 = 100 - 15 = \mu - \sigma$ and $115 = 100 + 15 = \mu + \sigma$.
$P(85 \leq X \leq 115) = P(-1 \leq Z \leq 1) \approx 0.6827$.
(b) What is $P(X > 130)$?
$z = (130 - 100)/15 = 2$.
$P(X > 130) = P(Z > 2) = 1 - P(Z \leq 2) \approx 1 - 0.9772 = 0.0228$.
(c) What is $P(70 \leq X \leq 100)$?
$z_1 = (70-100)/15 = -2$, $z_2 = (100-100)/15 = 0$.
$P(-2 \leq Z \leq 0) = P(Z \leq 0) - P(Z \leq -2) = 0.5 - 0.0228 = 0.4772$.
Boxed answers: (a) $\approx 0.6827$; (b) $\approx 0.0228$; (c) $\approx 0.4772$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using $\sigma^2$ where $\sigma$ is needed | Standardizing with $Z = (X-\mu)/\sigma^2$ | The correct standardization is $Z = (X-\mu)/\sigma$; use the standard deviation, not the variance |
| Forgetting symmetry: $P(Z \leq -z) = P(Z \geq z)$ | Computing $\Phi(-1.5)$ as $\Phi(1.5)$ | By symmetry of the standard normal, $\Phi(-z) = 1 - \Phi(z)$; use this when a table only gives positive $z$ values |
| Confusing $P(Z \leq z)$ with $P(Z = z)$ | Reading $\Phi(1) = P(Z = 1)$ | $\Phi(z) = P(Z \leq z)$ is the CDF (an area); $P(Z = z) = 0$ for any continuous distribution |
Common Misconceptions
the peak value of the normal pdf, $f(\mu) = 1/(\sigma\sqrt{2\pi})$, represents the probability that $X = \mu$.
This is the concept-image-conflicts-definition error. For a continuous random variable, $P(X = \mu) = 0$ regardless of how large $f(\mu)$ is. The peak value $1/(\sigma\sqrt{2\pi})$ is a density, describing how concentrated probability is near the mean. For the standard normal ($\sigma = 1$), $\phi(0) = 1/\sqrt{2\pi} \approx 0.399 < 1$, so the density value happens to be less than $1$ here, but this is coincidental; what matters is that probability always requires integrating $f$ over an interval, never reading off a point value.
$\sigma^2$ (the variance) and $\sigma$ (the standard deviation) can be used interchangeably in the standardization formula $Z = (X - \mu)/\sigma$.
This is the concept-image-conflicts-definition error applied to the normal parameters. The standardization formula uses $\sigma$, the standard deviation, which has the same units as $X$. For $X \sim N(100, 225)$, the standard deviation is $\sigma = \sqrt{225} = 15$, not $225$. Using $\sigma^2 = 225$ in the denominator of the $z$-score formula gives $z = (X - 100)/225$, which produces values 15 times smaller than the correct $z$-scores and leads to incorrect probability lookups. The variance $\sigma^2$ appears in the pdf exponent but not in the $z$-score formula.
Leveled Practice
Level 1 -- Applying the Empirical Rule
Problem 1. Scores on an exam are normally distributed with mean 75 and standard deviation 10. What percentage of scores fall between 55 and 95?
Show answer
$55 = 75 - 20 = \mu - 2\sigma$ and $95 = 75 + 20 = \mu + 2\sigma$.
By the 68-95-99.7 rule: approximately 95.45% of scores fall within two standard deviations of the mean.
Level 2 -- Computing a z-Score Probability
Problem 2. For $X \sim N(50, 16)$ ($\sigma = 4$), find $P(46 \leq X \leq 58)$. Use $\Phi(0.5) \approx 0.6915$ and $\Phi(2) \approx 0.9772$.
Show answer
$z_1 = (46-50)/4 = -1$. $z_2 = (58-50)/4 = 2$.
$P(46 \leq X \leq 58) = P(-1 \leq Z \leq 2) = \Phi(2) - \Phi(-1) = \Phi(2) - (1-\Phi(1))$.
$\Phi(1) \approx 0.8413$ (from $P(-1 \leq Z \leq 1) \approx 0.6827$: $\Phi(1) = 0.5 + 0.6827/2 \approx 0.8413$).
$P = 0.9772 - (1-0.8413) = 0.9772 - 0.1587 = 0.8185$.
Boxed answer: $P \approx 0.8185$.
Mastery Checklist
Mental Model
The normal distribution is the “default” distribution in nature and statistics. Its bell shape arises whenever a quantity is the sum of many small independent influences (Central Limit Theorem, studied in probability courses). The parameters $\mu$ (mean) and $\sigma$ (standard deviation) completely determine the shape: $\mu$ sets the center, $\sigma$ sets the spread. Standardization converts any normal to the standard normal, so one table suffices for all normal distributions. Every probability under a normal curve is an integral of $e^{-z^2/2}$, which cannot be computed in closed form -- so tables or calculators carry the tabulated values.
Connections
Looking back
- Improper integrals (Section 7.8): The normalization $\int_{-\infty}^{\infty} e^{-x^2/(2\sigma^2)}\,dx = \sigma\sqrt{2\pi}$ is an improper integral.
- Probability density functions (Section 8.5): The normal pdf is the most important example of a pdf.
- Expected value (Section 8.5): The mean $\mu$ is the expected value, and by symmetry the median as well.
Looking ahead
- Statistics courses: confidence intervals, hypothesis testing, and regression all build on the normal distribution and its CDF $\Phi$.