PDF, z-scores, empirical rule, and 2 fully solved probability examples
The normal distribution (also called the Gaussian distribution or "bell curve") is a continuous probability distribution describing how data clusters symmetrically around a central mean. This guide covers its probability density function, the standard normal distribution, z-scores, the empirical rule, and 2 fully solved examples.
Probability histograms visualize how data is spread out. Figure 1 shows three distribution shapes:
Figure 2 shows a symmetric histogram, where the mean and median are approximately equal:
\[ \mu \approx 3.5 \quad \text{and} \quad \text{median} \approx 3.5 \]
Consider a discrete dataset with mean \( \mu = 3.5 \) and standard deviation \( \sigma = 1 \). The near-equality of mean and median indicates strong symmetry (Figures 3 and 4).
Overlaying the continuous normal density with the same parameters (\( \mu = 3.5, \sigma = 1 \)) onto the histogram gives a close fit (Figure 5):
\[ f_X(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(x - 3.5)^2} \]
The general normal distribution, defined by its mean \( \mu \) and standard deviation \( \sigma \):
| Form | Formula |
|---|---|
| General normal \( N(\mu, \sigma) \) | \( f_X(x) = \dfrac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} \) |
| Standard normal \( N(0, 1) \) | \( \varphi(x) = \dfrac{1}{\sqrt{2\pi}}\, e^{-x^2/2} \) |
| Z-score | \( z = \dfrac{x - \mu}{\sigma} \) |
The normal curve's shape is completely determined by \( \mu \) (horizontal center) and \( \sigma \) (spread — smaller \( \sigma \) makes a taller, narrower peak).
For any normal distribution, the following fixed areas hold:
≈68% within 1 standard deviation of the mean:
\[ P(\mu - \sigma \le X \le \mu + \sigma) \approx 0.68268 \]
≈95% within 2 standard deviations:
\[ P(\mu - 2\sigma \le X \le \mu + 2\sigma) \approx 0.95449 \]
≈99.7% within 3 standard deviations:
\[ P(\mu - 3\sigma \le X \le \mu + 3\sigma) \approx 0.99730 \]
The normal density has no elementary antiderivative, so probabilities are not computed by direct integration. Instead we standardize any normal variable \( X \sim N(\mu, \sigma) \) via \( z = (x-\mu)/\sigma \) and look up cumulative areas in the standard normal distribution \( N(0, 1) \).
\[ P(X \le a) = \Phi\left(\frac{a - \mu}{\sigma}\right) \]where \( \Phi \) denotes the standard normal cumulative distribution function.
Compute it digitally. In Google Sheets, =NORM.S.DIST(z, TRUE) returns
\( \Phi(z) \). In Excel, the same function works. On a TI-84, use normalcdf(−1E99, z).
Step 1 — Z-score.
\[ z = \frac{1.2 - 2.2}{2.5} = \frac{-1.0}{2.5} = -0.4 \]Step 2 — Look up \( \Phi(-0.4) \).
\[ P(X \le 1.2) = P(Z \le -0.4) \approx 0.3446 \]Answer: about 34.46%.
Given \( X \sim N(-2.5, 2) \), find:
Z-scores for both boundaries:
\[ z_1 = \frac{0.5 - (-2.5)}{2} = 1.5, \qquad z_2 = \frac{3.1 - (-2.5)}{2} = 2.8 \]Cumulative areas:
\[ P(Z \le 2.8) \approx 0.99744, \qquad P(Z \le 1.5) \approx 0.93319 \]Difference:
\[ P(0.5 \le X \le 3.1) \approx 0.99744 - 0.93319 = 0.06425 \]Answer: about 6.43%.
Z-score:
\[ z_3 = \frac{0.8 - (-2.5)}{2} = 1.65 \]Cumulative area:
\[ P(X \le 0.8) = P(Z \le 1.65) \approx 0.95053 \]Complement:
\[ P(X \ge 0.8) = 1 - 0.95053 = 0.04947 \]Answer: about 4.95%.
Given \( X \sim N(50, 8) \), find \( P(X \le 42) \).
Given \( X \sim N(100, 15) \), find \( P(85 \le X \le 115) \).
(Consistent with the empirical rule's 68% rule.)
A dataset follows an approximately normal distribution with \( \mu = 500 \) g and \( \sigma = 20 \) g. What interval contains approximately 95% of observations?
95% of values lie within \( \mu \pm 2\sigma \):
\[ [500 - 40,\; 500 + 40] = [460, 540] \text{ g} \]A continuous, symmetric, bell-shaped distribution defined by mean \( \mu \) and standard deviation \( \sigma \). Its PDF is \( f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-(x-\mu)^2/(2\sigma^2)} \).
A measure of how many standard deviations a value is from the mean: \( z = (x - \mu)/\sigma \). It standardizes any normal distribution to \( N(0,1) \) so a single table can be used.
About 68% of values fall within \( \mu \pm \sigma \), 95% within \( \mu \pm 2\sigma \), and 99.7% within \( \mu \pm 3\sigma \). Also called the 68-95-99.7 rule.
The normal PDF has no elementary antiderivative, so probabilities can't be computed by direct integration. Standardizing to z-scores and looking up values in a standard normal table (or calculator) provides an easy universal method.
Last updated: October 20, 2026