Understanding Probability Distributions

Probability distributions

A probability distribution tells you how likely different outcomes are. On the one hand, in discrete outcomes (like "how many defective rods"), we list probabilities for each possible count. For continuous measurements (like length or weight), on the other hand, we describe density across a range. General discrete vs continuous formulas:

$$P(X \in A) = \sum_{x \in A}p(x)\quad(\text{discrete}) \\[6pt] P(a \le X \le b) = \int_a^b f(x)dx \quad (continious)$$

Example (quick check): If a process guarantees all lengths between 49.5 and 50.5 cm are equally likely, the probability a rod lies in a 0.4 cm sub-range will be the sub-range width divided by 1.0 cm (this is the uniform idea — below we show it in detail).

Binomial distribution

The binomial models the number of successes (e.g., defective rods) in a fixed number of independent trials (e.g., 100 rods), when each trial has the same probability of success.

Formula:

$$P(X = k) = \begin{pmatrix}n\\k\end{pmatrix}p^k(1-p)^{n-k}$$

Example:

In a batch of $n=100$ rods where each rod independently has probability $p=0.02$ of being defective, what is the probability of exactly $k=3$ defective rods?

Step 1 — compute the combination:

$$\begin{pmatrix}100 \\ 3\end{pmatrix} = \frac{100!}{3!97!} = 161700$$

Step 2 — compute powers:

$$p^3 = 0.02^3 = 0.000008 \\ (1-p)^{97} = 0.98^{97} \approx 0.1409059532$$

Step 3 — multiply all parts:

$$P(X = 3) = 161700 \times 0.000008 \times 0.1409059532 \approx 0.182275941$$

What this means: About 18.23% chance of exactly 3 defective rods in a 100-rod sample. If you see 3 defects, that is a plausible outcome.

Uniform distribution

The uniform distribution models a continuous measurement where every value within a range [a,b] is equally likely (e.g., a tolerance range for rod length).

Formula:

$$f(x) = \frac{1}{b-a},\quad a \le x \le b$$

Probability between two points:

$$P(l \le X \le u) = \frac{u - l}{b - a}$$

Example:

Parameters: a=49.5, b=50.5. What is the probability a rod length X lies between 49.8 and 50.2? Compute range width:

$$b-a = 50.5 - 49.5 = 1.0$$

Compute sub-interval:

$$u - l = 50.2 - 49.8 = 0.4$$

Probability:

$$P(49.8 \le X \le 50.2) = \frac{0.4}{1.0} = 0.4$$

Interpretation: There is a 40% chance a randomly measured rod will fall in this tighter tolerance.

Normal distribution

The normal distribution describes continuous measurements that cluster around a mean $μ$ with spread measured by standard deviation $σ$. Many measurement errors and natural variations follow this bell-shaped curve.

Formula:

$$f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$

Standardize with z-score:

$$z = \frac{x-\mu}{\sigma}$$

Probability between two values uses the cumulative distribution (CDF) or symmetry for standard cases:

$$P(a \le X \le b) = \Phi\left(\frac{b-\mu}{\sigma}\right) - \Phi\left(\frac{a-\mu}{\sigma}\right)$$

Here $\Phi$ is the standard normal CDF.

Example A:

Parameters: $μ=200$, $σ=5$, find $P(195≤X≤205)$.

Z-scores:

$$z_1 = \frac{195 - 200}{5} = -1 \\[6pt] z_2 = \frac{205 - 200}{5} = 1$$

Using the symmetry of the normal distribution, the probability between $−1$ and $+1$ standard deviation is the well-known:

$$P(195 \le X \le 205) \approx 0.6826894921$$

Interpretation: About 68.27% of rod weights fall within ±1 standard deviation of the mean — a classic "68% rule".