2.1 Probability Distributions, PMFs, and PDFs

2.1 Probability Distributions, PMFs, and PDFs

Probability and statistics provide the mathematical foundation for managing uncertainty, evaluating component reliability, performing quality control, and interpreting experimental data on the FE Other Disciplines exam. This section covers the fundamental rules of probability, conditional probability, Bayes' Theorem, and the primary discrete and continuous distributions specified in the NCEES FE Reference Handbook.

1. Fundamentals of Probability

An event $A$ is a subset of a sample space $S$. The probability of an event $P(A)$ satisfies the basic Kolmogorov axioms:

1. $0 \le P(A) \le 1$
2. $P(S) = 1$
3. For mutually exclusive (disjoint) events $A$ and $B$, $P(A \cup B) = P(A) + P(B)$

Union and Intersection Rules

For any two events $A$ and $B$, the general addition rule is: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

If $A$ and $B$ are independent, the occurrence of one does not alter the probability of the other, so: $P(A \cap B) = P(A) \times P(B)$

Conditional Probability and Bayes' Theorem

The conditional probability of event $A$ occurring given that event $B$ has already occurred is: $P(A|B) = \frac{P(A \cap B)}{P(B)} \quad \text{for } P(B) > 0$

By rearranging, the multiplication rule is $P(A \cap B) = P(A|B) P(B) = P(B|A) P(A)$.

Law of Total Probability

If $A_1, A_2, \dots, A_k$ form a partition of the sample space $S$ (mutually exclusive and exhaustive), then for any event $B$: $P(B) = \sum_{i=1}^k P(B|A_i) P(A_i)$

Bayes' Theorem

Bayes' Theorem allows updating prior probabilities $P(A_k)$ based on observed evidence $B$: $P(A_k|B) = \frac{P(B|A_k) P(A_k)}{\sum_{i=1}^k P(B|A_i) P(A_i)}$

Worked Engineering Example: Quality Inspection & Bayes' Theorem

Problem: A manufacturing plant operates two automated assembly lines. Line 1 produces 60% of all electronic control units (ECUs) and has a defect rate of 2.0%. Line 2 produces 40% of all ECUs and has a defect rate of 5.0%. If an ECU selected at random is found to be defective, what is the probability it was manufactured on Line 2?

Solution: Let $L_1$ and $L_2$ be the events that an ECU comes from Line 1 and Line 2, respectively. Let $D$ be the event that an ECU is defective. Given parameters:

• $P(L_1) = 0.60, \quad P(L_2) = 0.40$
• $P(D|L_1) = 0.020, \quad P(D|L_2) = 0.050$

Step 1: Compute overall probability of a defect $P(D)$ using Total Probability: $P(D) = P(D|L_1)P(L_1) + P(D|L_2)P(L_2)$ $P(D) = (0.020)(0.60) + (0.050)(0.40) = 0.012 + 0.020 = 0.032 \quad (3.2\%)$

Step 2: Apply Bayes' Theorem to find $P(L_2|D)$: $P(L_2|D) = \frac{P(D|L_2)P(L_2)}{P(D)} = \frac{0.020}{0.032} = 0.625 \quad (62.5\%)$ Conclusion: Although Line 2 produces fewer total units, a defective unit has a 62.5% probability of coming from Line 2 because of its higher defect rate.

2. Discrete Probability Distributions (PMFs)

A discrete random variable $X$ takes on distinct, countable values. Its probability distribution is defined by a Probability Mass Function (PMF) $p(x) = P(X = x)$, where: $\sum_{\text{all } x} p(x) = 1 \quad \text{and} \quad p(x) \ge 0$

A. Binomial Distribution

The Binomial distribution models the number of successes $X = k$ in $n$ independent Bernoulli trials, each having a constant probability of success $p$ (and failure probability $q = 1 - p$).

NCEES Formula

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} = \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k}$ $\text{Mean: } \mu = n p$ $\text{Variance: } \sigma^2 = n p (1-p)$ $\text{Standard Deviation: } \sigma = \sqrt{n p (1-p)}$

B. Poisson Distribution

The Poisson distribution models the number of rare events $X = x$ occurring randomly across a continuous interval of time, length, area, or volume at a constant average rate $\lambda$.

NCEES Formula

$P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}, \quad x = 0, 1, 2, \dots$ $\text{Mean: } \mu = \lambda$ $\text{Variance: } \sigma^2 = \lambda$

If the rate is given as $\mu_0$ per unit interval and the interval length is $t$, then $\lambda = \mu_0 t$.

Worked Engineering Example: Binomial & Poisson Calculations

Problem A (Binomial): A quality control engineer samples $n = 10$ structural bolts from a batch with a known defect rate of $p = 0.05$. Calculate the probability of finding exactly 2 defective bolts, and find the mean and variance of defective bolts per sample.

Solution A: $P(X = 2) = \binom{10}{2} (0.05)^2 (0.95)^{10-2} = \frac{10 \times 9}{2 \times 1} (0.0025)(0.66342) = 45 \times 0.0025 \times 0.66342 = 0.0746 \quad (7.46\%)$ Mean $\mu = np = 10 \times 0.05 = 0.50$ bolts. Variance $\sigma^2 = np(1-p) = 10 \times 0.05 \times 0.95 = 0.475$ bolts$^2$.

Problem B (Poisson): Microscopic surface cracks in a turbine blade occur at an average rate of $\mu_0 = 1.5$ cracks per meter. Find the probability that a 2-meter section of blade has exactly 3 cracks.

Solution B: For a length $t = 2\text{ m}$, the expected number of cracks is $\lambda = \mu_0 t = 1.5 \times 2 = 3.0$. $P(X = 3) = \frac{3.0^3 \cdot e^{-3.0}}{3!} = \frac{27 \times 0.049787}{6} = 0.2240 \quad (22.40\%)$.

3. Continuous Probability Distributions (PDFs)

A continuous random variable $X$ can take any real value within a continuous range. Its distribution is defined by a Probability Density Function (PDF) $f(x)$, where: $\int_{-\infty}^{\infty} f(x) dx = 1 \quad \text{and} \quad f(x) \ge 0$

For continuous variables, the probability at any single point is zero: $P(X = a) = 0$. Probabilities are evaluated over intervals using the Cumulative Distribution Function (CDF) $F(x)$: $P(a \le X \le b) = F(b) - F(a) = \int_a^b f(x) dx$

A. Continuous Uniform Distribution

A continuous random variable $X$ uniformly distributed over interval $[a, b]$ has equal probability density across its domain.

NCEES Formula

$f(x) = \frac{1}{b - a}, \quad a \le x \le b$ $\text{Mean: } \mu = \frac{a + b}{2}$ $\text{Variance: } \sigma^2 = \frac{(b - a)^2}{12}$

B. Normal (Gaussian) Distribution

The Normal distribution is symmetric and bell-shaped, parameterized by mean $\mu$ and variance $\sigma^2$.

NCEES Formula

$f(x) = \frac{1}{\sigma \sqrt{2\pi}} \exp \left( -\frac{(x - \mu)^2}{2\sigma^2} \right), \quad -\infty < x < \infty$ $\text{Standard Normal Transformation: } Z = \frac{X - \mu}{\sigma}$

To find probabilities for $X \sim N(\mu, \sigma^2)$, transform to the standard normal distribution $Z \sim N(0, 1)$ and look up probabilities in the standard normal cumulative table $\Phi(z) = P(Z \le z)$. Key standard normal probabilities:

• $P(-1 \le Z \le 1) \approx 68.27%$
• $P(-2 \le Z \le 2) \approx 95.45%$
• $P(-3 \le Z \le 3) \approx 99.73%$

C. Exponential Distribution

The Exponential distribution models the continuous time between Poisson events or component time-to-failure under a constant failure rate $\lambda$.

NCEES Formula

$f(x) = \lambda e^{-\lambda x}, \quad x \ge 0$ $\text{CDF: } F(x) = P(X \le x) = 1 - e^{-\lambda x}$ $\text{Reliability Function: } R(t) = P(X > t) = e^{-\lambda t}$ $\text{Mean (MTBF): } \mu = \frac{1}{\lambda}$ $\text{Variance: } \sigma^2 = \frac{1}{\lambda^2}$

The Exponential distribution is memoryless: $P(T > t + s \mid T > s) = P(T > t)$.

4. Worked Engineering Example: Normal Distribution & Reliability

Problem: Compressive strength tests of concrete batches yield a normal distribution with mean $\mu = 35.0\text{ MPa}$ and standard deviation $\sigma = 4.0\text{ MPa}$. Structural design standards require a minimum compressive strength of $30.0\text{ MPa}$.

1. What fraction of concrete batches will fail to meet the $30.0\text{ MPa}$ design specification?
2. If the project engineer wants no more than 1.0% of batches to fail specification, what minimum mean strength $\mu_{new}$ must the mixing plant achieve, assuming $\sigma$ remains $4.0\text{ MPa}$?

Solution:

Part 1: Calculate $P(X < 30.0\text{ MPa})$: Compute the standardized $Z$-score: $Z = \frac{X - \mu}{\sigma} = \frac{30.0 - 35.0}{4.0} = -1.25$ Using standard normal table symmetry $\Phi(-z) = 1 - \Phi(z)$: $\Phi(1.25) = 0.8944$ $P(X < 30.0) = P(Z < -1.25) = 1 - 0.8944 = 0.1056 \quad (10.56\%)\text{ failure rate}$

Part 2: Target failure rate is $P(X < 30.0) = 0.010$. From standard normal tables, $P(Z < z_{crit}) = 0.010 \implies z_{crit} = -2.326$. $Z = \frac{30.0 - \mu_{new}}{\sigma} = -2.326$ $30.0 - \mu_{new} = -2.326 \times 4.0 = -9.304$ $\mu_{new} = 30.0 + 9.304 = 39.304\text{ MPa} \approx 39.3\text{ MPa}$

5. Summary Table of NCEES Probability Distributions