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Probability and Statistics

6% of exam

Distributions and expected value, point and interval estimation, hypothesis testing, and goodness of fit.

4 concepts

A. Estimation

Probability Distributions and Estimation

Probability rules plus the binomial, Poisson, normal, exponential, and uniform models — choose the right one, then point-estimate and build a confidence interval for the mean.

Roughly one in fifteen questions on the FE Other Disciplines exam is probability or statistics, and the single most reliable points come from recognizing which standard distribution a word problem is describing, then turning the crank. The story tells you the model: counting successes in a fixed number of trials is binomial, counting events in a fixed interval is Poisson, a bell-shaped measurement is normal, a waiting time is exponential, and equally likely values are uniform. The FE Reference Handbook's "Engineering Probability and Statistics" chapter hands you every formula and a unit normal table — your job is selection, careful unit-of-the-interval bookkeeping, and not confusing a discrete sum (a PMF) with a continuous area (a PDF). This concept also closes the loop from data back to the population: point estimation and the confidence interval for the mean.

Probability rules you build everything on

Every probability lies in

[0,1]

. The addition rule (inclusion-exclusion) gives the chance that

A

or

B

happens, subtracting the overlap so it is not counted twice; for mutually exclusive events that overlap is zero. The multiplication (joint) rule gives the chance both happen,

P(A\cap B)=P(A)\,P(B\mid A)

, which collapses to

P(A)P(B)

only when the events are independent. The complement rule

P(\bar A)=1-P(A)

is the workhorse behind every "at least one" question — compute the probability of none and subtract.

P(A\cup B)=P(A)+P(B)-P(A\cap B)

PMF, PDF, and CDF

A discrete random variable is described by a probability mass function

f(x_k)=P(X=x_k)

and you sum probabilities. A continuous random variable is described by a probability density function

f(x)

, where probability is the area under the curve,

P(a\le X\le b)=\int_a^b f(x)\,dx

— so the probability of any exact value is zero. Either way the cumulative distribution function

F(x)=P(X\le x)

accumulates probability from the left and rises from

0

to

1

, and

P(a<X\le b)=F(b)-F(a)

. Nearly every table lookup is really a CDF difference.

F(x)=P(X\le x),\qquad P(a<X\le b)=F(b)-F(a)

Binomial: successes in a fixed number of trials

Use the binomial when you have a fixed number

n

of independent trials, each a success or failure with the same probability

p

, and you count successes

x

. The combination

C(n,x)

counts the orderings. Its mean is

np

and variance

npq

with

q=1-p

. The fastest route to "at least one" is the complement,

P(X\ge 1)=1-q^{\,n}

.

P(X=x)=C(n,x)\,p^{x}q^{\,n-x},\qquad \mu=np,\quad \sigma^2=npq

Poisson: events in a continuous interval

Use the Poisson when events occur independently at a constant average rate and you count how many land in a fixed span of time, length, or area — calls per hour, flaws per square meter, arrivals per minute. The one parameter

\lambda

is the expected count over that interval and equals both the mean and the variance. The trap is the interval: if a rate is given per hour but the question asks about 20 minutes, scale

\lambda

to that span first.

P(X=x)=\dfrac{\lambda^{x}e^{-\lambda}}{x!},\qquad \mu=\sigma^2=\lambda

Normal and the standard-normal z-transform

The normal (Gaussian) distribution is the symmetric bell with mean

\mu

and standard deviation

\sigma

; it models measurement scatter and, via the central limit theorem, sample means. You never integrate it — you standardize to

z

and read the handbook's unit normal table. In that table

F(z)

is the area to the left and

R(z)=1-F(z)

is the area to the right; by symmetry

F(-z)=1-F(z)

. Sketch the bell, shade what you want, and assemble it from table entries.

z=\dfrac{x-\mu}{\sigma},\qquad P(X\le x)=F(z)

Exponential and uniform: the two continuous shortcuts

The exponential models the waiting time until the next Poisson event, with rate

\lambda

(the same

\lambda

as the Poisson). Its CDF is clean,

F(t)=1-e^{-\lambda t}

, so

P(T>t)=e^{-\lambda t}

— and it is memoryless, meaning the past does not change future waiting odds. Its mean is

1/\lambda

and variance

1/\lambda^2

. The continuous uniform spreads probability evenly over

[a,b]

with constant density

1/(b-a)

, mean

(a+b)/2

, and variance

(b-a)^2/12

.

F_{\text{exp}}(t)=1-e^{-\lambda t},\qquad \mu_U=\dfrac{a+b}{2},\quad \sigma_U^2=\dfrac{(b-a)^2}{12}

From sample to population: point estimation

An estimator is a formula applied to sample data; the number it produces is the estimate. The sample mean

\bar X

estimates the population mean

\mu

, and the sample variance

s^2

— note the

n-1

divisor, which makes it unbiased — estimates

\sigma^2

. These are point estimates: single best guesses with no statement of how far off they might be. That missing piece is exactly what a confidence interval supplies.

\bar X=\dfrac{1}{n}\sum_{i=1}^{n}X_i,\qquad s^2=\dfrac{1}{n-1}\sum_{i=1}^{n}\left(X_i-\bar X\right)^2

Confidence interval for the mean

A confidence interval brackets

\mu

with a stated confidence (the long-run capture rate). When the population standard deviation

\sigma

is known (or

n

is large), use the normal multiplier

z_{\alpha/2}

— for example

1.96

for 95% and

1.645

for 90%, both straight from the handbook fractile table. When

\sigma

is unknown and you must use the sample

s

with a small sample, swap in

t_{\alpha/2,\,n-1}

from the t-table, which is wider to pay for the extra uncertainty. The half-width

z_{\alpha/2}\,\sigma/\sqrt n

is the margin of error.

\bar X - z_{\alpha/2}\,\dfrac{\sigma}{\sqrt n}\ \le\ \mu\ \le\ \bar X + z_{\alpha/2}\,\dfrac{\sigma}{\sqrt n}

Exam strategy

First classify the random variable from the story — discrete count vs. continuous measurement, fixed trials vs. a rate over an interval — and write down which distribution and which parameters apply before touching numbers. For normal problems, always sketch the bell, mark

\mu

, convert to

z

, and assemble the answer from

F(z)

and

R(z)

, remembering

F(-z)=1-F(z)

. For Poisson and exponential, scale

\lambda

to the interval the question actually asks about. For a confidence interval, decide

z

vs.

t

by whether

\sigma

is known, then plug into

\bar X \pm (\text{multiplier})\,\sigma/\sqrt n

.

Key equations

Addition rule (inclusion-exclusion)

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Probability that A or B occurs; the overlap term is zero for mutually exclusive events.

Conditional / joint probability

P(A\cap B)=P(A)\,P(B\mid A)=P(B)\,P(A\mid B)

Binomial PMF

P(X=x)=C(n,x)\,p^{x}(1-p)^{\,n-x}

Poisson PMF

P(X=x)=\dfrac{\lambda^{x}e^{-\lambda}}{x!}

Standard-normal transform

z=\dfrac{x-\mu}{\sigma}

Converts any normal value to standard-normal units so the unit normal table applies; $\mu$ , $\sigma$

Normal table relations

R(z)=1-F(z),\qquad F(-z)=1-F(z)

$F(z)$

Exponential distribution

f(t)=\lambda e^{-\lambda t},\qquad P(T>t)=e^{-\lambda t}

Continuous uniform

f(x)=\dfrac{1}{b-a},\quad \mu=\dfrac{a+b}{2},\quad \sigma^2=\dfrac{(b-a)^2}{12}

Sample mean and variance

\bar X=\dfrac{1}{n}\sum X_i,\qquad s^2=\dfrac{1}{n-1}\sum\left(X_i-\bar X\right)^2

Confidence interval, σ known

\bar X \pm z_{\alpha/2}\,\dfrac{\sigma}{\sqrt n}

Confidence interval, σ unknown

\bar X \pm t_{\alpha/2,\,n-1}\,\dfrac{s}{\sqrt n}

Worked examples

Binomial defectives — exactly one and at least one

Problem. A lot has a $5\%$ defective rate. You draw $8$ parts at random (large lot, so trials are effectively independent). Find the probability that exactly one is defective, and the probability that at least one is defective.

Solution. Binomial with

n=8

,

p=0.05

,

q=0.95

. Exactly one:

P(X=1)=C(8,1)(0.05)^1(0.95)^7 = 8(0.05)(0.6983)=0.279

. At least one via the complement:

P(X\ge1)=1-P(X=0)=1-(0.95)^8 = 1-0.663=0.337

. Sanity:

P(X=0)=0.663

is the largest single term and

P(X=1)=0.279

the next, summing to

0.942

for zero-or-one defectives — reasonable for a low rate. Answers:

P(X=1)=0.279

,

P(X\ge1)=0.337

.

P(X\ge 1)=1-(0.95)^{8}=0.337

Normal process — tail and band probabilities

Problem. A fill process is normal with mean $\mu=100\,\text{mL}$ and standard deviation $\sigma=4\,\text{mL}$ . Find $P(X>106)$ and $P(94<X<108)$

Poisson arrivals over a scaled interval

Problem. Calls arrive at a help desk at an average rate of $9$ per hour, independently. What is the probability that two or more calls arrive in a $20$ -minute window?

Solution. Scale the rate to the window:

20\,\text{min}=\tfrac13\,\text{hr}

95% confidence interval for the mean (σ known)

Problem. A bearing diameter has known process standard deviation $\sigma=6\,\mu\text{m}$ about its target. A sample of $n=25$ gives $\bar X=52.0\,\mu\text{m}$

Common pitfalls

•Confusing binomial with Poisson. Fixed number of trials with a per-trial probability $p$ is binomial; a rate over a continuous interval with no fixed trial count is Poisson. Read for $n$ .
•Forgetting to scale $\lambda$ to the asked interval. A rate of $9$ /hour over $20$ minutes means $\lambda=3$ , not $9$ — the most common Poisson error.
•Misreading the normal table: $F(z)$ is area to the LEFT, $R(z)$ to the RIGHT. For a negative z use $F(-z)=1-F(z)$ rather than looking up a nonexistent negative row.
•Treating $P(X=a)$ as nonzero for a continuous variable. For normal/exponential/uniform, single points have zero probability; only intervals carry probability.
•Dividing the sample variance by $n$ instead of $n-1$ . The FE uses the unbiased $n-1$ divisor for $s^2$ and $s$
•Using a t-multiplier when $\sigma$ is known (or n is large) — that is a z problem — or using z with a small sample and only $s$ available, which understates the interval width.
•Plugging $\sigma$ instead of the standard error $\sigma/\sqrt n$ into the confidence interval. The interval shrinks with sample size through the $\sqrt n$

References

NCEES FE Reference Handbook — Engineering Probability and Statistics
NCEES FE Reference Handbook — Unit Normal Distribution table and t-Distribution table
Montgomery & Runger, Applied Statistics and Probability for Engineers — Distribution selection and confidence-interval derivations

B. Expected value and expected error in decision making

Expected Value and Decision-Making

Compute the expected value and variance of a random variable, rank choices by expected monetary value, and propagate measurement uncertainty with the Kline-McClintock rule.

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C. Sample distributions and sizes

Sampling, Hypothesis Testing, and Sample Size

The sampling distribution of the mean and the central limit theorem, z- and t-tests with p-values, Type I/II error, and how large a sample you need.

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D. Goodness of fit

Goodness of Fit and Linear Regression

Fit a least-squares line, then judge how well it fits with the correlation coefficient R, the coefficient of determination R-squared, residuals, and the standard error of estimate.

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Reduces to $P(A)P(B)$ when A and B are independent; basis of the complement trick for "at least one."

Exactly $x$ successes in $n$ independent fixed trials, each with success probability $p$ . Mean $np$ , variance $np(1-p)$ .

Count of events in a fixed interval at constant rate; $\lambda$ = expected count = mean = variance. Scale $\lambda$ to the asked interval.

, and

x

share units.

= area to the left,

R(z)

= area to the right; symmetry lets you handle negative z with the positive-z table.

Waiting time to the next Poisson event; mean $1/\lambda$ , variance $1/\lambda^2$ , memoryless. $t\ge 0$ .

Constant density on $[a,b]$ ; everything outside the interval has zero density.

Unbiased point estimators of $\mu$ and $\sigma^2$ ; the $n-1$ (Bessel) divisor is what the FE expects.

Interval for $\mu$ when $\sigma$ is known or $n$ large; $z_{\alpha/2}=1.96$ (95%), $1.645$ (90%), $2.576$ (99%).

Interval for $\mu$ with small samples and estimated $s$ ; t-multiplier from the handbook table with $n-1$ degrees of freedom.

.

Solution. Standardize. For

106

:

z=(106-100)/4=1.5

, so

P(X>106)=R(1.5)=0.0668

. For the band:

z_{\text{low}}=(94-100)/4=-1.5

and

z_{\text{high}}=(108-100)/4=2.0

. Then

P=F(2.0)-F(-1.5)=F(2.0)-\big(1-F(1.5)\big)=0.9772-(1-0.9332)=0.9772-0.0668=0.910

. Sanity: the band spans about

-1.5\sigma

to

+2\sigma

, which should capture roughly

90\%

— it does. Answers:

P(X>106)=0.0668

,

P(94<X<108)=0.910

.

P(94<X<108)=F(2.0)-\big(1-F(1.5)\big)=0.910

, so

\lambda=9\times\tfrac13=3

calls. Use the complement:

P(X\ge2)=1-P(0)-P(1)

. With

\lambda=3

:

P(0)=e^{-3}=0.0498

,

P(1)=3e^{-3}=0.1494

.

P(X\ge2)=1-0.0498-0.1494=0.801

. Sanity: expecting

3

calls in the window, seeing fewer than two should be the minority —

0.801

for "two or more" fits. Answer:

0.801

.

P(X\ge 2)=1-e^{-3}\left(1+3\right)=0.801

. Construct the

95\%

confidence interval for the true mean.

Solution. Since

\sigma

is known, use

z_{\alpha/2}=1.96

. Margin of error

E=z_{\alpha/2}\,\sigma/\sqrt n = 1.96\,(6)/\sqrt{25}=1.96(6)/5=2.35\,\mu\text{m}

. Interval:

52.0\pm2.35 \Rightarrow (49.6,\ 54.4)\,\mu\text{m}

. Sanity: with

n=25

the standard error is

\sigma/\sqrt n=1.2

, and

1.96\times1.2\approx2.35

— consistent. Answer:

49.6\,\mu\text{m}\le\mu\le54.4\,\mu\text{m}

at

95\%

confidence.

52.0 \pm 1.96\,\dfrac{6}{\sqrt{25}} = (49.6,\ 54.4)\ \mu\text{m}

.