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Probability and Statistics

4% of exam

Central tendency & dispersion, probability distributions, expected value.

3 concepts

A. Measures of central tendency and dispersion

Central Tendency and Dispersion

Mean, median, and mode plus the variance, standard deviation, and coefficient of variation that quantify spread — and the population-versus-sample distinction the FE punishes.

Almost every probability and statistics question on the FE starts by collapsing a pile of numbers into one or two summary statistics: a measure of where the data sits (central tendency) and a measure of how scattered it is (dispersion). Get those two ideas exactly right and the rest of the topic — distributions, expected value, confidence intervals — is bookkeeping on top of them. The FE Reference Handbook collects all of this on one page, "Dispersion, Mean, Median, and Mode Values", so the formulas are handed to you; the points are lost in choosing the right one and in the population-versus-sample

N

-versus-

(n-1)

trap.

The three measures of center

The arithmetic mean

\bar{X}

is the balance point — the sum of the values divided by how many there are — and it is what "average" means unless a problem says otherwise. The median is the middle value once the data are sorted: for odd

n

it is the single center item, and for even

n

it is the average of the two center items. The mode is simply the most frequently occurring value. The mean uses every data point and is the basis for variance, but it is pulled toward outliers; the median ignores magnitude and so resists them. When a distribution is symmetric the three coincide; when it is right-skewed the mean exceeds the median.

\bar{X} = \frac{1}{n}\sum_{i=1}^{n} X_i

Weighted mean

When observations carry different importance — credit hours, sample sizes, reliability weights — the plain average is wrong and you must weight each value

X_i

by its weight

w_i

. The weighted mean divides the sum of

w_iX_i

by the sum of the weights, not by the count. A grade-point average, a blended interest rate, and a composite component reliability are all weighted means. If every weight is equal this reduces to the ordinary arithmetic mean.

\bar{X}_w = \frac{\sum_{i} w_i X_i}{\sum_{i} w_i}

Variance: the mean squared deviation

Dispersion is measured by how far, on average, the data fall from their mean — squared, so that positive and negative deviations do not cancel. The population variance

\sigma^2

divides the sum of squared deviations by the population size

N

, because

\mu

is the true mean of the whole population. Squaring puts variance in units

^2

(e.g.,

\text{V}^2

), which is why the standard deviation, its square root, is the more reportable number — it has the same units as the data.

\sigma^2 = \frac{1}{N}\sum_{i=1}^{N}\left(X_i - \mu\right)^2

Sample variance and the (n−1) divisor

When your numbers are a sample drawn from a larger population — the usual engineering case — you do not know

\mu

and must estimate it with

\bar{X}

. Doing so uses up one degree of freedom, so dividing by

n

would systematically underestimate the spread. The fix, baked into the handbook's sample variance formula, is to divide by

n-1

instead, which makes

s^2

an unbiased estimator of

\sigma^2

. The sample standard deviation

s

is its square root. Reading the problem to decide whether you have the whole population (

N

) or a sample (

n-1

) is the single most common point swing on this subtopic.

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

Coefficient of variation

The standard deviation alone cannot tell you whether spread is large or small — a

\pm 5\,\text{V}

scatter is trivial on a

480\,\text{V}

bus and enormous on a

3.3\,\text{V}

logic rail. The coefficient of variation normalizes the standard deviation by the mean, giving a dimensionless (or percent) measure of relative dispersion that lets you compare datasets with different units or scales. It is defined only for ratio data with a meaningful, nonzero mean.

CV = \frac{s}{\bar{X}} \quad\left(\text{or }\frac{\sigma}{\mu}\right)

A faster way to compute variance

Computing each deviation, squaring, and summing is slow under exam pressure. The algebraic identity

\sum(X_i-\bar{X})^2 = \sum X_i^2 - n\bar{X}^2

lets you accumulate

\sum X_i

and

\sum X_i^2

in a single pass, then form the variance at the end. It is the same expression your calculator's statistics mode uses, and it pairs with the expected-value identity

\operatorname{Var}(X)=E[X^2]-(E[X])^2

used elsewhere in this topic.

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

Exam strategy

First decide population or sample: the word "sample," or any mention of estimating a larger population, means divide variance by

n-1

. Use your calculator's statistics registers — enter the data once and read off

\bar{X}

,

\sigma_x

(population, divisor

n

), and

s_x

(sample, divisor

n-1

); knowing which key gives which divisor avoids the trap entirely. For median, always sort first and count carefully for even

n

. For

CV

, keep

s

and

\bar{X}

in the same units so they cancel, and multiply by

100

only if the answer choices are percentages.

Key equations

Arithmetic mean

\bar{X} = \frac{1}{n}\sum_{i=1}^{n} X_i

Balance point of the data. $X_i$ = ith observation, $n$ = number of observations; units match the data.

Weighted mean

\bar{X}_w = \frac{\sum_i w_i X_i}{\sum_i w_i}

Population variance

\sigma^2 = \frac{1}{N}\sum_{i=1}^{N}(X_i-\mu)^2

Sample variance

s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2

Standard deviation

\sigma=\sqrt{\sigma^2},\qquad s=\sqrt{s^2}

Coefficient of variation

CV = \frac{s}{\bar{X}}

Dimensionless relative dispersion; multiply by 100 for percent. Use $\sigma/\mu$

Computational sum-of-squares identity

\sum_{i=1}^{n}(X_i-\bar{X})^2=\sum_{i=1}^{n}X_i^2-n\bar{X}^2

Median (sorted data)

\tilde{X}=\begin{cases}X_{(n+1)/2}, & n\text{ odd}\\[2pt]\tfrac{1}{2}\!\left(X_{n/2}+X_{n/2+1}\right), & n\text{ even}\end{cases}

Sample range

R = X_{\max}-X_{\min}

Crudest dispersion measure: largest minus smallest value. Sensitive to a single outlier.

Worked examples

Mean, median, and sample standard deviation

Problem. A technician records seven repair times (minutes): $12, 15, 15, 18, 20, 22, 38$ . Treating these as a sample, find the mean, median, mode, and sample standard deviation.

Solution. Mean:

\bar{X}=(12+15+15+18+20+22+38)/7=140/7=20.0\,\text{min}

. Sorted,

n=7

is odd, so the median is the 4th item:

\tilde{X}=18\,\text{min}

. The mode is

15\,\text{min}

(occurs twice). The mean (

20

) exceeds the median (

18

), consistent with the right-skew from the

38

outlier. Sample variance:

\sum(X_i-20)^2 = 64+25+25+4+0+4+324=446

, so

s^2=446/(7-1)=74.33\,\text{min}^2

and

s=\sqrt{74.33}=8.62\,\text{min}

. Sanity check: a sample

s

(divisor 6) must exceed the population

\sigma

(divisor 7,

=7.98

), and it does.

Weighted mean of course grades

Problem. A course grade is $20\%$ homework, $30\%$ midterm, $50\%$ final. A student scores $78$ , $85$ , and $92$ respectively. Find the weighted average.

Comparing dispersion with the coefficient of variation

Problem. Resistor batch A has mean $\bar{X}=50\,\Omega$ with $s=6\,\Omega$ . Batch B has mean $\bar{X}=1{,}000\,\Omega$

Common pitfalls

•Dividing sample variance by $n$ instead of $n-1$ . The handbook's sample formula uses $n-1$ ; using $n$ underestimates spread. Population data (size $N$ ) divides by $N$ — read which one you have.
•Confusing your calculator's $\sigma_x$ (population, divisor $n$ ) with $s_x$ (sample, divisor $n-1$
•Forgetting to sort before finding the median, or mishandling even $n$ by taking one center value instead of averaging the two.
•Computing a weighted mean by dividing by the count instead of by $\sum w_i$ . Always divide by the sum of the weights (which is 1 only if they are already normalized).
•Reporting variance where the answer wants standard deviation (or vice versa). Variance carries units $^2$ ; take the square root for the same units as the data.
•Forming $CV$ with $s$ and $\bar{X}$ in different units, or reporting a percent when the choices are a fraction (or the reverse). Keep units consistent and match the answer format.

References

NCEES FE Reference Handbook — Engineering Probability and Statistics: Dispersion, Mean, Median, and Mode Values — Source of the mean, weighted mean, population/sample variance, standard deviation, CV, median, and range formulas.
Montgomery & Runger, Applied Statistics and Probability for Engineers — Ch. 6 — descriptive statistics and the rationale for the n−1 divisor (degrees of freedom).

B. Probability distributions

Probability Distributions

The probability rules plus the binomial, Poisson, normal/standard-normal, exponential, and uniform models — when each applies, and how to read PMFs, PDFs, and CDFs.

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C. Expected value (weighted average)

Expected Value and Decision-Making

The probability-weighted average of a random variable, its variance, and how expected monetary value drives engineering decisions under uncertainty.

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Use when observations have unequal importance. $w_i$ = weight on $X_i$ . Divide by the sum of weights, not the count.

Spread of a complete population of size $N$ about its true mean $\mu$ . Units are (data units) $^2$ .

Unbiased estimate of $\sigma^2$ from a sample. The $n-1$ divisor (Bessel's correction) accounts for estimating $\mu$ by $\bar{X}$ .

Square root of variance; same units as the data, which is why it is the reportable spread measure.

for a population. Requires a meaningful nonzero mean.

One-pass shortcut for variance using $\sum X_i$ and $\sum X_i^2$ ; divide by $N$ or $n-1$ as appropriate.

Middle value of the ascending-sorted data; for even $n$ average the two central items. Resistant to outliers.

Solution. Weights sum to

1

, so

\bar{X}_w=\sum w_iX_i = (0.20)(78)+(0.30)(85)+(0.50)(92)

.

\bar{X}_w = 15.6+25.5+46.0 = 87.1

. Sanity check:

87.1

lies between the lowest (

78

) and highest (

92

) scores and is pulled toward

92

because the final carries half the weight — as expected.

with

s=80\,\Omega

. Which batch is more consistent in relative terms?

Solution. Batch A:

CV_A = 6/50 = 0.120 = 12.0\%

. Batch B:

CV_B = 80/1000 = 0.080 = 8.00\%

. Batch B has the larger absolute spread (

80\,\Omega

vs

6\,\Omega

) but the smaller relative spread, so Batch B is more consistent. Sanity check:

CV

is dimensionless (ohms cancel), so the comparison across the two very different resistance scales is valid — which is exactly what the standard deviations alone could not tell us.

). Press the right key; they differ noticeably for small datasets.