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Probability Functions, Distributions, and Expected Values

Probability and Statistics · FE Reference Handbook section

Probability and Statistics
3 formulas
10 exam-style examples
~51 min
All Probability and Statistics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • A random variable X has a probability associated with each of its possible values. The probability is termed a discrete
  • probability if X can assume only discrete values, or

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
z-score — solve for z-score — Probability Functions, Distributions, and Expected Values

A probability and statistics problem uses z-score. Given mean (mu) = 3,460 psi; std deviation (sigma) = 150.0 psi; value (x) = 5,880 psi, determine the z-score (z).

Given

  • mean(mu)=3,460psimean (mu) = 3,460 psi
  • stddeviation(sigma)=150.0psistd deviation (sigma) = 150.0 psi
  • value(x)=5,880psivalue (x) = 5,880 psi

Find

z-score (z)

Start with the thinking

  • The governing relation printed in this handbook section is z-score.
  • Everything except z is given, so isolate z symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Probability and Statistics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    z=(x−μ)/σz = (x - \mu) / \sigma
  2. Step 2 — Rearrange the relation so that z stands alone on the left-hand side.

  3. Step 3

    Listthegivens:mean(mu)=3,460psi,stddeviation(sigma)=150.0psi,value(x)=5,880psiList the givens: mean (mu) = 3,460 psi, std deviation (sigma) = 150.0 psi, value (x) = 5,880 psi
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    z=16.1333z = 16.1333
  6. Step 6 — Check: returning z = 16.1333 to

    z=(x−μ)/σz = (x - \mu) / \sigma

    reproduces the given quantities, and both sides carry the same units.

Answer:
z=16.1333z = 16.1333

Why the other options are there

  • 32.2667 — kept a factor of two that cancels in the correct rearrangement.
  • 8.0667 — dropped that same factor in the other direction.
  • 17.7467 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 2
Expected value of a discrete random variable — solve for expected value — Probability Functions, Distributions, and Expected Values (2)

A student finds the expected value of a two-outcome gamble. Given outcome 1 (x1) = 2.5000; probability 1 (p1) = 0.3900; outcome 2 (x2) = 15.0000, determine the expected value (EX).

Given

  • outcome1(x1)=2.5000outcome 1 (x_{1}) = 2.5000
  • probability1(p1)=0.3900probability 1 (p_{1}) = 0.3900
  • outcome2(x2)=15.0000outcome 2 (x_{2}) = 15.0000

Find

expected value (EX)

Start with the thinking

  • The governing relation printed in this handbook section is Expected value of a discrete random variable.
  • Everything except EX is given, so isolate EX symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • The expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

  1. Step 1 — State the governing relation:

    E[X]=∑xipiE[X] = \sum x_i p_i
  2. Step 2 — Rearrange symbolically for EX:

    EX=x1p1+x2(1−p1)EX = x_1 p_1 + x_2(1-p_1)
  3. Step 3

    Listthegivens:outcome1(x1)=2.5000,probability1(p1)=0.3900,outcome2(x2)=15.0000List the givens: outcome 1 (x_{1}) = 2.5000, probability 1 (p_{1}) = 0.3900, outcome 2 (x_{2}) = 15.0000
  4. Step 4 — Substitute the given values:

    EX=x1p1+x2(1−p1)EX = x_1 p_1 + x_2(1-p_1)
  5. Step 5 — Evaluate:

    EX=10.1250EX = 10.1250
  6. Step 6 — Check: returning EX = 10.1250 to

    E[X]=∑xipiE[X] = \sum x_i p_i

    reproduces the given quantities, and both sides carry the same units.

Answer:
EX=10.1250EX = 10.1250

Why the other options are there

  • 20.2500 — kept a factor of two that cancels in the correct rearrangement.
  • 5.0625 — dropped that same factor in the other direction.
  • 11.1375 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

Example 3
z-score — solve for value — Probability Functions, Distributions, and Expected Values (3)

A probability and statistics problem uses z-score. Given mean (mu) = 2,890 psi; std deviation (sigma) = 260.0 psi; z-score (z) = -0.1400, determine the value (x) in psi.

Given

  • mean(mu)=2,890psimean (mu) = 2,890 psi
  • stddeviation(sigma)=260.0psistd deviation (sigma) = 260.0 psi
  • z−score(z)=−0.1400z-score (z) = -0.1400

Find

value (x), in psi

Start with the thinking

  • The governing relation printed in this handbook section is z-score.
  • Everything except x is given, so isolate x symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Probability and Statistics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    z=(x−μ)/σz = (x - \mu) / \sigma
  2. Step 2 — Rearrange the relation so that x stands alone on the left-hand side.

  3. Step 3

    Listthegivens:mean(mu)=2,890psi,stddeviation(sigma)=260.0psi,z−score(z)=−0.1400List the givens: mean (mu) = 2,890 psi, std deviation (sigma) = 260.0 psi, z-score (z) = -0.1400
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    x=2854 psix = 2854\ \text{psi}
  6. Step 6 — Check: returning x = 2,854 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

    reproduces the given quantities, and both sides carry the same units.

Answer:
x=2854 psix = 2854\ \text{psi}

Why the other options are there

  • 5,707 — kept a factor of two that cancels in the correct rearrangement.
  • 1,427 — dropped that same factor in the other direction.
  • 3,139 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 4
Expected value of a discrete random variable — solve for outcome 2 — Probability Functions, Distributions, and Expected Values (4)

The expected value of a production yield is estimated from probabilities. Given outcome 1 (x1) = 3.5000; probability 1 (p1) = 0.3500; expected value (EX) = 18.3700, determine the outcome 2 (x2).

Given

  • outcome1(x1)=3.5000outcome 1 (x_{1}) = 3.5000
  • probability1(p1)=0.3500probability 1 (p_{1}) = 0.3500
  • expectedvalue(EX)=18.3700expected value (EX) = 18.3700

Find

outcome 2 (x2)

Start with the thinking

  • The governing relation printed in this handbook section is Expected value of a discrete random variable.
  • Everything except x2 is given, so isolate x2 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • The expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

  1. Step 1 — State the governing relation:

    E[X]=∑xipiE[X] = \sum x_i p_i
  2. Step 2 — Rearrange symbolically for x2:

    x2=E[X]−x1p11−p1x_{2} = \dfrac{E[X] - x_1 p_1}{1-p_1}
  3. Step 3

    Listthegivens:outcome1(x1)=3.5000,probability1(p1)=0.3500,expectedvalue(EX)=18.3700List the givens: outcome 1 (x_{1}) = 3.5000, probability 1 (p_{1}) = 0.3500, expected value (EX) = 18.3700
  4. Step 4 — Substitute the given values:

    x2=E[X]−x1p11−p1x_{2} = \dfrac{E[X] - x_1 p_1}{1-p_1}
  5. Step 5 — Evaluate:

    x2=26.3769x_{2} = 26.3769
  6. Step 6 — Check: returning x2 = 26.3769 to

    E[X]=∑xipiE[X] = \sum x_i p_i

    reproduces the given quantities, and both sides carry the same units.

Answer:
x2=26.3769x_{2} = 26.3769

Why the other options are there

  • 52.7538 — kept a factor of two that cancels in the correct rearrangement.
  • 13.1885 — dropped that same factor in the other direction.
  • 29.0146 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

Example 5
z-score — solve for mean — Probability Functions, Distributions, and Expected Values (5)

A probability and statistics problem uses z-score. Given std deviation (sigma) = 350.0 psi; value (x) = 3,890 psi; z-score (z) = 2.9600, determine the mean (mu) in psi.

Given

  • stddeviation(sigma)=350.0psistd deviation (sigma) = 350.0 psi
  • value(x)=3,890psivalue (x) = 3,890 psi
  • z−score(z)=2.9600z-score (z) = 2.9600

Find

mean (mu), in psi

Start with the thinking

  • The governing relation printed in this handbook section is z-score.
  • Everything except mu is given, so isolate mu symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Probability and Statistics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    z=(x−μ)/σz = (x - \mu) / \sigma
  2. Step 2 — Rearrange the relation so that mu stands alone on the left-hand side.

  3. Step 3

    Listthegivens:stddeviation(sigma)=350.0psi,value(x)=3,890psi,z−score(z)=2.9600List the givens: std deviation (sigma) = 350.0 psi, value (x) = 3,890 psi, z-score (z) = 2.9600
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    μ=2854 psi\mu = 2854\ \text{psi}
  6. Step 6 — Check: returning mu = 2,854 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

    reproduces the given quantities, and both sides carry the same units.

Answer:
μ=2854 psi\mu = 2854\ \text{psi}

Why the other options are there

  • 5,708 — kept a factor of two that cancels in the correct rearrangement.
  • 1,427 — dropped that same factor in the other direction.
  • 3,139 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 6
Expected value of a discrete random variable — solve for probability 1 — Probability Functions, Distributions, and Expected Values (6)

An engineer computes the expected value of a warranty payout distribution. Given outcome 1 (x1) = 3.5000; outcome 2 (x2) = 15.0000; expected value (EX) = 7.3500, determine the probability 1 (p1).

Given

  • outcome1(x1)=3.5000outcome 1 (x_{1}) = 3.5000
  • outcome2(x2)=15.0000outcome 2 (x_{2}) = 15.0000
  • expectedvalue(EX)=7.3500expected value (EX) = 7.3500

Find

probability 1 (p1)

Start with the thinking

  • The governing relation printed in this handbook section is Expected value of a discrete random variable.
  • Everything except p1 is given, so isolate p1 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • The expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

  1. Step 1 — State the governing relation:

    E[X]=∑xipiE[X] = \sum x_i p_i
  2. Step 2 — Rearrange symbolically for p1:

    p1=x2−E[X]x2−x1p_{1} = \dfrac{x_2 - E[X]}{x_2 - x_1}
  3. Step 3

    Listthegivens:outcome1(x1)=3.5000,outcome2(x2)=15.0000,expectedvalue(EX)=7.3500List the givens: outcome 1 (x_{1}) = 3.5000, outcome 2 (x_{2}) = 15.0000, expected value (EX) = 7.3500
  4. Step 4 — Substitute the given values:

    p1=x2−E[X]x2−x1p_{1} = \dfrac{x_2 - E[X]}{x_2 - x_1}
  5. Step 5 — Evaluate:

    p1=0.6652p_{1} = 0.6652
  6. Step 6 — Check: returning p1 = 0.6652 to

    E[X]=∑xipiE[X] = \sum x_i p_i

    reproduces the given quantities, and both sides carry the same units.

Answer:
p1=0.6652p_{1} = 0.6652

Why the other options are there

  • 1.3304 — kept a factor of two that cancels in the correct rearrangement.
  • 0.3326 — dropped that same factor in the other direction.
  • 0.7317 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

Example 7
z-score — solve for std deviation — Probability Functions, Distributions, and Expected Values (7)

A probability and statistics problem uses z-score. Given mean (mu) = 4,500 psi; value (x) = 1,530 psi; z-score (z) = 1.0400, determine the std deviation (sigma) in psi.

Given

  • mean(mu)=4,500psimean (mu) = 4,500 psi
  • value(x)=1,530psivalue (x) = 1,530 psi
  • z−score(z)=1.0400z-score (z) = 1.0400

Find

std deviation (sigma), in psi

Start with the thinking

  • The governing relation printed in this handbook section is z-score.
  • Everything except sigma is given, so isolate sigma symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Probability and Statistics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    z=(x−μ)/σz = (x - \mu) / \sigma
  2. Step 2 — Rearrange the relation so that sigma stands alone on the left-hand side.

  3. Step 3

    Listthegivens:mean(mu)=4,500psi,value(x)=1,530psi,z−score(z)=1.0400List the givens: mean (mu) = 4,500 psi, value (x) = 1,530 psi, z-score (z) = 1.0400
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    σ=−2856 psi\sigma = -2856\ \text{psi}
  6. Step 6 — Check: returning sigma = -2,856 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

    reproduces the given quantities, and both sides carry the same units.

Answer:
σ=−2856 psi\sigma = -2856\ \text{psi}

Why the other options are there

  • -5,712 — kept a factor of two that cancels in the correct rearrangement.
  • -1,428 — dropped that same factor in the other direction.
  • -3,141 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 8
Expected value of a discrete random variable — solve for expected value (case 2) — Probability Functions, Distributions, and Expected Values (8)

A student finds the expected value of a two-outcome gamble. Given outcome 1 (x1) = 8.5000; probability 1 (p1) = 0.4400; outcome 2 (x2) = 20.0000, determine the expected value (EX).

Given

  • outcome1(x1)=8.5000outcome 1 (x_{1}) = 8.5000
  • probability1(p1)=0.4400probability 1 (p_{1}) = 0.4400
  • outcome2(x2)=20.0000outcome 2 (x_{2}) = 20.0000

Find

expected value (EX)

Start with the thinking

  • The governing relation printed in this handbook section is Expected value of a discrete random variable.
  • Everything except EX is given, so isolate EX symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • The expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

  1. Step 1 — State the governing relation:

    E[X]=∑xipiE[X] = \sum x_i p_i
  2. Step 2 — Rearrange symbolically for EX:

    EX=x1p1+x2(1−p1)EX = x_1 p_1 + x_2(1-p_1)
  3. Step 3

    Listthegivens:outcome1(x1)=8.5000,probability1(p1)=0.4400,outcome2(x2)=20.0000List the givens: outcome 1 (x_{1}) = 8.5000, probability 1 (p_{1}) = 0.4400, outcome 2 (x_{2}) = 20.0000
  4. Step 4 — Substitute the given values:

    EX=x1p1+x2(1−p1)EX = x_1 p_1 + x_2(1-p_1)
  5. Step 5 — Evaluate:

    EX=14.9400EX = 14.9400
  6. Step 6 — Check: returning EX = 14.9400 to

    E[X]=∑xipiE[X] = \sum x_i p_i

    reproduces the given quantities, and both sides carry the same units.

Answer:
EX=14.9400EX = 14.9400

Why the other options are there

  • 29.8800 — kept a factor of two that cancels in the correct rearrangement.
  • 7.4700 — dropped that same factor in the other direction.
  • 16.4340 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

Example 9
z-score — solve for z-score (case 2) — Probability Functions, Distributions, and Expected Values (9)

A probability and statistics problem uses z-score. Given mean (mu) = 4,060 psi; std deviation (sigma) = 510.0 psi; value (x) = 6,910 psi, determine the z-score (z).

Given

  • mean(mu)=4,060psimean (mu) = 4,060 psi
  • stddeviation(sigma)=510.0psistd deviation (sigma) = 510.0 psi
  • value(x)=6,910psivalue (x) = 6,910 psi

Find

z-score (z)

Start with the thinking

  • The governing relation printed in this handbook section is z-score.
  • Everything except z is given, so isolate z symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Probability and Statistics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    z=(x−μ)/σz = (x - \mu) / \sigma
  2. Step 2 — Rearrange the relation so that z stands alone on the left-hand side.

  3. Step 3

    Listthegivens:mean(mu)=4,060psi,stddeviation(sigma)=510.0psi,value(x)=6,910psiList the givens: mean (mu) = 4,060 psi, std deviation (sigma) = 510.0 psi, value (x) = 6,910 psi
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    z=5.5882z = 5.5882
  6. Step 6 — Check: returning z = 5.5882 to

    z=(x−μ)/σz = (x - \mu) / \sigma

    reproduces the given quantities, and both sides carry the same units.

Answer:
z=5.5882z = 5.5882

Why the other options are there

  • 11.1765 — kept a factor of two that cancels in the correct rearrangement.
  • 2.7941 — dropped that same factor in the other direction.
  • 6.1471 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Probability Functions, Distributions, and Expected Values

Example 10
Expected value of a discrete random variable — solve for outcome 2 (case 2) — Probability Functions, Distributions, and Expected Values (10)

The expected value of a production yield is estimated from probabilities. Given outcome 1 (x1) = 1.0000; probability 1 (p1) = 0.5700; expected value (EX) = 6.4900, determine the outcome 2 (x2).

Given

  • outcome1(x1)=1.0000outcome 1 (x_{1}) = 1.0000
  • probability1(p1)=0.5700probability 1 (p_{1}) = 0.5700
  • expectedvalue(EX)=6.4900expected value (EX) = 6.4900

Find

outcome 2 (x2)

Start with the thinking

  • The governing relation printed in this handbook section is Expected value of a discrete random variable.
  • Everything except x2 is given, so isolate x2 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • The expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

  1. Step 1 — State the governing relation:

    E[X]=∑xipiE[X] = \sum x_i p_i
  2. Step 2 — Rearrange symbolically for x2:

    x2=E[X]−x1p11−p1x_{2} = \dfrac{E[X] - x_1 p_1}{1-p_1}
  3. Step 3

    Listthegivens:outcome1(x1)=1.0000,probability1(p1)=0.5700,expectedvalue(EX)=6.4900List the givens: outcome 1 (x_{1}) = 1.0000, probability 1 (p_{1}) = 0.5700, expected value (EX) = 6.4900
  4. Step 4 — Substitute the given values:

    x2=E[X]−x1p11−p1x_{2} = \dfrac{E[X] - x_1 p_1}{1-p_1}
  5. Step 5 — Evaluate:

    x2=13.7674x_{2} = 13.7674
  6. Step 6 — Check: returning x2 = 13.7674 to

    E[X]=∑xipiE[X] = \sum x_i p_i

    reproduces the given quantities, and both sides carry the same units.

Answer:
x2=13.7674x_{2} = 13.7674

Why the other options are there

  • 27.5349 — kept a factor of two that cancels in the correct rearrangement.
  • 6.8837 — dropped that same factor in the other direction.
  • 15.1442 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

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