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Cumulative Distribution Functions

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.

  • The cumulative distribution function, F, of a discrete random variable X that has a probability distribution described by P(xi) is

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 — Cumulative Distribution Functions

A probability and statistics problem uses z-score. Given mean (mu) = 2,910 psi; std deviation (sigma) = 370.0 psi; value (x) = 5,920 psi, determine the z-score (z).

Given

  • mean(mu)=2,910psimean (mu) = 2,910 psi
  • stddeviation(sigma)=370.0psistd deviation (sigma) = 370.0 psi
  • value(x)=5,920psivalue (x) = 5,920 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)=2,910psi,stddeviation(sigma)=370.0psi,value(x)=5,920psiList the givens: mean (mu) = 2,910 psi, std deviation (sigma) = 370.0 psi, value (x) = 5,920 psi
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    z=8.1351z = 8.1351
  6. Step 6 — Check: returning z = 8.1351 to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
z=8.1351z = 8.1351

Why the other options are there

  • 16.2703 — kept a factor of two that cancels in the correct rearrangement.
  • 4.0676 — dropped that same factor in the other direction.
  • 8.9486 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 2
z-score — solve for value — Cumulative Distribution Functions (2)

A probability and statistics problem uses z-score. Given mean (mu) = 3,140 psi; std deviation (sigma) = 240.0 psi; z-score (z) = -0.8700, determine the value (x) in psi.

Given

  • mean(mu)=3,140psimean (mu) = 3,140 psi
  • stddeviation(sigma)=240.0psistd deviation (sigma) = 240.0 psi
  • z−score(z)=−0.8700z-score (z) = -0.8700

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)=3,140psi,stddeviation(sigma)=240.0psi,z−score(z)=−0.8700List the givens: mean (mu) = 3,140 psi, std deviation (sigma) = 240.0 psi, z-score (z) = -0.8700
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    x=2931 psix = 2931\ \text{psi}
  6. Step 6 — Check: returning x = 2,931 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
x=2931 psix = 2931\ \text{psi}

Why the other options are there

  • 5,862 — kept a factor of two that cancels in the correct rearrangement.
  • 1,466 — dropped that same factor in the other direction.
  • 3,224 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 3
z-score — solve for mean — Cumulative Distribution Functions (3)

A probability and statistics problem uses z-score. Given std deviation (sigma) = 530.0 psi; value (x) = 2,150 psi; z-score (z) = 0.1000, determine the mean (mu) in psi.

Given

  • stddeviation(sigma)=530.0psistd deviation (sigma) = 530.0 psi
  • value(x)=2,150psivalue (x) = 2,150 psi
  • z−score(z)=0.1000z-score (z) = 0.1000

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)=530.0psi,value(x)=2,150psi,z−score(z)=0.1000List the givens: std deviation (sigma) = 530.0 psi, value (x) = 2,150 psi, z-score (z) = 0.1000
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    μ=2097 psi\mu = 2097\ \text{psi}
  6. Step 6 — Check: returning mu = 2,097 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
μ=2097 psi\mu = 2097\ \text{psi}

Why the other options are there

  • 4,194 — kept a factor of two that cancels in the correct rearrangement.
  • 1,049 — dropped that same factor in the other direction.
  • 2,307 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 4
z-score — solve for std deviation — Cumulative Distribution Functions (4)

A probability and statistics problem uses z-score. Given mean (mu) = 5,650 psi; value (x) = 5,540 psi; z-score (z) = 0.1100, determine the std deviation (sigma) in psi.

Given

  • mean(mu)=5,650psimean (mu) = 5,650 psi
  • value(x)=5,540psivalue (x) = 5,540 psi
  • z−score(z)=0.1100z-score (z) = 0.1100

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)=5,650psi,value(x)=5,540psi,z−score(z)=0.1100List the givens: mean (mu) = 5,650 psi, value (x) = 5,540 psi, z-score (z) = 0.1100
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    σ=−1000 psi\sigma = -1000\ \text{psi}
  6. Step 6 — Check: returning sigma = -1,000 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
σ=−1000 psi\sigma = -1000\ \text{psi}

Why the other options are there

  • -2,000 — kept a factor of two that cancels in the correct rearrangement.
  • -500.0 — dropped that same factor in the other direction.
  • -1,100 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 5
z-score — solve for z-score (case 2) — Cumulative Distribution Functions (5)

A probability and statistics problem uses z-score. Given mean (mu) = 3,540 psi; std deviation (sigma) = 430.0 psi; value (x) = 3,940 psi, determine the z-score (z).

Given

  • mean(mu)=3,540psimean (mu) = 3,540 psi
  • stddeviation(sigma)=430.0psistd deviation (sigma) = 430.0 psi
  • value(x)=3,940psivalue (x) = 3,940 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)=3,540psi,stddeviation(sigma)=430.0psi,value(x)=3,940psiList the givens: mean (mu) = 3,540 psi, std deviation (sigma) = 430.0 psi, value (x) = 3,940 psi
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    z=0.9302z = 0.9302
  6. Step 6 — Check: returning z = 0.9302 to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
z=0.9302z = 0.9302

Why the other options are there

  • 1.8605 — kept a factor of two that cancels in the correct rearrangement.
  • 0.4651 — dropped that same factor in the other direction.
  • 1.0233 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 6
z-score — solve for value (case 2) — Cumulative Distribution Functions (6)

A probability and statistics problem uses z-score. Given mean (mu) = 5,900 psi; std deviation (sigma) = 190.0 psi; z-score (z) = 0.0200, determine the value (x) in psi.

Given

  • mean(mu)=5,900psimean (mu) = 5,900 psi
  • stddeviation(sigma)=190.0psistd deviation (sigma) = 190.0 psi
  • z−score(z)=0.0200z-score (z) = 0.0200

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)=5,900psi,stddeviation(sigma)=190.0psi,z−score(z)=0.0200List the givens: mean (mu) = 5,900 psi, std deviation (sigma) = 190.0 psi, z-score (z) = 0.0200
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    x=5904 psix = 5904\ \text{psi}
  6. Step 6 — Check: returning x = 5,904 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
x=5904 psix = 5904\ \text{psi}

Why the other options are there

  • 11,808 — kept a factor of two that cancels in the correct rearrangement.
  • 2,952 — dropped that same factor in the other direction.
  • 6,494 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 7
z-score — solve for mean (case 2) — Cumulative Distribution Functions (7)

A probability and statistics problem uses z-score. Given std deviation (sigma) = 370.0 psi; value (x) = 4,030 psi; z-score (z) = 2.5100, determine the mean (mu) in psi.

Given

  • stddeviation(sigma)=370.0psistd deviation (sigma) = 370.0 psi
  • value(x)=4,030psivalue (x) = 4,030 psi
  • z−score(z)=2.5100z-score (z) = 2.5100

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)=370.0psi,value(x)=4,030psi,z−score(z)=2.5100List the givens: std deviation (sigma) = 370.0 psi, value (x) = 4,030 psi, z-score (z) = 2.5100
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    μ=3101 psi\mu = 3101\ \text{psi}
  6. Step 6 — Check: returning mu = 3,101 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
μ=3101 psi\mu = 3101\ \text{psi}

Why the other options are there

  • 6,203 — kept a factor of two that cancels in the correct rearrangement.
  • 1,551 — dropped that same factor in the other direction.
  • 3,411 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 8
z-score — solve for std deviation (case 2) — Cumulative Distribution Functions (8)

A probability and statistics problem uses z-score. Given mean (mu) = 5,730 psi; value (x) = 3,440 psi; z-score (z) = 2.5200, determine the std deviation (sigma) in psi.

Given

  • mean(mu)=5,730psimean (mu) = 5,730 psi
  • value(x)=3,440psivalue (x) = 3,440 psi
  • z−score(z)=2.5200z-score (z) = 2.5200

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)=5,730psi,value(x)=3,440psi,z−score(z)=2.5200List the givens: mean (mu) = 5,730 psi, value (x) = 3,440 psi, z-score (z) = 2.5200
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    σ=−908.7 psi\sigma = -908.7\ \text{psi}
  6. Step 6 — Check: returning sigma = -908.7 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
σ=−908.7 psi\sigma = -908.7\ \text{psi}

Why the other options are there

  • -1,817 — kept a factor of two that cancels in the correct rearrangement.
  • -454.4 — dropped that same factor in the other direction.
  • -999.6 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 9
z-score — solve for z-score (case 3) — Cumulative Distribution Functions (9)

A probability and statistics problem uses z-score. Given mean (mu) = 4,550 psi; std deviation (sigma) = 450.0 psi; value (x) = 6,190 psi, determine the z-score (z).

Given

  • mean(mu)=4,550psimean (mu) = 4,550 psi
  • stddeviation(sigma)=450.0psistd deviation (sigma) = 450.0 psi
  • value(x)=6,190psivalue (x) = 6,190 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)=4,550psi,stddeviation(sigma)=450.0psi,value(x)=6,190psiList the givens: mean (mu) = 4,550 psi, std deviation (sigma) = 450.0 psi, value (x) = 6,190 psi
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    z=3.6444z = 3.6444
  6. Step 6 — Check: returning z = 3.6444 to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
z=3.6444z = 3.6444

Why the other options are there

  • 7.2889 — kept a factor of two that cancels in the correct rearrangement.
  • 1.8222 — dropped that same factor in the other direction.
  • 4.0089 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

Example 10
z-score — solve for value (case 3) — Cumulative Distribution Functions (10)

A probability and statistics problem uses z-score. Given mean (mu) = 2,790 psi; std deviation (sigma) = 400.0 psi; z-score (z) = 2.7400, determine the value (x) in psi.

Given

  • mean(mu)=2,790psimean (mu) = 2,790 psi
  • stddeviation(sigma)=400.0psistd deviation (sigma) = 400.0 psi
  • z−score(z)=2.7400z-score (z) = 2.7400

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)=2,790psi,stddeviation(sigma)=400.0psi,z−score(z)=2.7400List the givens: mean (mu) = 2,790 psi, std deviation (sigma) = 400.0 psi, z-score (z) = 2.7400
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    x=3886 psix = 3886\ \text{psi}
  6. Step 6 — Check: returning x = 3,886 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
x=3886 psix = 3886\ \text{psi}

Why the other options are there

  • 7,772 — kept a factor of two that cancels in the correct rearrangement.
  • 1,943 — dropped that same factor in the other direction.
  • 4,275 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Distribution Functions

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