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Standard Deviation Charts

Probability and Statistics · FE Reference Handbook section

Probability and Statistics
6 formulas
10 exam-style examples
~57 min
All Probability and Statistics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • Engineering Probability and Statistics

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 — Standard Deviation Charts

A probability and statistics problem uses z-score. Given mean (mu) = 4,370 psi; std deviation (sigma) = 460.0 psi; value (x) = 6,180 psi, determine the z-score (z).

Given

  • mean(mu)=4,370psimean (mu) = 4,370 psi
  • stddeviation(sigma)=460.0psistd deviation (sigma) = 460.0 psi
  • value(x)=6,180psivalue (x) = 6,180 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)=4,370psi,stddeviation(sigma)=460.0psi,value(x)=6,180psiList the givens: mean (mu) = 4,370 psi, std deviation (sigma) = 460.0 psi, value (x) = 6,180 psi
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    z=3.9348z = 3.9348
  6. Step 6 — Check: returning z = 3.9348 to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
z=3.9348z = 3.9348

Why the other options are there

  • 7.8696 — kept a factor of two that cancels in the correct rearrangement.
  • 1.9674 — dropped that same factor in the other direction.
  • 4.3283 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Standard Deviation Charts

Example 2
Standard deviation control chart limits — solve for upper control limit — Standard Deviation Charts (2)

An analyst computes the upper control limit for a standard deviation chart. Given control chart factor B4 (B4) = 1.7900; average sample std dev (sbar) = 4.8000, determine the upper control limit (UCL).

Given

  • controlchartfactorB4(B4)=1.7900control chart factor B_{4} (B_{4}) = 1.7900
  • averagesamplestddev(sbar)=4.8000average sample std dev (sbar) = 4.8000

Find

upper control limit (UCL)

Start with the thinking

  • The governing relation printed in this handbook section is Standard deviation control chart limits.
  • Everything except UCL is given, so isolate UCL 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.
  • Standard deviation charts monitor process variability using control limits based on the average sample standard deviation.
Standard deviation chart0.00.91.82.63.53.212.823.53342.95sample std dev

Figure 2 — schematic for Standard deviation control chart limits — solve for upper control limit — Standard Deviation Charts (2)

Step-by-step solution

  1. Step 1 — State the governing relation:

    UCLs=B4sˉUCL_s = B_4 \bar{s}
  2. Step 2 — Rearrange symbolically for UCL:

    UCL=B4sˉUCL = B_4 \bar{s}
  3. Step 3 — List the givens: control chart factor B4 (B4) = 1.7900, average sample std dev (sbar) = 4.8000.

  4. Step 4 — Substitute the given values:

    UCL=B4sˉUCL = B_4 \bar{s}
  5. Step 5 — Evaluate:

    UCL=8.5920UCL = 8.5920
  6. Step 6 — Check: returning UCL = 8.5920 to

    UCLs=B4sˉUCL_s = B_4 \bar{s}

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

Answer:
UCL=8.5920UCL = 8.5920

Why the other options are there

  • 17.1840 — kept a factor of two that cancels in the correct rearrangement.
  • 4.2960 — dropped that same factor in the other direction.
  • 9.4512 — rounded an intermediate value before the final step.

Reference: FE Handbook — Standard Deviation Charts

Example 3
z-score — solve for value — Standard Deviation Charts (3)

A probability and statistics problem uses z-score. Given mean (mu) = 2,860 psi; std deviation (sigma) = 150.0 psi; z-score (z) = 2.8900, determine the value (x) in psi.

Given

  • mean(mu)=2,860psimean (mu) = 2,860 psi
  • stddeviation(sigma)=150.0psistd deviation (sigma) = 150.0 psi
  • z−score(z)=2.8900z-score (z) = 2.8900

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)=2,860psi,stddeviation(sigma)=150.0psi,z−score(z)=2.8900List the givens: mean (mu) = 2,860 psi, std deviation (sigma) = 150.0 psi, z-score (z) = 2.8900
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    x=3294 psix = 3294\ \text{psi}
  6. Step 6 — Check: returning x = 3,294 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
x=3294 psix = 3294\ \text{psi}

Why the other options are there

  • 6,587 — kept a factor of two that cancels in the correct rearrangement.
  • 1,647 — dropped that same factor in the other direction.
  • 3,623 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Standard Deviation Charts

Example 4
Standard deviation control chart limits — solve for control chart factor B4 — Standard Deviation Charts (4)

The standard deviation chart shows increased variability beyond the control limit. Given average sample std dev (sbar) = 4.6500; upper control limit (UCL) = 21.0300, determine the control chart factor B4 (B4).

Given

  • averagesamplestddev(sbar)=4.6500average sample std dev (sbar) = 4.6500
  • uppercontrollimit(UCL)=21.0300upper control limit (UCL) = 21.0300

Find

control chart factor B4 (B4)

Start with the thinking

  • The governing relation printed in this handbook section is Standard deviation control chart limits.
  • Everything except B4 is given, so isolate B4 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.
  • Standard deviation charts monitor process variability using control limits based on the average sample standard deviation.
Standard deviation chart0.00.91.82.63.53.212.823.53342.95sample std dev

Figure 4 — schematic for Standard deviation control chart limits — solve for control chart factor B4 — Standard Deviation Charts (4)

Step-by-step solution

  1. Step 1 — State the governing relation:

    UCLs=B4sˉUCL_s = B_4 \bar{s}
  2. Step 2 — Rearrange symbolically for B4:

    B4=UCLsˉB_{4} = \dfrac{UCL}{\bar{s}}
  3. Step 3 — List the givens: average sample std dev (sbar) = 4.6500, upper control limit (UCL) = 21.0300.

  4. Step 4 — Substitute the given values:

    B4=21.0300sˉB_{4} = \dfrac{21.0300}{\bar{s}}
  5. Step 5 — Evaluate:

    B4=4.5226B_{4} = 4.5226
  6. Step 6 — Check: returning B4 = 4.5226 to

    UCLs=B4sˉUCL_s = B_4 \bar{s}

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

Answer:
B4=4.5226B_{4} = 4.5226

Why the other options are there

  • 9.0452 — kept a factor of two that cancels in the correct rearrangement.
  • 2.2613 — dropped that same factor in the other direction.
  • 4.9748 — rounded an intermediate value before the final step.

Reference: FE Handbook — Standard Deviation Charts

Example 5
z-score — solve for mean — Standard Deviation Charts (5)

A probability and statistics problem uses z-score. Given std deviation (sigma) = 280.0 psi; value (x) = 4,280 psi; z-score (z) = -2.0100, determine the mean (mu) in psi.

Given

  • stddeviation(sigma)=280.0psistd deviation (sigma) = 280.0 psi
  • value(x)=4,280psivalue (x) = 4,280 psi
  • z−score(z)=−2.0100z-score (z) = -2.0100

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)=280.0psi,value(x)=4,280psi,z−score(z)=−2.0100List the givens: std deviation (sigma) = 280.0 psi, value (x) = 4,280 psi, z-score (z) = -2.0100
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    μ=4843 psi\mu = 4843\ \text{psi}
  6. Step 6 — Check: returning mu = 4,843 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
μ=4843 psi\mu = 4843\ \text{psi}

Why the other options are there

  • 9,686 — kept a factor of two that cancels in the correct rearrangement.
  • 2,421 — dropped that same factor in the other direction.
  • 5,327 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Standard Deviation Charts

Example 6
Standard deviation control chart limits — solve for average sample std dev — Standard Deviation Charts (6)

A quality engineer builds a standard deviation chart for a filling process. Given control chart factor B4 (B4) = 1.6600; upper control limit (UCL) = 17.1900, determine the average sample std dev (sbar).

Given

  • controlchartfactorB4(B4)=1.6600control chart factor B_{4} (B_{4}) = 1.6600
  • uppercontrollimit(UCL)=17.1900upper control limit (UCL) = 17.1900

Find

average sample std dev (sbar)

Start with the thinking

  • The governing relation printed in this handbook section is Standard deviation control chart limits.
  • Everything except sbar is given, so isolate sbar 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.
  • Standard deviation charts monitor process variability using control limits based on the average sample standard deviation.
Standard deviation chart0.00.91.82.63.53.212.823.53342.95sample std dev

Figure 6 — schematic for Standard deviation control chart limits — solve for average sample std dev — Standard Deviation Charts (6)

Step-by-step solution

  1. Step 1 — State the governing relation:

    UCLs=B4sˉUCL_s = B_4 \bar{s}
  2. Step 2 — Rearrange symbolically for sbar:

    sbar=UCLB4sbar = \dfrac{UCL}{B_4}
  3. Step 3 — List the givens: control chart factor B4 (B4) = 1.6600, upper control limit (UCL) = 17.1900.

  4. Step 4 — Substitute the given values:

    sbar=17.1900B4sbar = \dfrac{17.1900}{B_4}
  5. Step 5 — Evaluate:

    sbar=10.3554sbar = 10.3554
  6. Step 6 — Check: returning sbar = 10.3554 to

    UCLs=B4sˉUCL_s = B_4 \bar{s}

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

Answer:
sbar=10.3554sbar = 10.3554

Why the other options are there

  • 20.7108 — kept a factor of two that cancels in the correct rearrangement.
  • 5.1777 — dropped that same factor in the other direction.
  • 11.3910 — rounded an intermediate value before the final step.

Reference: FE Handbook — Standard Deviation Charts

Example 7
z-score — solve for std deviation — Standard Deviation Charts (7)

A probability and statistics problem uses z-score. Given mean (mu) = 2,740 psi; value (x) = 1,570 psi; z-score (z) = -2.3400, determine the std deviation (sigma) in psi.

Given

  • mean(mu)=2,740psimean (mu) = 2,740 psi
  • value(x)=1,570psivalue (x) = 1,570 psi
  • z−score(z)=−2.3400z-score (z) = -2.3400

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)=2,740psi,value(x)=1,570psi,z−score(z)=−2.3400List the givens: mean (mu) = 2,740 psi, value (x) = 1,570 psi, z-score (z) = -2.3400
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    σ=500.0 psi\sigma = 500.0\ \text{psi}
  6. Step 6 — Check: returning sigma = 500.0 psi to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
σ=500.0 psi\sigma = 500.0\ \text{psi}

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Standard Deviation Charts

Example 8
Standard deviation control chart limits — solve for upper control limit (case 2) — Standard Deviation Charts (8)

An analyst computes the upper control limit for a standard deviation chart. Given control chart factor B4 (B4) = 2.1100; average sample std dev (sbar) = 8.9500, determine the upper control limit (UCL).

Given

  • controlchartfactorB4(B4)=2.1100control chart factor B_{4} (B_{4}) = 2.1100
  • averagesamplestddev(sbar)=8.9500average sample std dev (sbar) = 8.9500

Find

upper control limit (UCL)

Start with the thinking

  • The governing relation printed in this handbook section is Standard deviation control chart limits.
  • Everything except UCL is given, so isolate UCL 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.
  • Standard deviation charts monitor process variability using control limits based on the average sample standard deviation.
Standard deviation chart0.00.91.82.63.53.212.823.53342.95sample std dev

Figure 8 — schematic for Standard deviation control chart limits — solve for upper control limit (case 2) — Standard Deviation Charts (8)

Step-by-step solution

  1. Step 1 — State the governing relation:

    UCLs=B4sˉUCL_s = B_4 \bar{s}
  2. Step 2 — Rearrange symbolically for UCL:

    UCL=B4sˉUCL = B_4 \bar{s}
  3. Step 3 — List the givens: control chart factor B4 (B4) = 2.1100, average sample std dev (sbar) = 8.9500.

  4. Step 4 — Substitute the given values:

    UCL=B4sˉUCL = B_4 \bar{s}
  5. Step 5 — Evaluate:

    UCL=18.8845UCL = 18.8845
  6. Step 6 — Check: returning UCL = 18.8845 to

    UCLs=B4sˉUCL_s = B_4 \bar{s}

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

Answer:
UCL=18.8845UCL = 18.8845

Why the other options are there

  • 37.7690 — kept a factor of two that cancels in the correct rearrangement.
  • 9.4422 — dropped that same factor in the other direction.
  • 20.7729 — rounded an intermediate value before the final step.

Reference: FE Handbook — Standard Deviation Charts

Example 9
z-score — solve for z-score (case 2) — Standard Deviation Charts (9)

A probability and statistics problem uses z-score. Given mean (mu) = 5,390 psi; std deviation (sigma) = 300.0 psi; value (x) = 1,910 psi, determine the z-score (z).

Given

  • mean(mu)=5,390psimean (mu) = 5,390 psi
  • stddeviation(sigma)=300.0psistd deviation (sigma) = 300.0 psi
  • value(x)=1,910psivalue (x) = 1,910 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)=5,390psi,stddeviation(sigma)=300.0psi,value(x)=1,910psiList the givens: mean (mu) = 5,390 psi, std deviation (sigma) = 300.0 psi, value (x) = 1,910 psi
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    z=−11.6000z = -11.6000
  6. Step 6 — Check: returning z = -11.6000 to

    z=(x−μ)/σz = (x - \mu) / \sigma

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

Answer:
z=−11.6000z = -11.6000

Why the other options are there

  • -23.2000 — kept a factor of two that cancels in the correct rearrangement.
  • -5.8000 — dropped that same factor in the other direction.
  • -12.7600 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Standard Deviation Charts

Example 10
Standard deviation control chart limits — solve for control chart factor B4 (case 2) — Standard Deviation Charts (10)

The standard deviation chart shows increased variability beyond the control limit. Given average sample std dev (sbar) = 5.4000; upper control limit (UCL) = 15.5300, determine the control chart factor B4 (B4).

Given

  • averagesamplestddev(sbar)=5.4000average sample std dev (sbar) = 5.4000
  • uppercontrollimit(UCL)=15.5300upper control limit (UCL) = 15.5300

Find

control chart factor B4 (B4)

Start with the thinking

  • The governing relation printed in this handbook section is Standard deviation control chart limits.
  • Everything except B4 is given, so isolate B4 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.
  • Standard deviation charts monitor process variability using control limits based on the average sample standard deviation.
Standard deviation chart0.00.91.82.63.53.212.823.53342.95sample std dev

Figure 10 — schematic for Standard deviation control chart limits — solve for control chart factor B4 (case 2) — Standard Deviation Charts (10)

Step-by-step solution

  1. Step 1 — State the governing relation:

    UCLs=B4sˉUCL_s = B_4 \bar{s}
  2. Step 2 — Rearrange symbolically for B4:

    B4=UCLsˉB_{4} = \dfrac{UCL}{\bar{s}}
  3. Step 3 — List the givens: average sample std dev (sbar) = 5.4000, upper control limit (UCL) = 15.5300.

  4. Step 4 — Substitute the given values:

    B4=15.5300sˉB_{4} = \dfrac{15.5300}{\bar{s}}
  5. Step 5 — Evaluate:

    B4=2.8759B_{4} = 2.8759
  6. Step 6 — Check: returning B4 = 2.8759 to

    UCLs=B4sˉUCL_s = B_4 \bar{s}

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

Answer:
B4=2.8759B_{4} = 2.8759

Why the other options are there

  • 5.7519 — kept a factor of two that cancels in the correct rearrangement.
  • 1.4380 — dropped that same factor in the other direction.
  • 3.1635 — rounded an intermediate value before the final step.

Reference: FE Handbook — Standard Deviation Charts

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