sampling and monitoring plan for a groundwater monitoring well network Given standard normal deviate (z) = 2.2400; standard deviation of measurements (sigma) = 26.0000 mg/L; allowable error (E) = 6.4000 mg/L, determine the required number of samples (n).

Start with the thinking

• The governing relation printed in this handbook section is Sampling and Monitoring.
• Everything except n is given, so isolate n 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.
• Sampling and monitoring programs use statistical sample size equations to determine the number of samples needed for a target error.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \left(\dfrac{z\sigma}{E}\right)^2$$

3. Step 3 — List the givens: standard normal deviate (z) = 2.2400, standard deviation of measurements (sigma) = 26.0000 mg/L, allowable error (E) = 6.4000 mg/L.

4. Step 4 — Substitute the given values:

   $$n = \left(\dfrac{2.240026.0000}{6.4000}\right)^2$$

5. Step 5 — Evaluate:

   $$n = 82.8100$$

6. Step 6 — Check: returning n = 82.8100 to

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

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

Why the other options are there

• 165.6 — kept a factor of two that cancels in the correct rearrangement.
• 41.4050 — dropped that same factor in the other direction.
• 91.0910 — rounded an intermediate value before the final step.

Reference: FE Handbook — Sampling and Monitoring

sampling and monitoring number of samples required for compliance testing Given standard normal deviate (z) = 1.4300; allowable error (E) = 4.7000 mg/L; required number of samples (n) = 348.0, determine the standard deviation of measurements (sigma) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Sampling and Monitoring.
• 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.
• Sampling and monitoring programs use statistical sample size equations to determine the number of samples needed for a target error.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

2. Step 2 — Rearrange symbolically for sigma:

   $$\sigma = \dfrac{E\sqrt{n}}{z}$$

3. Step 3 — List the givens: standard normal deviate (z) = 1.4300, allowable error (E) = 4.7000 mg/L, required number of samples (n) = 348.0.

4. Step 4 — Substitute the given values:

   $$\sigma = \dfrac{4.7000\sqrt{348.0}}{1.4300}$$

5. Step 5 — Evaluate:

   $$\sigma = 61.3128\ \text{mg/L}$$

6. Step 6 — Check: returning sigma = 61.3128 mg/L to

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

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

Why the other options are there

• 122.6 — kept a factor of two that cancels in the correct rearrangement.
• 30.6564 — dropped that same factor in the other direction.
• 67.4441 — rounded an intermediate value before the final step.

Reference: FE Handbook — Sampling and Monitoring

environmental sampling and monitoring program sample size determination Given standard normal deviate (z) = 1.4100; standard deviation of measurements (sigma) = 24.1000 mg/L; required number of samples (n) = 123.0, determine the allowable error (E) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Sampling and Monitoring.
• Everything except E is given, so isolate E 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.
• Sampling and monitoring programs use statistical sample size equations to determine the number of samples needed for a target error.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \dfrac{z\sigma}{\sqrt{n}}$$

3. Step 3 — List the givens: standard normal deviate (z) = 1.4100, standard deviation of measurements (sigma) = 24.1000 mg/L, required number of samples (n) = 123.0.

4. Step 4 — Substitute the given values:

   $$E = \dfrac{1.410024.1000}{\sqrt{123.0}}$$

5. Step 5 — Evaluate:

   $$E = 3.0640\ \text{mg/L}$$

6. Step 6 — Check: returning E = 3.0640 mg/L to

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

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

Why the other options are there

• 6.1279 — kept a factor of two that cancels in the correct rearrangement.
• 1.5320 — dropped that same factor in the other direction.
• 3.3704 — rounded an intermediate value before the final step.

Reference: FE Handbook — Sampling and Monitoring

sampling and monitoring plan for a groundwater monitoring well network Given standard normal deviate (z) = 2.5500; standard deviation of measurements (sigma) = 23.0000 mg/L; allowable error (E) = 5.8000 mg/L, determine the required number of samples (n).

Start with the thinking

• The governing relation printed in this handbook section is Sampling and Monitoring.
• Everything except n is given, so isolate n 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.
• Sampling and monitoring programs use statistical sample size equations to determine the number of samples needed for a target error.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \left(\dfrac{z\sigma}{E}\right)^2$$

3. Step 3 — List the givens: standard normal deviate (z) = 2.5500, standard deviation of measurements (sigma) = 23.0000 mg/L, allowable error (E) = 5.8000 mg/L.

4. Step 4 — Substitute the given values:

   $$n = \left(\dfrac{2.550023.0000}{5.8000}\right)^2$$

5. Step 5 — Evaluate:

   $$n = 102.3$$

6. Step 6 — Check: returning n = 102.3 to

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

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

Why the other options are there

• 204.5 — kept a factor of two that cancels in the correct rearrangement.
• 51.1270 — dropped that same factor in the other direction.
• 112.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Sampling and Monitoring

sampling and monitoring number of samples required for compliance testing Given standard normal deviate (z) = 2.0400; allowable error (E) = 7.6000 mg/L; required number of samples (n) = 3.0000, determine the standard deviation of measurements (sigma) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Sampling and Monitoring.
• 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.
• Sampling and monitoring programs use statistical sample size equations to determine the number of samples needed for a target error.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

2. Step 2 — Rearrange symbolically for sigma:

   $$\sigma = \dfrac{E\sqrt{n}}{z}$$

3. Step 3 — List the givens: standard normal deviate (z) = 2.0400, allowable error (E) = 7.6000 mg/L, required number of samples (n) = 3.0000.

4. Step 4 — Substitute the given values:

   $$\sigma = \dfrac{7.6000\sqrt{3.0000}}{2.0400}$$

5. Step 5 — Evaluate:

   $$\sigma = 6.4527\ \text{mg/L}$$

6. Step 6 — Check: returning sigma = 6.4527 mg/L to

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

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

Why the other options are there

• 12.9055 — kept a factor of two that cancels in the correct rearrangement.
• 3.2264 — dropped that same factor in the other direction.
• 7.0980 — rounded an intermediate value before the final step.

Reference: FE Handbook — Sampling and Monitoring

environmental sampling and monitoring program sample size determination Given standard normal deviate (z) = 1.8900; standard deviation of measurements (sigma) = 13.5000 mg/L; required number of samples (n) = 170.0, determine the allowable error (E) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Sampling and Monitoring.
• Everything except E is given, so isolate E 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.
• Sampling and monitoring programs use statistical sample size equations to determine the number of samples needed for a target error.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \dfrac{z\sigma}{\sqrt{n}}$$

3. Step 3 — List the givens: standard normal deviate (z) = 1.8900, standard deviation of measurements (sigma) = 13.5000 mg/L, required number of samples (n) = 170.0.

4. Step 4 — Substitute the given values:

   $$E = \dfrac{1.890013.5000}{\sqrt{170.0}}$$

5. Step 5 — Evaluate:

   $$E = 1.9569\ \text{mg/L}$$

6. Step 6 — Check: returning E = 1.9569 mg/L to

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

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

Why the other options are there

• 3.9138 — kept a factor of two that cancels in the correct rearrangement.
• 0.9785 — dropped that same factor in the other direction.
• 2.1526 — rounded an intermediate value before the final step.

Reference: FE Handbook — Sampling and Monitoring

sampling and monitoring plan for a groundwater monitoring well network Given standard normal deviate (z) = 1.6000; standard deviation of measurements (sigma) = 5.0000 mg/L; allowable error (E) = 8.8000 mg/L, determine the required number of samples (n).

Start with the thinking

• The governing relation printed in this handbook section is Sampling and Monitoring.
• Everything except n is given, so isolate n 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.
• Sampling and monitoring programs use statistical sample size equations to determine the number of samples needed for a target error.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \left(\dfrac{z\sigma}{E}\right)^2$$

3. Step 3 — List the givens: standard normal deviate (z) = 1.6000, standard deviation of measurements (sigma) = 5.0000 mg/L, allowable error (E) = 8.8000 mg/L.

4. Step 4 — Substitute the given values:

   $$n = \left(\dfrac{1.60005.0000}{8.8000}\right)^2$$

5. Step 5 — Evaluate:

   $$n = 0.8264$$

6. Step 6 — Check: returning n = 0.8264 to

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

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

Why the other options are there

• 1.6529 — kept a factor of two that cancels in the correct rearrangement.
• 0.4132 — dropped that same factor in the other direction.
• 0.9091 — rounded an intermediate value before the final step.

Reference: FE Handbook — Sampling and Monitoring

sampling and monitoring number of samples required for compliance testing Given standard normal deviate (z) = 2.0200; allowable error (E) = 1.2000 mg/L; required number of samples (n) = 332.0, determine the standard deviation of measurements (sigma) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Sampling and Monitoring.
• 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.
• Sampling and monitoring programs use statistical sample size equations to determine the number of samples needed for a target error.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

2. Step 2 — Rearrange symbolically for sigma:

   $$\sigma = \dfrac{E\sqrt{n}}{z}$$

3. Step 3 — List the givens: standard normal deviate (z) = 2.0200, allowable error (E) = 1.2000 mg/L, required number of samples (n) = 332.0.

4. Step 4 — Substitute the given values:

   $$\sigma = \dfrac{1.2000\sqrt{332.0}}{2.0200}$$

5. Step 5 — Evaluate:

   $$\sigma = 10.8243\ \text{mg/L}$$

6. Step 6 — Check: returning sigma = 10.8243 mg/L to

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

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

Why the other options are there

• 21.6486 — kept a factor of two that cancels in the correct rearrangement.
• 5.4121 — dropped that same factor in the other direction.
• 11.9067 — rounded an intermediate value before the final step.

Reference: FE Handbook — Sampling and Monitoring

environmental sampling and monitoring program sample size determination Given standard normal deviate (z) = 1.3600; standard deviation of measurements (sigma) = 22.9000 mg/L; required number of samples (n) = 169.0, determine the allowable error (E) in mg/L.

Start with the thinking

• The governing relation printed in this handbook section is Sampling and Monitoring.
• Everything except E is given, so isolate E 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.
• Sampling and monitoring programs use statistical sample size equations to determine the number of samples needed for a target error.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \dfrac{z\sigma}{\sqrt{n}}$$

3. Step 3 — List the givens: standard normal deviate (z) = 1.3600, standard deviation of measurements (sigma) = 22.9000 mg/L, required number of samples (n) = 169.0.

4. Step 4 — Substitute the given values:

   $$E = \dfrac{1.360022.9000}{\sqrt{169.0}}$$

5. Step 5 — Evaluate:

   $$E = 2.3957\ \text{mg/L}$$

6. Step 6 — Check: returning E = 2.3957 mg/L to

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

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

Why the other options are there

• 4.7914 — kept a factor of two that cancels in the correct rearrangement.
• 1.1978 — dropped that same factor in the other direction.
• 2.6353 — rounded an intermediate value before the final step.

Reference: FE Handbook — Sampling and Monitoring

sampling and monitoring plan for a groundwater monitoring well network Given standard normal deviate (z) = 2.2000; standard deviation of measurements (sigma) = 21.7000 mg/L; allowable error (E) = 2.4000 mg/L, determine the required number of samples (n).

Start with the thinking

• The governing relation printed in this handbook section is Sampling and Monitoring.
• Everything except n is given, so isolate n 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.
• Sampling and monitoring programs use statistical sample size equations to determine the number of samples needed for a target error.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \left(\dfrac{z\sigma}{E}\right)^2$$

3. Step 3 — List the givens: standard normal deviate (z) = 2.2000, standard deviation of measurements (sigma) = 21.7000 mg/L, allowable error (E) = 2.4000 mg/L.

4. Step 4 — Substitute the given values:

   $$n = \left(\dfrac{2.200021.7000}{2.4000}\right)^2$$

5. Step 5 — Evaluate:

   $$n = 395.7$$

6. Step 6 — Check: returning n = 395.7 to

   $$n = \left(\dfrac{z \sigma}{E}\right)^2$$

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

Why the other options are there

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

Reference: FE Handbook — Sampling and Monitoring