A probability and statistics problem uses Binomial mean. Given trials (n) = 31.0000; success probability (p) = 0.1100, determine the mean (mu).

Start with the thinking

• The governing relation printed in this handbook section is Binomial mean.
• 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:

   $$\mu = n p$$

2. Step 2 — Rearrange the relation so that mu stands alone on the left-hand side.

3. Step 3

   $$List the givens: trials (n) = 31.0000, success probability (p) = 0.1100$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\mu = 3.4100$$

6. Step 6 — Check: returning mu = 3.4100 to

   $$\mu = n p$$

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

Why the other options are there

• 6.8200 — kept a factor of two that cancels in the correct rearrangement.
• 1.7050 — dropped that same factor in the other direction.
• 3.7510 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Binomial Probabilities P(X ≤ x)

A student computes the mean and variance of a binomial distribution. Given number of trials (n) = 125.0; success probability (p) = 0.6900; variance (var) = 23.1600, determine the mean (mu).

Start with the thinking

• The governing relation printed in this handbook section is Binomial distribution mean and variance.
• 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.
• The binomial distribution describes the number of successes in n independent trials with probability p.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\mu = n p, \quad \sigma^2 = n p (1-p)$$

2. Step 2 — Rearrange symbolically for mu:

   $$\mu = np$$

3. Step 3 — List the givens: number of trials (n) = 125.0, success probability (p) = 0.6900, variance (var) = 23.1600.

4. Step 4 — Substitute the given values:

   $$\mu = n0.6900$$

5. Step 5 — Evaluate:

   $$\mu = 86.2500$$

6. Step 6 — Check: returning mu = 86.2500 to

   $$\mu = n p, \quad \sigma^2 = n p (1-p)$$

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

Why the other options are there

• 172.5 — kept a factor of two that cancels in the correct rearrangement.
• 43.1250 — dropped that same factor in the other direction.
• 94.8750 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given success probability (p) = 0.6900; mean (mu) = 87.0000, determine the trials (n).

Start with the thinking

• The governing relation printed in this handbook section is Binomial mean.
• 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.
• Probability and Statistics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\mu = n p$$

2. Step 2 — Rearrange the relation so that n stands alone on the left-hand side.

3. Step 3

   $$List the givens: success probability (p) = 0.6900, mean (mu) = 87.0000$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$n = 126.1$$

6. Step 6 — Check: returning n = 126.1 to

   $$\mu = n p$$

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

Why the other options are there

• 252.2 — kept a factor of two that cancels in the correct rearrangement.
• 63.0435 — dropped that same factor in the other direction.
• 138.7 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Binomial Probabilities P(X ≤ x)

The binomial distribution predicts the number of failures in a batch of tests. Given number of trials (n) = 68.0000; success probability (p) = 0.3300; mean (mu) = 124.0, determine the variance (var).

Start with the thinking

• The governing relation printed in this handbook section is Binomial distribution mean and variance.
• Everything except var is given, so isolate var 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 binomial distribution describes the number of successes in n independent trials with probability p.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\mu = n p, \quad \sigma^2 = n p (1-p)$$

2. Step 2 — Rearrange symbolically for var:

   $$var = np(1-p)$$

3. Step 3 — List the givens: number of trials (n) = 68.0000, success probability (p) = 0.3300, mean (mu) = 124.0.

4. Step 4 — Substitute the given values:

   $$var = n0.3300(1-0.3300)$$

5. Step 5 — Evaluate:

   $$var = 15.0348$$

6. Step 6 — Check: returning var = 15.0348 to

   $$\mu = n p, \quad \sigma^2 = n p (1-p)$$

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

Why the other options are there

• 30.0696 — kept a factor of two that cancels in the correct rearrangement.
• 7.5174 — dropped that same factor in the other direction.
• 16.5383 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given trials (n) = 96.0000; mean (mu) = 94.9000, determine the success probability (p).

Start with the thinking

• The governing relation printed in this handbook section is Binomial mean.
• Everything except p is given, so isolate p 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:

   $$\mu = n p$$

2. Step 2 — Rearrange the relation so that p stands alone on the left-hand side.

3. Step 3

   $$List the givens: trials (n) = 96.0000, mean (mu) = 94.9000$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$p = 0.9885$$

6. Step 6 — Check: returning p = 0.9885 to

   $$\mu = n p$$

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

Why the other options are there

• 1.9771 — kept a factor of two that cancels in the correct rearrangement.
• 0.4943 — dropped that same factor in the other direction.
• 1.0874 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Binomial Probabilities P(X ≤ x)

An engineer models the binomial distribution of defective parts in a lot. Given success probability (p) = 0.5600; mean (mu) = 170.7; variance (var) = 25.8300, determine the number of trials (n).

Start with the thinking

• The governing relation printed in this handbook section is Binomial distribution mean and variance.
• 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.
• The binomial distribution describes the number of successes in n independent trials with probability p.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\mu = n p, \quad \sigma^2 = n p (1-p)$$

2. Step 2 — Rearrange symbolically for n:

   $$n = \dfrac{\mu}{p}$$

3. Step 3

   $$List the givens: success probability (p) = 0.5600, mean (mu) = 170.7, variance (var) = 25.8300$$

4. Step 4 — Substitute the given values:

   $$n = \dfrac{170.7}{0.5600}$$

5. Step 5 — Evaluate:

   $$n = 304.9$$

6. Step 6 — Check: returning n = 304.9 to

   $$\mu = n p, \quad \sigma^2 = n p (1-p)$$

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

Why the other options are there

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

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given trials (n) = 189.0; success probability (p) = 0.5600, determine the mean (mu).

Start with the thinking

• The governing relation printed in this handbook section is Binomial mean.
• 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:

   $$\mu = n p$$

2. Step 2 — Rearrange the relation so that mu stands alone on the left-hand side.

3. Step 3

   $$List the givens: trials (n) = 189.0, success probability (p) = 0.5600$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\mu = 105.8$$

6. Step 6 — Check: returning mu = 105.8 to

   $$\mu = n p$$

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

Why the other options are there

• 211.7 — kept a factor of two that cancels in the correct rearrangement.
• 52.9200 — dropped that same factor in the other direction.
• 116.4 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Binomial Probabilities P(X ≤ x)

A student computes the mean and variance of a binomial distribution. Given number of trials (n) = 151.0; success probability (p) = 0.0800; variance (var) = 20.6700, determine the mean (mu).

Start with the thinking

• The governing relation printed in this handbook section is Binomial distribution mean and variance.
• 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.
• The binomial distribution describes the number of successes in n independent trials with probability p.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\mu = n p, \quad \sigma^2 = n p (1-p)$$

2. Step 2 — Rearrange symbolically for mu:

   $$\mu = np$$

3. Step 3 — List the givens: number of trials (n) = 151.0, success probability (p) = 0.0800, variance (var) = 20.6700.

4. Step 4 — Substitute the given values:

   $$\mu = n0.0800$$

5. Step 5 — Evaluate:

   $$\mu = 12.0800$$

6. Step 6 — Check: returning mu = 12.0800 to

   $$\mu = n p, \quad \sigma^2 = n p (1-p)$$

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

Why the other options are there

• 24.1600 — kept a factor of two that cancels in the correct rearrangement.
• 6.0400 — dropped that same factor in the other direction.
• 13.2880 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given success probability (p) = 0.8400; mean (mu) = 8.8000, determine the trials (n).

Start with the thinking

• The governing relation printed in this handbook section is Binomial mean.
• 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.
• Probability and Statistics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\mu = n p$$

2. Step 2 — Rearrange the relation so that n stands alone on the left-hand side.

3. Step 3

   $$List the givens: success probability (p) = 0.8400, mean (mu) = 8.8000$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$n = 10.4762$$

6. Step 6 — Check: returning n = 10.4762 to

   $$\mu = n p$$

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

Why the other options are there

• 20.9524 — kept a factor of two that cancels in the correct rearrangement.
• 5.2381 — dropped that same factor in the other direction.
• 11.5238 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Cumulative Binomial Probabilities P(X ≤ x)

The binomial distribution predicts the number of failures in a batch of tests. Given number of trials (n) = 162.0; success probability (p) = 0.8200; mean (mu) = 72.1900, determine the variance (var).

Start with the thinking

• The governing relation printed in this handbook section is Binomial distribution mean and variance.
• Everything except var is given, so isolate var 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 binomial distribution describes the number of successes in n independent trials with probability p.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$\mu = n p, \quad \sigma^2 = n p (1-p)$$

2. Step 2 — Rearrange symbolically for var:

   $$var = np(1-p)$$

3. Step 3 — List the givens: number of trials (n) = 162.0, success probability (p) = 0.8200, mean (mu) = 72.1900.

4. Step 4 — Substitute the given values:

   $$var = n0.8200(1-0.8200)$$

5. Step 5 — Evaluate:

   $$var = 23.9112$$

6. Step 6 — Check: returning var = 23.9112 to

   $$\mu = n p, \quad \sigma^2 = n p (1-p)$$

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

Why the other options are there

• 47.8224 — kept a factor of two that cancels in the correct rearrangement.
• 11.9556 — dropped that same factor in the other direction.
• 26.3023 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial