A probability and statistics problem uses Binomial mean. Given trials (n) = 199.0; success probability (p) = 0.4600, 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

   $$List the givens: trials (n) = 199.0, success probability (p) = 0.4600$$

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

5. Step 5 — Evaluate:

   $$\mu = 91.5400$$

6. Step 6 — Check: returning mu = 91.5400 to

   $$\mu = n p$$

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

Why the other options are there

• 183.1 — kept a factor of two that cancels in the correct rearrangement.
• 45.7700 — dropped that same factor in the other direction.
• 100.7 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Binomial

A student computes the mean and variance of a binomial distribution. Given number of trials (n) = 99.0000; success probability (p) = 0.8800; variance (var) = 38.7000, 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) = 99.0000, success probability (p) = 0.8800, variance (var) = 38.7000.

4. Step 4 — Substitute the given values:

   $$\mu = n0.8800$$

5. Step 5 — Evaluate:

   $$\mu = 87.1200$$

6. Step 6 — Check: returning mu = 87.1200 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

• 174.2 — kept a factor of two that cancels in the correct rearrangement.
• 43.5600 — dropped that same factor in the other direction.
• 95.8320 — 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) = 94.0000, 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

   $$List the givens: success probability (p) = 0.6900, mean (mu) = 94.0000$$

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

5. Step 5 — Evaluate:

   $$n = 136.2$$

6. Step 6 — Check: returning n = 136.2 to

   $$\mu = n p$$

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

Why the other options are there

• 272.5 — kept a factor of two that cancels in the correct rearrangement.
• 68.1159 — dropped that same factor in the other direction.
• 149.9 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Binomial

The binomial distribution predicts the number of failures in a batch of tests. Given number of trials (n) = 147.0; success probability (p) = 0.8300; mean (mu) = 140.4, 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) = 147.0, success probability (p) = 0.8300, mean (mu) = 140.4.

4. Step 4 — Substitute the given values:

   $$var = n0.8300(1-0.8300)$$

5. Step 5 — Evaluate:

   $$var = 20.7417$$

6. Step 6 — Check: returning var = 20.7417 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

• 41.4834 — kept a factor of two that cancels in the correct rearrangement.
• 10.3709 — dropped that same factor in the other direction.
• 22.8159 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given trials (n) = 100.0; mean (mu) = 7.4000, 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

   $$List the givens: trials (n) = 100.0, mean (mu) = 7.4000$$

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

5. Step 5 — Evaluate:

   $$p = 0.0740$$

6. Step 6 — Check: returning p = 0.0740 to

   $$\mu = n p$$

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

Why the other options are there

• 0.1480 — kept a factor of two that cancels in the correct rearrangement.
• 0.0370 — dropped that same factor in the other direction.
• 0.0814 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Binomial

An engineer models the binomial distribution of defective parts in a lot. Given success probability (p) = 0.1900; mean (mu) = 141.9; variance (var) = 37.1100, 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

   $$List the givens: success probability (p) = 0.1900, mean (mu) = 141.9, variance (var) = 37.1100$$

4. Step 4 — Substitute the given values:

   $$n = \dfrac{141.9}{0.1900}$$

5. Step 5 — Evaluate:

   $$n = 746.8$$

6. Step 6 — Check: returning n = 746.8 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

• 1,494 — kept a factor of two that cancels in the correct rearrangement.
• 373.4 — dropped that same factor in the other direction.
• 821.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given trials (n) = 184.0; success probability (p) = 0.1400, 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

   $$List the givens: trials (n) = 184.0, success probability (p) = 0.1400$$

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

5. Step 5 — Evaluate:

   $$\mu = 25.7600$$

6. Step 6 — Check: returning mu = 25.7600 to

   $$\mu = n p$$

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

Why the other options are there

• 51.5200 — kept a factor of two that cancels in the correct rearrangement.
• 12.8800 — dropped that same factor in the other direction.
• 28.3360 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Binomial

A student computes the mean and variance of a binomial distribution. Given number of trials (n) = 31.0000; success probability (p) = 0.6900; variance (var) = 7.2500, 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) = 31.0000, success probability (p) = 0.6900, variance (var) = 7.2500.

4. Step 4 — Substitute the given values:

   $$\mu = n0.6900$$

5. Step 5 — Evaluate:

   $$\mu = 21.3900$$

6. Step 6 — Check: returning mu = 21.3900 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

• 42.7800 — kept a factor of two that cancels in the correct rearrangement.
• 10.6950 — dropped that same factor in the other direction.
• 23.5290 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given success probability (p) = 0.1200; mean (mu) = 32.7000, 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

   $$List the givens: success probability (p) = 0.1200, mean (mu) = 32.7000$$

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

5. Step 5 — Evaluate:

   $$n = 272.5$$

6. Step 6 — Check: returning n = 272.5 to

   $$\mu = n p$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Probability and Statistics → Binomial

The binomial distribution predicts the number of failures in a batch of tests. Given number of trials (n) = 139.0; success probability (p) = 0.5800; mean (mu) = 20.8600, 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) = 139.0, success probability (p) = 0.5800, mean (mu) = 20.8600.

4. Step 4 — Substitute the given values:

   $$var = n0.5800(1-0.5800)$$

5. Step 5 — Evaluate:

   $$var = 33.8604$$

6. Step 6 — Check: returning var = 33.8604 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

• 67.7208 — kept a factor of two that cancels in the correct rearrangement.
• 16.9302 — dropped that same factor in the other direction.
• 37.2464 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial