A probability and statistics problem uses Binomial mean. Given trials (n) = 186.0; success probability (p) = 0.2900, 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

   $$List the givens: trials (n) = 186.0, success probability (p) = 0.2900$$

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

5. Step 5 — Evaluate:

   $$\mu = 53.9400$$

6. Step 6 — Check: returning mu = 53.9400 to

   $$\mu = n p$$

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

Why the other options are there

• 107.9 — kept a factor of two that cancels in the correct rearrangement.
• 26.9700 — dropped that same factor in the other direction.
• 59.3340 — rounded an intermediate value before the final step.

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

A student computes the mean and variance of a binomial distribution. Given number of trials (n) = 179.0; success probability (p) = 0.0700; variance (var) = 3.5700, 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) = 179.0, success probability (p) = 0.0700, variance (var) = 3.5700.

4. Step 4 — Substitute the given values:

   $$\mu = n0.0700$$

5. Step 5 — Evaluate:

   $$\mu = 12.5300$$

6. Step 6 — Check: returning mu = 12.5300 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

• 25.0600 — kept a factor of two that cancels in the correct rearrangement.
• 6.2650 — dropped that same factor in the other direction.
• 13.7830 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given success probability (p) = 0.0700; mean (mu) = 33.3000, 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

   $$List the givens: success probability (p) = 0.0700, mean (mu) = 33.3000$$

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

5. Step 5 — Evaluate:

   $$n = 475.7$$

6. Step 6 — Check: returning n = 475.7 to

   $$\mu = n p$$

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

Why the other options are there

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

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

The binomial distribution predicts the number of failures in a batch of tests. Given number of trials (n) = 67.0000; success probability (p) = 0.8300; mean (mu) = 138.2, 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) = 67.0000, success probability (p) = 0.8300, mean (mu) = 138.2.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$var = 9.4537$$

6. Step 6 — Check: returning var = 9.4537 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

• 18.9074 — kept a factor of two that cancels in the correct rearrangement.
• 4.7269 — dropped that same factor in the other direction.
• 10.3991 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given trials (n) = 156.0; mean (mu) = 33.6000, 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

   $$List the givens: trials (n) = 156.0, mean (mu) = 33.6000$$

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

5. Step 5 — Evaluate:

   $$p = 0.2154$$

6. Step 6 — Check: returning p = 0.2154 to

   $$\mu = n p$$

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

Why the other options are there

• 0.4308 — kept a factor of two that cancels in the correct rearrangement.
• 0.1077 — dropped that same factor in the other direction.
• 0.2369 — rounded an intermediate value before the final step.

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

An engineer models the binomial distribution of defective parts in a lot. Given success probability (p) = 0.8000; mean (mu) = 9.7800; variance (var) = 9.9000, 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

   $$List the givens: success probability (p) = 0.8000, mean (mu) = 9.7800, variance (var) = 9.9000$$

4. Step 4 — Substitute the given values:

   $$n = \dfrac{9.7800}{0.8000}$$

5. Step 5 — Evaluate:

   $$n = 12.2250$$

6. Step 6 — Check: returning n = 12.2250 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

• 24.4500 — kept a factor of two that cancels in the correct rearrangement.
• 6.1125 — dropped that same factor in the other direction.
• 13.4475 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given trials (n) = 128.0; success probability (p) = 0.4700, 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

   $$List the givens: trials (n) = 128.0, success probability (p) = 0.4700$$

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

5. Step 5 — Evaluate:

   $$\mu = 60.1600$$

6. Step 6 — Check: returning mu = 60.1600 to

   $$\mu = n p$$

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

Why the other options are there

• 120.3 — kept a factor of two that cancels in the correct rearrangement.
• 30.0800 — dropped that same factor in the other direction.
• 66.1760 — rounded an intermediate value before the final step.

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

A student computes the mean and variance of a binomial distribution. Given number of trials (n) = 50.0000; success probability (p) = 0.7700; variance (var) = 42.9500, 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) = 50.0000, success probability (p) = 0.7700, variance (var) = 42.9500.

4. Step 4 — Substitute the given values:

   $$\mu = n0.7700$$

5. Step 5 — Evaluate:

   $$\mu = 38.5000$$

6. Step 6 — Check: returning mu = 38.5000 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

• 77.0000 — kept a factor of two that cancels in the correct rearrangement.
• 19.2500 — dropped that same factor in the other direction.
• 42.3500 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given success probability (p) = 0.5900; mean (mu) = 6.5000, 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

   $$List the givens: success probability (p) = 0.5900, mean (mu) = 6.5000$$

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

5. Step 5 — Evaluate:

   $$n = 11.0169$$

6. Step 6 — Check: returning n = 11.0169 to

   $$\mu = n p$$

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

Why the other options are there

• 22.0339 — kept a factor of two that cancels in the correct rearrangement.
• 5.5085 — dropped that same factor in the other direction.
• 12.1186 — rounded an intermediate value before the final step.

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

The binomial distribution predicts the number of failures in a batch of tests. Given number of trials (n) = 38.0000; success probability (p) = 0.5800; mean (mu) = 129.8, 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) = 38.0000, success probability (p) = 0.5800, mean (mu) = 129.8.

4. Step 4 — Substitute the given values:

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

5. Step 5 — Evaluate:

   $$var = 9.2568$$

6. Step 6 — Check: returning var = 9.2568 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

• 18.5136 — kept a factor of two that cancels in the correct rearrangement.
• 4.6284 — dropped that same factor in the other direction.
• 10.1825 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial