Each weld passes inspection with probability 0.90 independently. In a lot of 8 welds, what is the probability exactly 2 fail?

Start with the thinking

• Define success as the event you are counting — here, failure.
• Use the binomial pmf, not the normal approximation, for n = 8.

Step-by-step solution

1. Binomial pmf

   $$P(X = k) = C(n,k) p^k (1 - p)^(n-k)$$

2. Combinations

   $$C(8,2) = 8!/(2!6!) = 28$$

3. Probability terms

   $$(0.10)^{2} = 0.0100 and (0.90)^{6} = 0.5314$$

4. Substitute

   $$P = 28(0.0100)(0.5314)$$

5. Evaluate

   $$P = 0.149$$

Why the other options are there

• 0.0100 (combinations omitted)
• 0.383 (P(X ≤ 2) computed)

Reference: FE Reference Handbook — Probability and Statistics — Binomial distribution

A probability and statistics problem uses Binomial mean. Given trials (n) = 124.0; success probability (p) = 0.7000, 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

   $$List the givens: trials (n) = 124.0, success probability (p) = 0.7000$$

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

5. Step 5 — Evaluate:

   $$\mu = 86.8000$$

6. Step 6 — Check: returning mu = 86.8000 to

   $$\mu = n p$$

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

Why the other options are there

• 173.6 — kept a factor of two that cancels in the correct rearrangement.
• 43.4000 — dropped that same factor in the other direction.
• 95.4800 — 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) = 115.0; success probability (p) = 0.1000; variance (var) = 13.6100, 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) = 115.0, success probability (p) = 0.1000, variance (var) = 13.6100.

4. Step 4 — Substitute the given values:

   $$\mu = n0.1000$$

5. Step 5 — Evaluate:

   $$\mu = 11.5000$$

6. Step 6 — Check: returning mu = 11.5000 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

• 23.0000 — kept a factor of two that cancels in the correct rearrangement.
• 5.7500 — dropped that same factor in the other direction.
• 12.6500 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given success probability (p) = 0.5100; mean (mu) = 49.2000, 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

   $$List the givens: success probability (p) = 0.5100, mean (mu) = 49.2000$$

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

5. Step 5 — Evaluate:

   $$n = 96.4706$$

6. Step 6 — Check: returning n = 96.4706 to

   $$\mu = n p$$

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

Why the other options are there

• 192.9 — kept a factor of two that cancels in the correct rearrangement.
• 48.2353 — dropped that same factor in the other direction.
• 106.1 — 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) = 179.0; success probability (p) = 0.8000; mean (mu) = 62.5700, 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) = 179.0, success probability (p) = 0.8000, mean (mu) = 62.5700.

4. Step 4 — Substitute the given values:

   $$var = n0.8000(1-0.8000)$$

5. Step 5 — Evaluate:

   $$var = 28.6400$$

6. Step 6 — Check: returning var = 28.6400 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

• 57.2800 — kept a factor of two that cancels in the correct rearrangement.
• 14.3200 — dropped that same factor in the other direction.
• 31.5040 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given trials (n) = 185.0; 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

   $$List the givens: trials (n) = 185.0, mean (mu) = 33.6000$$

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

5. Step 5 — Evaluate:

   $$p = 0.1816$$

6. Step 6 — Check: returning p = 0.1816 to

   $$\mu = n p$$

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

Why the other options are there

• 0.3632 — kept a factor of two that cancels in the correct rearrangement.
• 0.0908 — dropped that same factor in the other direction.
• 0.1998 — 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) = 0.2200; mean (mu) = 174.0; variance (var) = 36.3000, 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

   $$List the givens: success probability (p) = 0.2200, mean (mu) = 174.0, variance (var) = 36.3000$$

4. Step 4 — Substitute the given values:

   $$n = \dfrac{174.0}{0.2200}$$

5. Step 5 — Evaluate:

   $$n = 790.8$$

6. Step 6 — Check: returning n = 790.8 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

• 1,582 — kept a factor of two that cancels in the correct rearrangement.
• 395.4 — dropped that same factor in the other direction.
• 869.9 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given trials (n) = 165.0; success probability (p) = 0.3700, 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

   $$List the givens: trials (n) = 165.0, success probability (p) = 0.3700$$

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

5. Step 5 — Evaluate:

   $$\mu = 61.0500$$

6. Step 6 — Check: returning mu = 61.0500 to

   $$\mu = n p$$

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

Why the other options are there

• 122.1 — kept a factor of two that cancels in the correct rearrangement.
• 30.5250 — dropped that same factor in the other direction.
• 67.1550 — 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) = 49.0000; success probability (p) = 0.2500; variance (var) = 20.5200, 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) = 49.0000, success probability (p) = 0.2500, variance (var) = 20.5200.

4. Step 4 — Substitute the given values:

   $$\mu = n0.2500$$

5. Step 5 — Evaluate:

   $$\mu = 12.2500$$

6. Step 6 — Check: returning mu = 12.2500 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

• 24.5000 — kept a factor of two that cancels in the correct rearrangement.
• 6.1250 — dropped that same factor in the other direction.
• 13.4750 — rounded an intermediate value before the final step.

Reference: FE Handbook — Binomial

A probability and statistics problem uses Binomial mean. Given success probability (p) = 0.8600; mean (mu) = 48.5000, 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

   $$List the givens: success probability (p) = 0.8600, mean (mu) = 48.5000$$

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

5. Step 5 — Evaluate:

   $$n = 56.3953$$

6. Step 6 — Check: returning n = 56.3953 to

   $$\mu = n p$$

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

Why the other options are there

• 112.8 — kept a factor of two that cancels in the correct rearrangement.
• 28.1977 — dropped that same factor in the other direction.
• 62.0349 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Probability and Statistics → Binomial