An engineer computes the expected value of a warranty payout distribution. Given outcome 1 (x1) = 8.0000; probability 1 (p1) = 0.4800; outcome 2 (x2) = 14.5000, determine the expected value (EX).

Start with the thinking

• The governing relation printed in this handbook section is Expected value of a discrete random variable.
• Everything except EX is given, so isolate EX 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 expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E[X] = \sum x_i p_i$$

2. Step 2 — Rearrange symbolically for EX:

   $$EX = x_1 p_1 + x_2(1-p_1)$$

3. Step 3

   $$List the givens: outcome 1 (x_{1}) = 8.0000, probability 1 (p_{1}) = 0.4800, outcome 2 (x_{2}) = 14.5000$$

4. Step 4 — Substitute the given values:

   $$EX = x_1 p_1 + x_2(1-p_1)$$

5. Step 5 — Evaluate:

   $$EX = 11.3800$$

6. Step 6 — Check: returning EX = 11.3800 to

   $$E[X] = \sum x_i p_i$$

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

Why the other options are there

• 22.7600 — kept a factor of two that cancels in the correct rearrangement.
• 5.6900 — dropped that same factor in the other direction.
• 12.5180 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

A student finds the expected value of a two-outcome gamble. Given outcome 1 (x1) = 4.0000; probability 1 (p1) = 0.5000; expected value (EX) = 4.8800, determine the outcome 2 (x2).

Start with the thinking

• The governing relation printed in this handbook section is Expected value of a discrete random variable.
• Everything except x2 is given, so isolate x2 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 expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E[X] = \sum x_i p_i$$

2. Step 2 — Rearrange symbolically for x2:

   $$x_{2} = \dfrac{E[X] - x_1 p_1}{1-p_1}$$

3. Step 3

   $$List the givens: outcome 1 (x_{1}) = 4.0000, probability 1 (p_{1}) = 0.5000, expected value (EX) = 4.8800$$

4. Step 4 — Substitute the given values:

   $$x_{2} = \dfrac{E[X] - x_1 p_1}{1-p_1}$$

5. Step 5 — Evaluate:

   $$x_{2} = 5.7600$$

6. Step 6 — Check: returning x2 = 5.7600 to

   $$E[X] = \sum x_i p_i$$

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

Why the other options are there

• 11.5200 — kept a factor of two that cancels in the correct rearrangement.
• 2.8800 — dropped that same factor in the other direction.
• 6.3360 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

The expected value of a production yield is estimated from probabilities. Given outcome 1 (x1) = 5.0000; outcome 2 (x2) = 19.0000; expected value (EX) = 8.9600, determine the probability 1 (p1).

Start with the thinking

• The governing relation printed in this handbook section is Expected value of a discrete random variable.
• Everything except p1 is given, so isolate p1 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 expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E[X] = \sum x_i p_i$$

2. Step 2 — Rearrange symbolically for p1:

   $$p_{1} = \dfrac{x_2 - E[X]}{x_2 - x_1}$$

3. Step 3

   $$List the givens: outcome 1 (x_{1}) = 5.0000, outcome 2 (x_{2}) = 19.0000, expected value (EX) = 8.9600$$

4. Step 4 — Substitute the given values:

   $$p_{1} = \dfrac{x_2 - E[X]}{x_2 - x_1}$$

5. Step 5 — Evaluate:

   $$p_{1} = 0.7171$$

6. Step 6 — Check: returning p1 = 0.7171 to

   $$E[X] = \sum x_i p_i$$

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

Why the other options are there

• 1.4343 — kept a factor of two that cancels in the correct rearrangement.
• 0.3586 — dropped that same factor in the other direction.
• 0.7889 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

An engineer computes the expected value of a warranty payout distribution. Given outcome 1 (x1) = 7.0000; probability 1 (p1) = 0.3800; outcome 2 (x2) = 11.0000, determine the expected value (EX).

Start with the thinking

• The governing relation printed in this handbook section is Expected value of a discrete random variable.
• Everything except EX is given, so isolate EX 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 expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E[X] = \sum x_i p_i$$

2. Step 2 — Rearrange symbolically for EX:

   $$EX = x_1 p_1 + x_2(1-p_1)$$

3. Step 3

   $$List the givens: outcome 1 (x_{1}) = 7.0000, probability 1 (p_{1}) = 0.3800, outcome 2 (x_{2}) = 11.0000$$

4. Step 4 — Substitute the given values:

   $$EX = x_1 p_1 + x_2(1-p_1)$$

5. Step 5 — Evaluate:

   $$EX = 9.4800$$

6. Step 6 — Check: returning EX = 9.4800 to

   $$E[X] = \sum x_i p_i$$

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

Why the other options are there

• 18.9600 — kept a factor of two that cancels in the correct rearrangement.
• 4.7400 — dropped that same factor in the other direction.
• 10.4280 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

A student finds the expected value of a two-outcome gamble. Given outcome 1 (x1) = 10.0000; probability 1 (p1) = 0.4000; expected value (EX) = 8.3500, determine the outcome 2 (x2).

Start with the thinking

• The governing relation printed in this handbook section is Expected value of a discrete random variable.
• Everything except x2 is given, so isolate x2 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 expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E[X] = \sum x_i p_i$$

2. Step 2 — Rearrange symbolically for x2:

   $$x_{2} = \dfrac{E[X] - x_1 p_1}{1-p_1}$$

3. Step 3

   $$List the givens: outcome 1 (x_{1}) = 10.0000, probability 1 (p_{1}) = 0.4000, expected value (EX) = 8.3500$$

4. Step 4 — Substitute the given values:

   $$x_{2} = \dfrac{E[X] - x_1 p_1}{1-p_1}$$

5. Step 5 — Evaluate:

   $$x_{2} = 7.2500$$

6. Step 6 — Check: returning x2 = 7.2500 to

   $$E[X] = \sum x_i p_i$$

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

Why the other options are there

• 14.5000 — kept a factor of two that cancels in the correct rearrangement.
• 3.6250 — dropped that same factor in the other direction.
• 7.9750 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

The expected value of a production yield is estimated from probabilities. Given outcome 1 (x1) = 5.0000; outcome 2 (x2) = 17.5000; expected value (EX) = 2.8600, determine the probability 1 (p1).

Start with the thinking

• The governing relation printed in this handbook section is Expected value of a discrete random variable.
• Everything except p1 is given, so isolate p1 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 expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E[X] = \sum x_i p_i$$

2. Step 2 — Rearrange symbolically for p1:

   $$p_{1} = \dfrac{x_2 - E[X]}{x_2 - x_1}$$

3. Step 3

   $$List the givens: outcome 1 (x_{1}) = 5.0000, outcome 2 (x_{2}) = 17.5000, expected value (EX) = 2.8600$$

4. Step 4 — Substitute the given values:

   $$p_{1} = \dfrac{x_2 - E[X]}{x_2 - x_1}$$

5. Step 5 — Evaluate:

   $$p_{1} = 1.1712$$

6. Step 6 — Check: returning p1 = 1.1712 to

   $$E[X] = \sum x_i p_i$$

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

Why the other options are there

• 2.3424 — kept a factor of two that cancels in the correct rearrangement.
• 0.5856 — dropped that same factor in the other direction.
• 1.2883 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

An engineer computes the expected value of a warranty payout distribution. Given outcome 1 (x1) = 2.5000; probability 1 (p1) = 0.5500; outcome 2 (x2) = 12.0000, determine the expected value (EX).

Start with the thinking

• The governing relation printed in this handbook section is Expected value of a discrete random variable.
• Everything except EX is given, so isolate EX 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 expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E[X] = \sum x_i p_i$$

2. Step 2 — Rearrange symbolically for EX:

   $$EX = x_1 p_1 + x_2(1-p_1)$$

3. Step 3

   $$List the givens: outcome 1 (x_{1}) = 2.5000, probability 1 (p_{1}) = 0.5500, outcome 2 (x_{2}) = 12.0000$$

4. Step 4 — Substitute the given values:

   $$EX = x_1 p_1 + x_2(1-p_1)$$

5. Step 5 — Evaluate:

   $$EX = 6.7750$$

6. Step 6 — Check: returning EX = 6.7750 to

   $$E[X] = \sum x_i p_i$$

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

Why the other options are there

• 13.5500 — kept a factor of two that cancels in the correct rearrangement.
• 3.3875 — dropped that same factor in the other direction.
• 7.4525 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

A student finds the expected value of a two-outcome gamble. Given outcome 1 (x1) = 1.5000; probability 1 (p1) = 0.3000; expected value (EX) = 9.4800, determine the outcome 2 (x2).

Start with the thinking

• The governing relation printed in this handbook section is Expected value of a discrete random variable.
• Everything except x2 is given, so isolate x2 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 expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E[X] = \sum x_i p_i$$

2. Step 2 — Rearrange symbolically for x2:

   $$x_{2} = \dfrac{E[X] - x_1 p_1}{1-p_1}$$

3. Step 3

   $$List the givens: outcome 1 (x_{1}) = 1.5000, probability 1 (p_{1}) = 0.3000, expected value (EX) = 9.4800$$

4. Step 4 — Substitute the given values:

   $$x_{2} = \dfrac{E[X] - x_1 p_1}{1-p_1}$$

5. Step 5 — Evaluate:

   $$x_{2} = 12.9000$$

6. Step 6 — Check: returning x2 = 12.9000 to

   $$E[X] = \sum x_i p_i$$

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

Why the other options are there

• 25.8000 — kept a factor of two that cancels in the correct rearrangement.
• 6.4500 — dropped that same factor in the other direction.
• 14.1900 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

The expected value of a production yield is estimated from probabilities. Given outcome 1 (x1) = 2.0000; outcome 2 (x2) = 20.0000; expected value (EX) = 12.0500, determine the probability 1 (p1).

Start with the thinking

• The governing relation printed in this handbook section is Expected value of a discrete random variable.
• Everything except p1 is given, so isolate p1 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 expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E[X] = \sum x_i p_i$$

2. Step 2 — Rearrange symbolically for p1:

   $$p_{1} = \dfrac{x_2 - E[X]}{x_2 - x_1}$$

3. Step 3

   $$List the givens: outcome 1 (x_{1}) = 2.0000, outcome 2 (x_{2}) = 20.0000, expected value (EX) = 12.0500$$

4. Step 4 — Substitute the given values:

   $$p_{1} = \dfrac{x_2 - E[X]}{x_2 - x_1}$$

5. Step 5 — Evaluate:

   $$p_{1} = 0.4417$$

6. Step 6 — Check: returning p1 = 0.4417 to

   $$E[X] = \sum x_i p_i$$

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

Why the other options are there

• 0.8833 — kept a factor of two that cancels in the correct rearrangement.
• 0.2208 — dropped that same factor in the other direction.
• 0.4858 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values

An engineer computes the expected value of a warranty payout distribution. Given outcome 1 (x1) = 7.5000; probability 1 (p1) = 0.3800; outcome 2 (x2) = 18.5000, determine the expected value (EX).

Start with the thinking

• The governing relation printed in this handbook section is Expected value of a discrete random variable.
• Everything except EX is given, so isolate EX 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 expected value of a discrete random variable weights each outcome by its probability.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E[X] = \sum x_i p_i$$

2. Step 2 — Rearrange symbolically for EX:

   $$EX = x_1 p_1 + x_2(1-p_1)$$

3. Step 3

   $$List the givens: outcome 1 (x_{1}) = 7.5000, probability 1 (p_{1}) = 0.3800, outcome 2 (x_{2}) = 18.5000$$

4. Step 4 — Substitute the given values:

   $$EX = x_1 p_1 + x_2(1-p_1)$$

5. Step 5 — Evaluate:

   $$EX = 14.3200$$

6. Step 6 — Check: returning EX = 14.3200 to

   $$E[X] = \sum x_i p_i$$

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

Why the other options are there

• 28.6400 — kept a factor of two that cancels in the correct rearrangement.
• 7.1600 — dropped that same factor in the other direction.
• 15.7520 — rounded an intermediate value before the final step.

Reference: FE Handbook — Expected Values