Expected Value of a Random Variable: Formula and Interpretation
Compute expected value as a probability-weighted mean and interpret it as a long-run average rather than a guaranteed single outcome.
Expected Value Random Variable: direct answer
Expected value random variable calculations use E(X)=ΣxP(X=x), a probability-weighted mean. The result describes a long-run average across repetitions and can differ from every single possible outcome.
This page keeps the lesson centered on expected value random variable. Practice is included only after the method, assumptions, interpretation, and common decision points are explained.
Quick reference: Expected Value of a Random Variable: Formula and Interpretation
| Expected value | E(X)=ΣxP(x) |
|---|---|
| Variance | Var(X)=Σ(x−μ)2P(x) |
Concept mastery: Expected Value of a Random Variable: Formula and Interpretation
Expected value is a probability-weighted mean
For a discrete random variable X, μₓ=E(X)=ΣxP(X=x). Each possible value is weighted by how often it occurs in the long run. The ordinary unweighted average of the listed outcomes is correct only when those outcomes are equally likely.
The expected value can be an impossible single outcome
A game may have expected profit $1.35 even though no play can produce exactly $1.35. Expectation summarizes the long-run average across many repetitions; it is not a prediction that the next outcome will equal μ.
Variance weights squared deviations from the mean
Var(X)=Σ(x−μ)²P(X=x), and SD(X)=√Var(X). Squaring keeps positive and negative deviations from canceling and emphasizes larger departures. Standard deviation returns to the original units of X, which makes it easier to interpret than variance.
Linear transformations change center and spread predictably
If Y=a+bX, then E(Y)=a+bE(X) and SD(Y)=|b|SD(X). Adding a constant shifts every outcome without changing spread; multiplying by b scales all distances from the mean by |b|.
Fair-game language refers to expected net value
A game is fair to a player when expected net gain is zero under the specified probabilities and payoffs. A positive expected value favors the player in the long run; a negative value represents a long-run expected loss. Risk still matters because games with the same expectation can have very different variability.
Expected value supports decisions but does not erase context
Insurance, warranties, queues, and repeated investments often use expected values, but a decision can also depend on variability, constraints, risk tolerance, and one-time consequences. AP questions usually ask for the statistical expectation and its interpretation, not an economic recommendation beyond the information supplied.
12 worked expected value random variable cases
Worked case 1: Lottery ticket
Scenario. Net gains are −$2 with probability .98 and $98 with probability .02.
Reasoning. E(X)=−2(.98)+98(.02)=0, so the game is fair in expected net value despite highly variable outcomes.
Case 1: Lottery ticket check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 1; do not replace it with a memorized generic sentence.
Worked case 2: Warranty cost
Scenario. A repair costs $300 with probability .05 and $0 otherwise.
Reasoning. Expected repair cost is 300(.05)= $15 per contract in the long run.
Case 2: Warranty cost check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 2; do not replace it with a memorized generic sentence.
Worked case 3: Delivery count
Scenario. A courier makes 2, 3, or 4 deliveries with probabilities .2, .5, .3.
Reasoning. E(X)=2(.2)+3(.5)+4(.3)=3.1 deliveries.
Case 3: Delivery count check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 3; do not replace it with a memorized generic sentence.
Worked case 4: Impossible mean
Scenario. A die outcome is always an integer but E(X)=3.5.
Reasoning. The expected value can be 3.5 because it is a long-run average, not a required individual outcome.
Case 4: Impossible mean check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 4; do not replace it with a memorized generic sentence.
Worked case 5: Shift
Scenario. Y=X+10.
Reasoning. E(Y)=E(X)+10 while SD(Y)=SD(X); adding a constant shifts center but not spread.
Case 5: Shift check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 5; do not replace it with a memorized generic sentence.
Worked case 6: Scale
Scenario. Y=2X.
Reasoning. E(Y)=2E(X) and SD(Y)=2SD(X); multiplying doubles both center and standard deviation magnitude.
Case 6: Scale check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 6; do not replace it with a memorized generic sentence.
Worked case 7: Negative scale
Scenario. Y=5−3X.
Reasoning. E(Y)=5−3E(X) and SD(Y)=3SD(X); standard deviation uses the absolute scale factor.
Case 7: Negative scale check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 7; do not replace it with a memorized generic sentence.
Worked case 8: Variance units
Scenario. X is measured in dollars.
Reasoning. Var(X) is measured in dollars squared, while SD(X) returns to dollars.
Case 8: Variance units check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 8; do not replace it with a memorized generic sentence.
Worked case 9: Risk comparison
Scenario. Two games both have E(X)=0, but one has much larger SD.
Reasoning. Both are fair in expected value, but the higher-SD game has greater outcome variability.
Case 9: Risk comparison check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 9; do not replace it with a memorized generic sentence.
Worked case 10: Probability check
Scenario. A proposed distribution has probabilities .2, .5, and .4.
Reasoning. It is invalid because the probabilities sum to 1.1 rather than 1.
Case 10: Probability check check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 10; do not replace it with a memorized generic sentence.
Worked case 11: Weighted versus unweighted
Scenario. Outcomes 0 and 10 have probabilities .9 and .1.
Reasoning. The expected value is 1, not the simple midpoint 5, because outcomes are not equally likely.
Case 11: Weighted versus unweighted check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 11; do not replace it with a memorized generic sentence.
Worked case 12: Long-run interpretation
Scenario. A service process has expected delay 4.2 minutes.
Reasoning. Across many comparable cases the average delay tends toward about 4.2 minutes; an individual delay need not equal 4.2.
Case 12: Long-run interpretation check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 12; do not replace it with a memorized generic sentence.
Deep-dive notes for Expected Value of a Random Variable: Formula and Interpretation
Expectation is linear even when outcomes are not equally likely
The expected value operator averages with probability weights. If one outcome is common and another rare, the common outcome contributes more to the long-run mean. This is why simply averaging listed payoff values can be badly wrong. Always pair each possible value with its probability before summing.
Expected value describes repeated systems
A warranty company, casino, insurer, or service system cares about expectation because many similar cases accumulate. An expected cost of $12.40 does not mean any customer produces exactly $12.40 of cost. It means average cost across a large collection of comparable contracts tends toward that value when the probability model is appropriate.
Standard deviation answers a different question from expectation
Two games can have the same expected net gain and radically different risk. Standard deviation summarizes how far outcomes typically vary around the expected value. A decision maker who cares about uncertainty should therefore inspect both center and spread rather than treating equal expected values as identical experiences.
Variance uses squared deviations because signs cancel
If raw deviations from the mean were averaged, positive and negative differences would cancel to zero. Squaring prevents cancellation and gives larger departures more weight. The square root at the end converts variance back to the original measurement units, producing standard deviation and a more interpretable scale of random variation.
Adding a constant does not change uncertainty
If every payoff receives an extra $5, the entire distribution shifts upward by $5. Expected value rises by $5, but distances between outcomes and the mean remain unchanged, so standard deviation is unchanged. This transformation rule is a powerful self-check when costs, bonuses, or fixed fees are added to a random variable.
Multiplying rescales both center and spread
If all outcomes are converted from dollars to cents by multiplying by 100, expected value multiplies by 100 and standard deviation also multiplies by 100. More generally, SD(a+bX)=|b|SD(X). The absolute value matters because spread cannot be negative even when the transformation reverses the order of outcomes.
Fairness refers to net expectation under the model
A game is called fair to the player when expected net gain is zero. Be careful to subtract entry cost before deciding. A lottery may have a positive expected gross payout but a negative expected net value after the ticket price. Fairness in expected value also does not guarantee low risk or desirable participation.
Validate the probability distribution first
Before computing expectation or variance, confirm that every probability lies between 0 and 1 and that the probabilities sum to 1, allowing for minor rounding. An invalid distribution makes every downstream summary meaningless. This quick structural check should come before calculator entry, especially when probabilities were constructed from a table.
Additional method notes
Expected profit requires net outcomes, not prize amounts
When a game charges an entry fee, convert every listed prize into net gain before calculating expectation. A $100 prize from a $5 ticket contributes a net outcome of $95, while losing contributes −$5 rather than 0. Mixing gross and net values can reverse the fairness conclusion even when all probabilities are entered correctly.
Long-run expectation needs a stable probability model
The law-of-large-numbers intuition behind expected value assumes repetitions are comparable enough that the same distribution remains relevant. If probabilities drift over time, player strategy changes, or the population generating outcomes changes, a historical expectation may not describe future repetitions. The formula is exact for the specified model; model relevance is a separate question.
Expected Value of a Random Variable: Formula and Interpretation: 36 multiple-choice questions
These questions stay within this page’s topic. Work them after the concept and worked-case sections so practice reinforces the method rather than replacing instruction.
Question 1. Expected Value and Standard Deviation of a Random Variable
A county election office in Lakeside district during a follow-up evaluation period outcome pays $21 with probability 0.193 and loses $6 otherwise. Find and interpret the expected net value.
Answer: D
E(X)=0.193($21)+0.807(−$6)=$-0.79. Over many independent repetitions under the same conditions, the average net result approaches about $-0.79 per repetition; it is not a guaranteed result on one trial.
Question 2. Expected Value and Standard Deviation of a Random Variable
A municipal emergency dispatch center in Sunbelt district during a pre-exam training cycle outcome pays $27 with probability 0.395 and loses $19 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.395($27)+0.605(−$19)=$-0.83. Over many independent repetitions under the same conditions, the average net result approaches about $-0.83 per repetition; it is not a guaranteed result on one trial.
Question 3. Expected Value and Standard Deviation of a Random Variable
A county election office in Midwest consortium during a school-year data collection outcome pays $63 with probability 0.208 and loses $14 otherwise. Find and interpret the expected net value.
Answer: A
E(X)=0.208($63)+0.792(−$14)=$2.02. Over many independent repetitions under the same conditions, the average net result approaches about $2.02 per repetition; it is not a guaranteed result on one trial.
Question 4. Expected Value and Standard Deviation of a Random Variable
A regional airport authority in Sunbelt district during a follow-up evaluation period outcome pays $33 with probability 0.162 and loses $21 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.162($33)+0.838(−$21)=$-12.25. Over many independent repetitions under the same conditions, the average net result approaches about $-12.25 per repetition; it is not a guaranteed result on one trial.
Question 5. Expected Value and Standard Deviation of a Random Variable
A solar installer in Pacific Northwest during a follow-up evaluation period outcome pays $55 with probability 0.353 and loses $22 otherwise. Find and interpret the expected net value.
Answer: C
E(X)=0.353($55)+0.647(−$22)=$5.18. Over many independent repetitions under the same conditions, the average net result approaches about $5.18 per repetition; it is not a guaranteed result on one trial.
Question 6. Expected Value and Standard Deviation of a Random Variable
A city transit agency in Westview during a multiweek validation study outcome pays $66 with probability 0.238 and loses $20 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.238($66)+0.762(−$20)=$0.47. Over many independent repetitions under the same conditions, the average net result approaches about $0.47 per repetition; it is not a guaranteed result on one trial.
Question 7. Expected Value and Standard Deviation of a Random Variable
A recycling program in North Valley during a service-improvement study outcome pays $20 with probability 0.44 and loses $14 otherwise. Find and interpret the expected net value.
Answer: A
E(X)=0.44($20)+0.56(−$14)=$0.96. Over many independent repetitions under the same conditions, the average net result approaches about $0.96 per repetition; it is not a guaranteed result on one trial.
Question 8. Expected Value and Standard Deviation of a Random Variable
A regional manufacturer in Sunbelt district during a quarterly performance study outcome pays $16 with probability 0.234 and loses $21 otherwise. Find and interpret the expected net value.
Answer: D
E(X)=0.234($16)+0.766(−$21)=$-12.34. Over many independent repetitions under the same conditions, the average net result approaches about $-12.34 per repetition; it is not a guaranteed result on one trial.
Question 9. Expected Value and Standard Deviation of a Random Variable
A solar installer in South Harbor during a summer implementation review outcome pays $44 with probability 0.201 and loses $19 otherwise. Find and interpret the expected net value.
Answer: D
E(X)=0.201($44)+0.799(−$19)=$-6.34. Over many independent repetitions under the same conditions, the average net result approaches about $-6.34 per repetition; it is not a guaranteed result on one trial.
Question 10. Expected Value and Standard Deviation of a Random Variable
A university advising center in Desert County during a two-month observation window outcome pays $55 with probability 0.203 and loses $18 otherwise. Find and interpret the expected net value.
Answer: A
E(X)=0.203($55)+0.797(−$18)=$-3.18. Over many independent repetitions under the same conditions, the average net result approaches about $-3.18 per repetition; it is not a guaranteed result on one trial.
Question 11. Expected Value and Standard Deviation of a Random Variable
A school district in South Harbor during a fall 2026 audit outcome pays $72 with probability 0.395 and loses $6 otherwise. Find and interpret the expected net value.
Answer: C
E(X)=0.395($72)+0.605(−$6)=$24.81. Over many independent repetitions under the same conditions, the average net result approaches about $24.81 per repetition; it is not a guaranteed result on one trial.
Question 12. Expected Value and Standard Deviation of a Random Variable
A city recreation department in Sunbelt district during a community outreach cycle outcome pays $54 with probability 0.266 and loses $7 otherwise. Find and interpret the expected net value.
Answer: A
E(X)=0.266($54)+0.734(−$7)=$9.23. Over many independent repetitions under the same conditions, the average net result approaches about $9.23 per repetition; it is not a guaranteed result on one trial.
Question 13. Expected Value and Standard Deviation of a Random Variable
A university advising center in Cedar Grove during a school-year data collection outcome pays $42 with probability 0.354 and loses $12 otherwise. Find and interpret the expected net value.
Answer: A
E(X)=0.354($42)+0.646(−$12)=$7.12. Over many independent repetitions under the same conditions, the average net result approaches about $7.12 per repetition; it is not a guaranteed result on one trial.
Question 14. Expected Value and Standard Deviation of a Random Variable
A wildlife clinic in Cedar Grove during a quarterly performance study outcome pays $77 with probability 0.274 and loses $13 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.274($77)+0.726(−$13)=$11.66. Over many independent repetitions under the same conditions, the average net result approaches about $11.66 per repetition; it is not a guaranteed result on one trial.
Question 15. Expected Value and Standard Deviation of a Random Variable
A solar installer in Lakeside district during a community outreach cycle outcome pays $25 with probability 0.201 and loses $20 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.201($25)+0.799(−$20)=$-10.95. Over many independent repetitions under the same conditions, the average net result approaches about $-10.95 per repetition; it is not a guaranteed result on one trial.
Question 16. Expected Value and Standard Deviation of a Random Variable
A county election office in Midwest consortium during a fall 2026 audit outcome pays $62 with probability 0.229 and loses $16 otherwise. Find and interpret the expected net value.
Answer: D
E(X)=0.229($62)+0.771(−$16)=$1.86. Over many independent repetitions under the same conditions, the average net result approaches about $1.86 per repetition; it is not a guaranteed result on one trial.
Question 17. Expected Value and Standard Deviation of a Random Variable
A food safety laboratory in North Valley during a service-improvement study outcome pays $51 with probability 0.399 and loses $25 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.399($51)+0.601(−$25)=$5.32. Over many independent repetitions under the same conditions, the average net result approaches about $5.32 per repetition; it is not a guaranteed result on one trial.
Question 18. Expected Value and Standard Deviation of a Random Variable
A regional hospital in South Harbor during a yearly program evaluation outcome pays $63 with probability 0.169 and loses $7 otherwise. Find and interpret the expected net value.
Answer: D
E(X)=0.169($63)+0.831(−$7)=$4.83. Over many independent repetitions under the same conditions, the average net result approaches about $4.83 per repetition; it is not a guaranteed result on one trial.
Question 19. Expected Value and Standard Deviation of a Random Variable
A school district in Central County during a school-year data collection outcome pays $50 with probability 0.415 and loses $15 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.415($50)+0.585(−$15)=$11.98. Over many independent repetitions under the same conditions, the average net result approaches about $11.98 per repetition; it is not a guaranteed result on one trial.
Question 20. Expected Value and Standard Deviation of a Random Variable
A solar installer in Midwest consortium during a community outreach cycle outcome pays $43 with probability 0.198 and loses $9 otherwise. Find and interpret the expected net value.
Answer: D
E(X)=0.198($43)+0.802(−$9)=$1.3. Over many independent repetitions under the same conditions, the average net result approaches about $1.3 per repetition; it is not a guaranteed result on one trial.
Question 21. Expected Value and Standard Deviation of a Random Variable
A regional manufacturer in Great Lakes during a six-week field trial outcome pays $18 with probability 0.26 and loses $9 otherwise. Find and interpret the expected net value.
Answer: D
E(X)=0.26($18)+0.74(−$9)=$-1.98. Over many independent repetitions under the same conditions, the average net result approaches about $-1.98 per repetition; it is not a guaranteed result on one trial.
Question 22. Expected Value and Standard Deviation of a Random Variable
A municipal water office in Great Lakes during a fall 2026 audit outcome pays $69 with probability 0.433 and loses $9 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.433($69)+0.567(−$9)=$24.77. Over many independent repetitions under the same conditions, the average net result approaches about $24.77 per repetition; it is not a guaranteed result on one trial.
Question 23. Expected Value and Standard Deviation of a Random Variable
A school district in Cedar Grove during a two-month observation window outcome pays $41 with probability 0.383 and loses $12 otherwise. Find and interpret the expected net value.
Answer: C
E(X)=0.383($41)+0.617(−$12)=$8.3. Over many independent repetitions under the same conditions, the average net result approaches about $8.3 per repetition; it is not a guaranteed result on one trial.
Question 24. Expected Value and Standard Deviation of a Random Variable
A regional hospital in Prairie District during a fall 2026 audit outcome pays $29 with probability 0.152 and loses $8 otherwise. Find and interpret the expected net value.
Answer: D
E(X)=0.152($29)+0.848(−$8)=$-2.38. Over many independent repetitions under the same conditions, the average net result approaches about $-2.38 per repetition; it is not a guaranteed result on one trial.
Question 25. Expected Value and Standard Deviation of a Random Variable
A regional airport authority in New England network during a winter readiness review outcome pays $59 with probability 0.2 and loses $18 otherwise. Find and interpret the expected net value.
Answer: C
E(X)=0.2($59)+0.8(−$18)=$-2.6. Over many independent repetitions under the same conditions, the average net result approaches about $-2.6 per repetition; it is not a guaranteed result on one trial.
Question 26. Expected Value and Standard Deviation of a Random Variable
A county election office in Mountain Region during a regional benchmarking study outcome pays $45 with probability 0.152 and loses $8 otherwise. Find and interpret the expected net value.
Answer: A
E(X)=0.152($45)+0.848(−$8)=$0.06. Over many independent repetitions under the same conditions, the average net result approaches about $0.06 per repetition; it is not a guaranteed result on one trial.
Question 27. Expected Value and Standard Deviation of a Random Variable
A wildlife clinic in Midwest consortium during a school-year data collection outcome pays $39 with probability 0.175 and loses $18 otherwise. Find and interpret the expected net value.
Answer: D
E(X)=0.175($39)+0.825(−$18)=$-8.03. Over many independent repetitions under the same conditions, the average net result approaches about $-8.03 per repetition; it is not a guaranteed result on one trial.
Question 28. Expected Value and Standard Deviation of a Random Variable
A housing authority in Pine Ridge during a community outreach cycle outcome pays $18 with probability 0.251 and loses $23 otherwise. Find and interpret the expected net value.
Answer: C
E(X)=0.251($18)+0.749(−$23)=$-12.71. Over many independent repetitions under the same conditions, the average net result approaches about $-12.71 per repetition; it is not a guaranteed result on one trial.
Question 29. Expected Value and Standard Deviation of a Random Variable
A food safety laboratory in North Valley during a yearly program evaluation outcome pays $80 with probability 0.181 and loses $15 otherwise. Find and interpret the expected net value.
Answer: A
E(X)=0.181($80)+0.819(−$15)=$2.2. Over many independent repetitions under the same conditions, the average net result approaches about $2.2 per repetition; it is not a guaranteed result on one trial.
Question 30. Expected Value and Standard Deviation of a Random Variable
A food safety laboratory in Coastal Plains during a quarterly performance study outcome pays $16 with probability 0.341 and loses $24 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.341($16)+0.659(−$24)=$-10.36. Over many independent repetitions under the same conditions, the average net result approaches about $-10.36 per repetition; it is not a guaranteed result on one trial.
Question 31. Expected Value and Standard Deviation of a Random Variable
A municipal emergency dispatch center in Atlantic Corridor during a two-month observation window outcome pays $78 with probability 0.171 and loses $15 otherwise. Find and interpret the expected net value.
Answer: C
E(X)=0.171($78)+0.829(−$15)=$0.9. Over many independent repetitions under the same conditions, the average net result approaches about $0.9 per repetition; it is not a guaranteed result on one trial.
Question 32. Expected Value and Standard Deviation of a Random Variable
A recycling program in Riverbend during a service-improvement study outcome pays $56 with probability 0.246 and loses $18 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.246($56)+0.754(−$18)=$0.2. Over many independent repetitions under the same conditions, the average net result approaches about $0.2 per repetition; it is not a guaranteed result on one trial.
Question 33. Expected Value and Standard Deviation of a Random Variable
A public health department in Great Lakes during a multiweek validation study outcome pays $33 with probability 0.356 and loses $11 otherwise. Find and interpret the expected net value.
Answer: D
E(X)=0.356($33)+0.644(−$11)=$4.66. Over many independent repetitions under the same conditions, the average net result approaches about $4.66 per repetition; it is not a guaranteed result on one trial.
Question 34. Expected Value and Standard Deviation of a Random Variable
A community bank in Capital Region during a quarterly performance study outcome pays $44 with probability 0.179 and loses $23 otherwise. Find and interpret the expected net value.
Answer: C
E(X)=0.179($44)+0.821(−$23)=$-11.01. Over many independent repetitions under the same conditions, the average net result approaches about $-11.01 per repetition; it is not a guaranteed result on one trial.
Question 35. Expected Value and Standard Deviation of a Random Variable
A regional manufacturer in Metro East during a service-improvement study outcome pays $43 with probability 0.232 and loses $12 otherwise. Find and interpret the expected net value.
Answer: C
E(X)=0.232($43)+0.768(−$12)=$0.76. Over many independent repetitions under the same conditions, the average net result approaches about $0.76 per repetition; it is not a guaranteed result on one trial.
Question 36. Expected Value and Standard Deviation of a Random Variable
A municipal emergency dispatch center in Sunbelt district during a monthly quality review outcome pays $54 with probability 0.437 and loses $6 otherwise. Find and interpret the expected net value.
Answer: B
E(X)=0.437($54)+0.563(−$6)=$20.22. Over many independent repetitions under the same conditions, the average net result approaches about $20.22 per repetition; it is not a guaranteed result on one trial.
Expected Value of a Random Variable: Formula and Interpretation: 14 free-response questions
For each response, show the statistical reasoning, use the scenario’s language, and state only the conclusion supported by the design or probability model.
FRQ set 1: Expected Value and Standard Deviation of a Random Variable
Scenario. A university advising center in Mountain Region during a baseline measurement week outcome pays $19 with probability 0.317 and loses $25 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.317($19)+0.683(−$25)=$-11.05. Over many independent repetitions under the same conditions, the average net result approaches about $-11.05 per repetition; it is not a guaranteed result on one trial.
FRQ set 2: Expected Value and Standard Deviation of a Random Variable
Scenario. A digital learning platform in Pacific Northwest during a winter readiness review outcome pays $76 with probability 0.239 and loses $7 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.239($76)+0.761(−$7)=$12.84. Over many independent repetitions under the same conditions, the average net result approaches about $12.84 per repetition; it is not a guaranteed result on one trial.
FRQ set 3: Expected Value and Standard Deviation of a Random Variable
Scenario. A regional hospital in South Harbor during a yearly program evaluation outcome pays $61 with probability 0.388 and loses $17 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.388($61)+0.612(−$17)=$13.26. Over many independent repetitions under the same conditions, the average net result approaches about $13.26 per repetition; it is not a guaranteed result on one trial.
FRQ set 4: Expected Value and Standard Deviation of a Random Variable
Scenario. A recycling program in Pacific Northwest during a summer implementation review outcome pays $21 with probability 0.214 and loses $9 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.214($21)+0.786(−$9)=$-2.58. Over many independent repetitions under the same conditions, the average net result approaches about $-2.58 per repetition; it is not a guaranteed result on one trial.
FRQ set 5: Expected Value and Standard Deviation of a Random Variable
Scenario. A state park in Mountain Region during a weekday operations study outcome pays $37 with probability 0.398 and loses $12 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.398($37)+0.602(−$12)=$7.5. Over many independent repetitions under the same conditions, the average net result approaches about $7.5 per repetition; it is not a guaranteed result on one trial.
FRQ set 6: Expected Value and Standard Deviation of a Random Variable
Scenario. A city transit agency in Capital Region during a fall 2026 audit outcome pays $28 with probability 0.342 and loses $19 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.342($28)+0.658(−$19)=$-2.93. Over many independent repetitions under the same conditions, the average net result approaches about $-2.93 per repetition; it is not a guaranteed result on one trial.
FRQ set 7: Expected Value and Standard Deviation of a Random Variable
Scenario. A grocery cooperative in Central County during a two-month observation window outcome pays $62 with probability 0.215 and loses $13 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.215($62)+0.785(−$13)=$3.12. Over many independent repetitions under the same conditions, the average net result approaches about $3.12 per repetition; it is not a guaranteed result on one trial.
FRQ set 8: Expected Value and Standard Deviation of a Random Variable
Scenario. A municipal emergency dispatch center in North Valley during a school-year data collection outcome pays $72 with probability 0.28 and loses $18 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.28($72)+0.72(−$18)=$7.2. Over many independent repetitions under the same conditions, the average net result approaches about $7.2 per repetition; it is not a guaranteed result on one trial.
FRQ set 9: Expected Value and Standard Deviation of a Random Variable
Scenario. A solar installer in Cedar Grove during a community outreach cycle outcome pays $44 with probability 0.334 and loses $20 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.334($44)+0.666(−$20)=$1.38. Over many independent repetitions under the same conditions, the average net result approaches about $1.38 per repetition; it is not a guaranteed result on one trial.
FRQ set 10: Expected Value and Standard Deviation of a Random Variable
Scenario. A public high school in Central County during a community outreach cycle outcome pays $80 with probability 0.244 and loses $21 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.244($80)+0.756(−$21)=$3.64. Over many independent repetitions under the same conditions, the average net result approaches about $3.64 per repetition; it is not a guaranteed result on one trial.
FRQ set 11: Expected Value and Standard Deviation of a Random Variable
Scenario. A state park in Pacific Northwest during a fall 2026 audit outcome pays $71 with probability 0.198 and loses $12 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.198($71)+0.802(−$12)=$4.43. Over many independent repetitions under the same conditions, the average net result approaches about $4.43 per repetition; it is not a guaranteed result on one trial.
FRQ set 12: Expected Value and Standard Deviation of a Random Variable
Scenario. A digital learning platform in Pine Ridge during a school-year data collection outcome pays $61 with probability 0.423 and loses $10 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.423($61)+0.577(−$10)=$20.03. Over many independent repetitions under the same conditions, the average net result approaches about $20.03 per repetition; it is not a guaranteed result on one trial.
FRQ set 13: Expected Value and Standard Deviation of a Random Variable
Scenario. A grocery cooperative in Atlantic Corridor during a two-month observation window outcome pays $30 with probability 0.342 and loses $13 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.342($30)+0.658(−$13)=$1.71. Over many independent repetitions under the same conditions, the average net result approaches about $1.71 per repetition; it is not a guaranteed result on one trial.
FRQ set 14: Expected Value and Standard Deviation of a Random Variable
Scenario. A state park in Mountain Region during a regional benchmarking study outcome pays $33 with probability 0.429 and loses $8 otherwise. Find and interpret the expected net value.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
E(X)=0.429($33)+0.571(−$8)=$9.59. Over many independent repetitions under the same conditions, the average net result approaches about $9.59 per repetition; it is not a guaranteed result on one trial.
Continue with a connected AP Statistics skill
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