Random Variables: Discrete, Continuous, Mean, and SD
Build probability distributions, calculate expected value and standard deviation, and interpret a random variable in context.
Direct answer
random variables: Random variables assign numerical values to outcomes of chance processes. A valid discrete distribution has nonnegative probabilities summing to 1; its expected value is a probability-weighted long-run mean and its standard deviation measures spread.
Quick reference: Random Variables: Discrete, Continuous, Mean, and SD
| Mean | μX = ΣxP(x) |
|---|---|
| SD | σX = √Σ(x−μX)2P(x) |
Random Variables: Discrete, Continuous, Mean, and SD: complete lesson
A random variable turns chance outcomes into numerical values
Random variables provide the bridge between a probability model and numerical summaries. A random variable X assigns a number to each outcome of a random process. In a discrete model, X takes a countable list of values with probabilities that add to 1. In a continuous model, probabilities are represented by area under a density curve; the probability of any single exact point is 0 even though intervals can have positive probability.
The letter X is not the outcome itself. If two coins are tossed, the elementary outcomes HH, HT, TH, and TT are not numerical. Defining X as the number of heads maps those outcomes to 2, 1, 1, and 0. That mapping can combine different elementary outcomes into the same random-variable value, so P(X=1)=P(HT or TH).
A discrete probability distribution must satisfy two validity conditions
For every possible x, P(X=x) must be between 0 and 1, inclusive, and the probabilities across all possible values must sum to 1. These checks come before expected value or standard deviation. A table with probabilities 0.18, 0.27, 0.31, and 0.30 is invalid because the total is 1.06. A negative probability is invalid even if the total happens to equal 1.
A distribution can be valid even when its expected value is not one of the possible outcomes. If a game pays $0, $2, or $8, its long-run mean payoff might be $2.74. That does not mean a player can receive exactly $2.74 on one play. Expected value is a long-run balance point across repeated independent repetitions of the same random process.
Expected value is a probability-weighted average
For a discrete random variable, the mean is μX=ΣxP(X=x). Each value is weighted by how often it is expected to occur in the long run. Merely averaging the possible x-values gives the correct mean only in the special case of equal probabilities. The expected value should be interpreted in the units of X and in long-run context, not as a guarantee for a single trial.
When X represents net gain, use net values rather than prize amounts. If a ticket costs $3 and pays $10 with probability 0.1 and $0 otherwise, the net random variable is +$7 with probability 0.1 and −$3 with probability 0.9. Its mean is 0.1(7)+0.9(−3)=−$2.00 per ticket. Confusing gross prize with net gain changes the question being answered.
Variance and standard deviation describe spread around the expected value
For a discrete random variable, Var(X)=Σ(x−μX)²P(X=x), and σX is the square root of that variance. The deviations are squared before weighting, so observations farther from the mean contribute disproportionately. Standard deviation returns to the original units of X, which makes it easier to interpret than variance.
A common shortcut computes E(X²)−[E(X)]². First calculate Σx²P(X=x), then subtract μ². This is algebraically equivalent to the deviation formula. Whichever route is used, a negative variance signals an arithmetic error because variance cannot be below zero.
Linear transformations change center and spread predictably
If Y=a+bX, then μY=a+bμX. Adding a shifts every outcome by the same amount and does not change standard deviation. Multiplying by b scales distances from the mean, so σY=|b|σX. A negative multiplier reverses the scale but standard deviation remains nonnegative.
Suppose X is temperature in Celsius with mean 18 and standard deviation 4. Converting to Fahrenheit gives Y=32+1.8X. The mean becomes 64.4°F and the standard deviation becomes 7.2°F. The +32 affects the mean only; the ×1.8 affects both center and spread.
Discrete and continuous models answer probability questions differently
For a discrete variable such as number of defectives in a sample, P(X=2) can be positive. For a continuous variable such as fill volume, P(X=500.000 exactly)=0 under an ideal continuous model. Therefore P(X≤500) and P(X<500) are the same in a continuous model, while those events can differ in a discrete model when P(X=500)>0.
This distinction matters when translating words into events. “At most 3” means X≤3; “fewer than 3” means X<3; “at least 3” means X≥3; and “more than 3” means X>3. In discrete count models, changing < to ≤ can add an entire probability mass at the boundary.
Expected value is a model property, not a fairness judgment by itself
A game with expected net value zero is fair in the mathematical sense that neither side has a long-run expected advantage, but individual outcomes can still be highly variable. Two games can have the same expected value and radically different risk. Comparing both μ and σ gives a more complete description of long-run payoff behavior.
Context also determines whether higher variability is desirable. A lottery player may seek variability because rare large wins are attractive, whereas a hospital may prefer low variability in treatment time even at the same mean. Statistics describes the distribution; a decision rule determines how the distribution is valued.
Worked random-variable distributions
Worked distribution 1: Warranty claims
Let X be the number of claims per appliance. The possible values are 0, 1, 2, 3 with probabilities 0.52, 0.28, 0.15, 0.05. The validity check gives 0.52 + 0.28 + 0.15 + 0.05=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=0(0.52) + 1(0.28) + 2(0.15) + 3(0.05)=0.730. Interpreted correctly, repeated observations of this process would average about 0.730 number of claims per appliance per repetition; the number is not promised on any single repetition.
The variance is Σ(x−0.730)²P(x)=0.797, so σX=√0.797=0.893. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 2: Late buses
Let X be the late buses observed in a morning. The possible values are 0, 1, 2, 3, 4 with probabilities 0.18, 0.31, 0.27, 0.16, 0.08. The validity check gives 0.18 + 0.31 + 0.27 + 0.16 + 0.08=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=0(0.18) + 1(0.31) + 2(0.27) + 3(0.16) + 4(0.08)=1.650. Interpreted correctly, repeated observations of this process would average about 1.650 late buses observed in a morning per repetition; the number is not promised on any single repetition.
The variance is Σ(x−1.650)²P(x)=1.387, so σX=√1.387=1.178. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 3: Support tickets
Let X be the tickets opened by one account in a week. The possible values are 0, 1, 2, 3 with probabilities 0.40, 0.34, 0.19, 0.07. The validity check gives 0.40 + 0.34 + 0.19 + 0.07=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=0(0.40) + 1(0.34) + 2(0.19) + 3(0.07)=0.930. Interpreted correctly, repeated observations of this process would average about 0.930 tickets opened by one account in a week per repetition; the number is not promised on any single repetition.
The variance is Σ(x−0.930)²P(x)=0.865, so σX=√0.865=0.930. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 4: Goals scored
Let X be the goals by a team in a match. The possible values are 0, 1, 2, 3, 4 with probabilities 0.14, 0.29, 0.30, 0.18, 0.09. The validity check gives 0.14 + 0.29 + 0.30 + 0.18 + 0.09=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=0(0.14) + 1(0.29) + 2(0.30) + 3(0.18) + 4(0.09)=1.790. Interpreted correctly, repeated observations of this process would average about 1.790 goals by a team in a match per repetition; the number is not promised on any single repetition.
The variance is Σ(x−1.790)²P(x)=1.346, so σX=√1.346=1.160. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 5: Returned items
Let X be the items returned by a customer order. The possible values are 0, 1, 2, 3 with probabilities 0.61, 0.24, 0.11, 0.04. The validity check gives 0.61 + 0.24 + 0.11 + 0.04=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=0(0.61) + 1(0.24) + 2(0.11) + 3(0.04)=0.580. Interpreted correctly, repeated observations of this process would average about 0.580 items returned by a customer order per repetition; the number is not promised on any single repetition.
The variance is Σ(x−0.580)²P(x)=0.704, so σX=√0.704=0.839. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 6: Power interruptions
Let X be the interruptions in a month. The possible values are 0, 1, 2, 3, 4 with probabilities 0.36, 0.29, 0.20, 0.10, 0.05. The validity check gives 0.36 + 0.29 + 0.20 + 0.10 + 0.05=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=0(0.36) + 1(0.29) + 2(0.20) + 3(0.10) + 4(0.05)=1.190. Interpreted correctly, repeated observations of this process would average about 1.190 interruptions in a month per repetition; the number is not promised on any single repetition.
The variance is Σ(x−1.190)²P(x)=1.374, so σX=√1.374=1.172. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 7: Free-throw points
Let X be the points scored from a two-shot trip. The possible values are 0, 1, 2 with probabilities 0.08, 0.24, 0.68. The validity check gives 0.08 + 0.24 + 0.68=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=0(0.08) + 1(0.24) + 2(0.68)=1.600. Interpreted correctly, repeated observations of this process would average about 1.600 points scored from a two-shot trip per repetition; the number is not promised on any single repetition.
The variance is Σ(x−1.600)²P(x)=0.400, so σX=√0.400=0.632. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 8: Clinic no-shows
Let X be the no-shows in a four-appointment block. The possible values are 0, 1, 2, 3 with probabilities 0.33, 0.38, 0.21, 0.08. The validity check gives 0.33 + 0.38 + 0.21 + 0.08=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=0(0.33) + 1(0.38) + 2(0.21) + 3(0.08)=1.040. Interpreted correctly, repeated observations of this process would average about 1.040 no-shows in a four-appointment block per repetition; the number is not promised on any single repetition.
The variance is Σ(x−1.040)²P(x)=0.858, so σX=√0.858=0.926. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 9: Machine faults
Let X be the faults recorded in one shift. The possible values are 0, 1, 2, 3, 4 with probabilities 0.47, 0.26, 0.15, 0.08, 0.04. The validity check gives 0.47 + 0.26 + 0.15 + 0.08 + 0.04=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=0(0.47) + 1(0.26) + 2(0.15) + 3(0.08) + 4(0.04)=0.960. Interpreted correctly, repeated observations of this process would average about 0.960 faults recorded in one shift per repetition; the number is not promised on any single repetition.
The variance is Σ(x−0.960)²P(x)=1.298, so σX=√1.298=1.139. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 10: Delivery attempts
Let X be the attempts required for a package. The possible values are 1, 2, 3, 4 with probabilities 0.70, 0.20, 0.07, 0.03. The validity check gives 0.70 + 0.20 + 0.07 + 0.03=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=1(0.70) + 2(0.20) + 3(0.07) + 4(0.03)=1.430. Interpreted correctly, repeated observations of this process would average about 1.430 attempts required for a package per repetition; the number is not promised on any single repetition.
The variance is Σ(x−1.430)²P(x)=0.565, so σX=√0.565=0.752. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 11: Quiz bonuses
Let X be the bonus points earned. The possible values are 0, 2, 5, 10 with probabilities 0.45, 0.30, 0.20, 0.05. The validity check gives 0.45 + 0.30 + 0.20 + 0.05=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=0(0.45) + 2(0.30) + 5(0.20) + 10(0.05)=2.100. Interpreted correctly, repeated observations of this process would average about 2.100 bonus points earned per repetition; the number is not promised on any single repetition.
The variance is Σ(x−2.100)²P(x)=6.790, so σX=√6.790=2.606. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Worked distribution 12: Insurance net payout
Let X be the hundreds of dollars of net payout. The possible values are -2, 3, 8, 18 with probabilities 0.78, 0.14, 0.06, 0.02. The validity check gives 0.78 + 0.14 + 0.06 + 0.02=1.00, and every probability is nonnegative.
The expected value is μX=ΣxP(x)=-2(0.78) + 3(0.14) + 8(0.06) + 18(0.02)=-0.300. Interpreted correctly, repeated observations of this process would average about -0.300 hundreds of dollars of net payout per repetition; the number is not promised on any single repetition.
The variance is Σ(x−-0.300)²P(x)=14.610, so σX=√14.610=3.822. This standard deviation describes a typical scale of variation around the long-run mean in the same units as X.
Distribution audit laboratory
Distribution audit laboratory 1: school survey
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this school survey, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 1: school survey,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 1: school survey,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 2: public-health study
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this public-health study, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 2: public-health study,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 2: public-health study,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 3: manufacturing process
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this manufacturing process, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 3: manufacturing process,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 3: manufacturing process,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 4: transportation system
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this transportation system, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 4: transportation system,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 4: transportation system,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 5: consumer study
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this consumer study, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 5: consumer study,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 5: consumer study,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 6: environmental monitoring
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this environmental monitoring, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 6: environmental monitoring,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 6: environmental monitoring,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 7: sports analysis
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this sports analysis, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 7: sports analysis,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 7: sports analysis,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 8: education program
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this education program, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 8: education program,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 8: education program,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 9: service operation
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this service operation, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 9: service operation,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 9: service operation,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 10: technology experiment
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this technology experiment, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 10: technology experiment,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 10: technology experiment,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 11: community poll
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this community poll, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 11: community poll,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 11: community poll,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Distribution audit laboratory 12: quality-control review
For each probability table, verify nonnegative masses and total probability before computing μ and σ. Then compare the weighted mean with the unweighted mean of possible outcomes and explain why only the probability-weighted value has the correct long-run interpretation. In this quality-control review, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Distribution audit laboratory 12: quality-control review,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Distribution audit laboratory 12: quality-control review,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Verify probability mass and units before using a weighted mean or spread calculation; the long-run interpretation belongs to repeated realizations of this variable.
Random Variables: Discrete, Continuous, Mean, and SD: 22 multiple-choice questions
These items focus on probability-distribution validity, expected value, standard deviation, transformations, and discrete-versus-continuous reasoning.
Question 1. Random Variables
A random variable X for a housing authority in Midwest consortium during a randomized pilot period takes values [0, 1, 2, 3] with probabilities [0.191, 0.17, 0.21, 0.429]. Verify validity and find E(X).
Answer: A
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.191)+1(0.17)+2(0.21)+3(0.429)=1.877.
Question 2. Random Variables
A random variable X for a wildlife clinic in South Harbor during a multiweek validation study takes values [0, 1, 2, 3] with probabilities [0.142, 0.116, 0.116, 0.626]. Verify validity and find E(X).
Answer: A
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.142)+1(0.116)+2(0.116)+3(0.626)=2.226.
Question 3. Random Variables
A random variable X for a regional airport authority in Desert County during a regional benchmarking study takes values [0, 1, 2, 3] with probabilities [0.137, 0.122, 0.153, 0.588]. Verify validity and find E(X).
Answer: D
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.137)+1(0.122)+2(0.153)+3(0.588)=2.192.
Question 4. Random Variables
A random variable X for a food safety laboratory in Pine Ridge during a multiweek validation study takes values [0, 1, 2, 3] with probabilities [0.093, 0.195, 0.097, 0.615]. Verify validity and find E(X).
Answer: A
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.093)+1(0.195)+2(0.097)+3(0.615)=2.234.
Question 5. Random Variables
A random variable X for a university advising center in North Valley during a pre-exam training cycle takes values [0, 1, 2, 3] with probabilities [0.137, 0.193, 0.193, 0.477]. Verify validity and find E(X).
Answer: B
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.137)+1(0.193)+2(0.193)+3(0.477)=2.01.
Question 6. Random Variables
A random variable X for a county election office in Pacific Northwest during a pre-exam training cycle takes values [0, 1, 2, 3] with probabilities [0.207, 0.158, 0.158, 0.477]. Verify validity and find E(X).
Answer: A
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.207)+1(0.158)+2(0.158)+3(0.477)=1.905.
Question 7. Random Variables
A random variable X for a recycling program in Westview during a weekday operations study takes values [0, 1, 2, 3] with probabilities [0.21, 0.109, 0.154, 0.527]. Verify validity and find E(X).
Answer: C
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.21)+1(0.109)+2(0.154)+3(0.527)=1.998.
Question 8. Random Variables
A random variable X for a regional manufacturer in Coastal Plains during a multiweek validation study takes values [0, 1, 2, 3] with probabilities [0.214, 0.14, 0.166, 0.48]. Verify validity and find E(X).
Answer: A
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.214)+1(0.14)+2(0.166)+3(0.48)=1.912.
Question 9. Random Variables
A random variable X for a regional airport authority in North Valley during a randomized pilot period takes values [0, 1, 2, 3] with probabilities [0.186, 0.175, 0.174, 0.465]. Verify validity and find E(X).
Answer: D
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.186)+1(0.175)+2(0.174)+3(0.465)=1.918.
Question 10. Random Variables
A random variable X for a regional hospital in Pine Ridge during a summer implementation review takes values [0, 1, 2, 3] with probabilities [0.156, 0.126, 0.119, 0.599]. Verify validity and find E(X).
Answer: D
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.156)+1(0.126)+2(0.119)+3(0.599)=2.161.
Question 11. Random Variables
A random variable X for a county library in Midwest consortium during a semester-long cohort study takes values [0, 1, 2, 3] with probabilities [0.083, 0.129, 0.195, 0.593]. Verify validity and find E(X).
Answer: A
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.083)+1(0.129)+2(0.195)+3(0.593)=2.298.
Question 12. Random Variables
A random variable X for a food safety laboratory in Prairie District during a six-week field trial takes values [0, 1, 2, 3] with probabilities [0.157, 0.189, 0.122, 0.532]. Verify validity and find E(X).
Answer: D
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.157)+1(0.189)+2(0.122)+3(0.532)=2.029.
Question 13. Random Variables
A random variable X for a digital learning platform in Westview during a multiweek validation study takes values [0, 1, 2, 3] with probabilities [0.117, 0.201, 0.213, 0.469]. Verify validity and find E(X).
Answer: A
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.117)+1(0.201)+2(0.213)+3(0.469)=2.034.
Question 14. Random Variables
A random variable X for a community college in Desert County during a yearly program evaluation takes values [0, 1, 2, 3] with probabilities [0.131, 0.125, 0.118, 0.626]. Verify validity and find E(X).
Answer: D
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.131)+1(0.125)+2(0.118)+3(0.626)=2.239.
Question 15. Random Variables
A random variable X for a regional airport authority in Midwest consortium during a school-year data collection takes values [0, 1, 2, 3] with probabilities [0.216, 0.127, 0.123, 0.534]. Verify validity and find E(X).
Answer: A
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.216)+1(0.127)+2(0.123)+3(0.534)=1.975.
Question 16. Random Variables
A random variable X for a city recreation department in North Valley during a weekday operations study takes values [0, 1, 2, 3] with probabilities [0.088, 0.217, 0.142, 0.553]. Verify validity and find E(X).
Answer: B
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.088)+1(0.217)+2(0.142)+3(0.553)=2.16.
Question 17. Random Variables
A random variable X for a municipal emergency dispatch center in Westview during a summer implementation review takes values [0, 1, 2, 3] with probabilities [0.197, 0.162, 0.083, 0.558]. Verify validity and find E(X).
Answer: D
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.197)+1(0.162)+2(0.083)+3(0.558)=2.002.
Question 18. Random Variables
A random variable X for a regional hospital in Mountain Region during a yearly program evaluation takes values [0, 1, 2, 3] with probabilities [0.142, 0.151, 0.193, 0.514]. Verify validity and find E(X).
Answer: B
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.142)+1(0.151)+2(0.193)+3(0.514)=2.079.
Question 19. Random Variables
A random variable X for a regional airport authority in Central County during a fall 2026 audit takes values [0, 1, 2, 3] with probabilities [0.151, 0.193, 0.195, 0.461]. Verify validity and find E(X).
Answer: B
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.151)+1(0.193)+2(0.195)+3(0.461)=1.966.
Question 20. Random Variables
A random variable X for a grocery cooperative in Westview during a yearly program evaluation takes values [0, 1, 2, 3] with probabilities [0.15, 0.181, 0.129, 0.54]. Verify validity and find E(X).
Answer: A
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.15)+1(0.181)+2(0.129)+3(0.54)=2.059.
Question 21. Random Variables
A random variable X for a community bank in Prairie District during a spring 2027 pilot takes values [0, 1, 2, 3] with probabilities [0.206, 0.15, 0.186, 0.458]. Verify validity and find E(X).
Answer: B
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.206)+1(0.15)+2(0.186)+3(0.458)=1.896.
Question 22. Random Variables
A random variable X for a community college in Pacific Northwest during a semester-long cohort study takes values [0, 1, 2, 3] with probabilities [0.092, 0.203, 0.128, 0.577]. Verify validity and find E(X).
Answer: A
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.092)+1(0.203)+2(0.128)+3(0.577)=2.19.
Random Variables: Discrete, Continuous, Mean, and SD: 6 free-response questions
For each free-response prompt, define X in context, verify or construct the probability model, show the weighted calculation, and interpret expected value or spread in the units of X.
FRQ set 1: Random Variables
Scenario. A random variable X for a food safety laboratory in Riverbend during a follow-up evaluation period takes values [0, 1, 2, 3] with probabilities [0.108, 0.11, 0.199, 0.583]. Verify validity and find E(X).
- 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
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.108)+1(0.11)+2(0.199)+3(0.583)=2.257.
FRQ set 2: Random Variables
Scenario. A random variable X for a university advising center in Midwest consortium during a community outreach cycle takes values [0, 1, 2, 3] with probabilities [0.173, 0.214, 0.147, 0.466]. Verify validity and find E(X).
- 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
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.173)+1(0.214)+2(0.147)+3(0.466)=1.906.
FRQ set 3: Random Variables
Scenario. A random variable X for a public high school in Coastal Plains during a baseline measurement week takes values [0, 1, 2, 3] with probabilities [0.214, 0.089, 0.094, 0.603]. Verify validity and find E(X).
- 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
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.214)+1(0.089)+2(0.094)+3(0.603)=2.086.
FRQ set 4: Random Variables
Scenario. A random variable X for a county election office in Metro East during a semester-long cohort study takes values [0, 1, 2, 3] with probabilities [0.179, 0.204, 0.087, 0.53]. Verify validity and find E(X).
- 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
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.179)+1(0.204)+2(0.087)+3(0.53)=1.968.
FRQ set 5: Random Variables
Scenario. A random variable X for a school district in Coastal Plains during a service-improvement study takes values [0, 1, 2, 3] with probabilities [0.163, 0.106, 0.134, 0.597]. Verify validity and find E(X).
- 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
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.163)+1(0.106)+2(0.134)+3(0.597)=2.165.
FRQ set 6: Random Variables
Scenario. A random variable X for a city transit agency in Mountain Region during a school-year data collection takes values [0, 1, 2, 3] with probabilities [0.088, 0.164, 0.1, 0.648]. Verify validity and find E(X).
- 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
Every probability is nonnegative and the sum is 1, so the distribution is valid. E(X)=ΣxP(x)=0(0.088)+1(0.164)+2(0.1)+3(0.648)=2.308.
Continue with the next connected AP Statistics skill
The most useful next step is to connect this topic to a neighboring method rather than repeating the same question type indefinitely.
