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CSPA practice questions

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Question 1Property-casualty insurance quantities and foundations

A collision occurs in December 2024 and its first payment is made in February 2025. For an accident-year target, where is the event assigned?

  1. Accident year 2025 solely because payment occurred then
  2. Accident year 2024 Correct answer.
  3. Neither year until the claim is closed
  4. Both years as separate accidents

Explanation

Accident-year grouping follows the occurrence date rather than the payment date. Payment timing is relevant to other views but does not create another accident.

Question 2Property-casualty insurance quantities and foundations

A portfolio records 100 claims over 1,250 policy-years. What is its claim frequency per policy-year?

  1. 0.125
  2. 8.0
  3. 0.08 Correct answer.
  4. 12.5

Explanation

Frequency uses claim count divided by matching exposure. One hundred divided by 1,250 equals 0.08 claims per policy-year for the supplied population.

Question 3Property-casualty insurance quantities and foundations

The loss total for 120 included claims is $360,000. What average severity follows those definitions?

  1. $120 per policy-year
  2. $3,000 per claim Correct answer.
  3. $360,120 per claim
  4. $432 per claim

Explanation

Severity divides the defined loss amount by the corresponding included claim count. Three hundred sixty thousand divided by 120 gives three thousand dollars per claim.

Question 4Property-casualty insurance quantities and foundations

Frequency is 0.08 claims per exposure and matching severity is $3,000 per claim. What pure premium results?

  1. $240 per exposure Correct answer.
  2. $37,500 per exposure
  3. $3,000 per exposure
  4. $24 per exposure

Explanation

Compatible frequency and severity multiply to expected loss per exposure. The claim unit cancels, leaving 0.08 times three thousand, or 240 dollars per exposure.

Question 5Property-casualty insurance quantities and foundations

An analyst multiplies net-of-deductible claim frequency by gross severity measured on a different claim population. What should be resolved first?

  1. Whether the count, loss, exposure, and coverage definitions are compatible Correct answer.
  2. Whether a single unweighted average of the two measures can replace their product
  3. Whether severity has been rounded before multiplying
  4. Whether using the same calendar label alone makes the loss bases compatible

Explanation

The frequency-severity identity requires compatible definitions. Mixing coverage bases or populations can produce a number that does not represent the intended loss cost.

Question 6Property-casualty insurance quantities and foundations

A stated model gives 800 exposures, frequency 0.10 per exposure, and severity $4,500. What aggregate expected loss follows?

  1. $36,000
  2. $45,000
  3. $3,600,000
  4. $360,000 Correct answer.

Explanation

Expected count is eighty from 800 times 0.10. Multiplying that count by the supplied severity gives 360,000 dollars of aggregate expected loss.

Question 7Property-casualty insurance quantities and foundations

An annual premium is $1,200 and the problem specifies uniform pro-rata earning with no cancellation. What is earned after six of twelve months?

  1. $720
  2. $1,200
  3. $100
  4. $600 Correct answer.

Explanation

Under the explicit uniform earning assumption, six months represents half the annual period. Half of twelve hundred dollars is six hundred dollars of earned premium.

Question 8Property-casualty insurance quantities and foundations

For this exercise, loss ratio includes $350,000 losses plus $50,000 loss-adjustment expense over $800,000 earned premium. What is the ratio?

  1. 6.25%
  2. 43.75%
  3. 200%
  4. 50% Correct answer.

Explanation

The defined numerator is four hundred thousand dollars. Dividing it by eight hundred thousand gives a fifty-percent ratio under the problem's explicit expense definition.

Question 9Property-casualty insurance quantities and foundations

A simplified combined ratio is defined as loss ratio plus expense ratio. With 65% and 30%, respectively, what is it?

  1. 35%
  2. 105%
  3. 95% Correct answer.
  4. 19.5%

Explanation

Apply the supplied additive definition: sixty-five percent plus thirty percent equals ninety-five percent. This is not a multiplication or a complete statement about investment income.

Question 10Property-casualty insurance quantities and foundations

A pricing exercise requires expected loss $300, fixed expense $30, variable expense 20% of premium, and profit provision 5% of premium. What premium solves these assumptions?

  1. $412.50
  2. $330
  3. $440 Correct answer.
  4. $264

Explanation

Premium must satisfy P = 330 + 0.25P. Dividing 330 by 0.75 gives 440 dollars under the stated simplified pricing assumptions.

Question 11Property-casualty insurance quantities and foundations

Reported incurred loss is defined here as paid loss plus case reserve. Paid is $200,000 and case reserve $80,000. What reported incurred amount results?

  1. $120,000
  2. $280,000 Correct answer.
  3. $200,000
  4. $16,000,000

Explanation

Use the stated definition and add the two matching amounts. Two hundred thousand plus eighty thousand equals 280,000 dollars of reported incurred loss.

Question 12Property-casualty insurance quantities and foundations

A supplied development factor to ultimate is 1.25 and reported loss is $400,000. What ultimate estimate follows that factor?

  1. $400,001.25
  2. $100,000
  3. $320,000
  4. $500,000 Correct answer.

Explanation

Multiply the reported loss by the given factor. Four hundred thousand times 1.25 gives a five-hundred-thousand-dollar estimate, subject to the factor's assumptions.

Question 13Property-casualty insurance quantities and foundations

In a simplified reserving exercise, ultimate loss is $600,000 and paid loss is $450,000. What total unpaid amount is implied, excluding all other components?

  1. $450,000
  2. $150,000 Correct answer.
  3. $600,000
  4. $1,050,000

Explanation

Under the stated definition, unpaid loss is ultimate less already paid. Subtracting 450,000 from 600,000 gives 150,000 dollars remaining unpaid.

Question 14Property-casualty insurance quantities and foundations

A problem defines IBNR as ultimate less reported incurred. Ultimate is $100 million and reported incurred is $80 million. What IBNR results?

  1. $20 million Correct answer.
  2. $80 million
  3. $180 million
  4. $1.25 million

Explanation

Apply the problem's stated definition: one hundred million minus eighty million gives twenty million. Different real-world terminology should not be silently substituted for that definition.

Question 15Property-casualty insurance quantities and foundations

A cost of $1,000 is trended for two years at a supplied annual multiplicative factor of 1.05. What is the trended amount?

  1. $1,050.00
  2. $1,102.50 Correct answer.
  3. $1,100.00
  4. $2,100.00

Explanation

Two years of multiplicative trend means multiplying by 1.05 twice. One thousand times 1.1025 equals 1,102.50 dollars under the stated assumption.

Question 16Property-casualty insurance quantities and foundations

A simplified policy pays loss above a $1,000 deductible with no limit. For a covered $5,000 loss, what payment follows those terms?

  1. $6,000
  2. $4,000 Correct answer.
  3. $1,000
  4. $5,000

Explanation

The stated payment is the covered loss minus the deductible when positive. Five thousand less one thousand gives four thousand dollars in this exercise.

Question 17Property-casualty insurance quantities and foundations

For this exercise, payment is min(max(loss - $2,000, 0), $10,000). What payment results from a $25,000 loss?

  1. $25,000
  2. $8,000
  3. $23,000
  4. $10,000 Correct answer.

Explanation

The deductible first leaves 23,000 dollars, but the supplied payment limit caps that at ten thousand. Apply the explicit order in the stated formula.

Question 18Property-casualty insurance quantities and foundations

A loss measure is $1,000 gross and the matching recovery is $250. What net amount follows the simple gross-minus-recovery definition?

  1. $250
  2. $4
  3. $1,250
  4. $750 Correct answer.

Explanation

Subtract the recovery from the matching gross loss. One thousand minus 250 gives 750 dollars, assuming no other components in the supplied definition.

Question 19Property-casualty insurance quantities and foundations

Group A has 20 claims and 100 exposures; group B has 10 claims and 900 exposures. What pooled frequency uses total claims over total exposure?

  1. 0.20
  2. 0.1056
  3. 0.0111
  4. 0.03 Correct answer.

Explanation

Add thirty claims and one thousand exposures before dividing. The pooled rate is 0.03, not the unweighted average of the two different-exposure rates.

Question 20Property-casualty insurance quantities and foundations

A portfolio's overall severity rises after its mix shifts toward a higher-severity coverage, while within-coverage means are stable. What is a plausible explanation?

  1. Composition change rather than necessarily rising severity within each coverage Correct answer.
  2. The increase proves the loss data are corrupted
  3. Coverage mix cannot affect an overall average
  4. Every coverage must have experienced identical cost inflation

Explanation

An aggregate average depends on the mix of observations. A shift toward a higher-mean group can raise the overall mean without changing either group's conditional mean.

Question 21Property-casualty insurance quantities and foundations

Claim count N is Poisson with mean 0.20. Using P(N = 0) = exp(-0.20), what is the probability of at least one claim, approximately?

  1. 0.1813 Correct answer.
  2. 0.8187
  3. 0.2000
  4. 1.2000

Explanation

At least one claim is the complement of zero claims. One minus exp of negative 0.20 is approximately 0.1813 under the supplied Poisson model.

Question 22Property-casualty insurance quantities and foundations

Use Var(S) = E[N]Var(X) + Var(N)E[X]^2 for independent claim count and severities. If N is Poisson with mean 2, E[X] = 100, and Var(X) = 2,500, what is Var(S)?

  1. 10,000
  2. 20,000
  3. 25,000 Correct answer.
  4. 5,000

Explanation

For the supplied Poisson count, variance also equals two. The formula gives two times 2,500 plus two times 10,000, totaling 25,000.

Question 23Property-casualty insurance quantities and foundations

A one-event simplified model has a 0.10 probability of a $1,000 loss and otherwise zero loss. What is its expected loss?

  1. $100 Correct answer.
  2. $10,000
  3. $900
  4. $1,000

Explanation

Expected loss weights the positive amount by its probability and adds the zero outcome. One tenth times one thousand equals one hundred dollars.

Question 24Property-casualty insurance quantities and foundations

A territory variable improves prediction in historical claims data. What does that fact alone establish?

  1. Automatic permission to use it in any rating jurisdiction
  2. Proof that no proxy or fairness issue can exist
  3. A causal effect of moving every policyholder
  4. A predictive association in that setting, not automatic causal or regulatory approval Correct answer.

Explanation

Predictive performance is evidence about the fitted setting. Causation, lawful rating use, and proxy concerns require additional reasoning beyond an observed association.

Question 25Property-casualty insurance quantities and foundations

A learner treats this multiple-choice bank as the entire CSPA qualification. What important scope correction is needed?

  1. CSPA is simply another name for the separate CAS PCPA assessment
  2. CSPA includes multiple component assessments and practical and professional requirements Correct answer.
  3. A high score here automatically awards the credential
  4. All official requirements are replaced by one generic question count

Explanation

The iCAS CSPA pathway includes component learning and a practical project with professional responsibilities. This original bank cannot replace the actual current credential requirements.

Question 26Data preparation, features, and visualization

A policy table has one exposure row per policy, while a payment table has several rows per claim. What must be checked before summing a joined table?

  1. Only whether the joined table has more columns
  2. Whether join multiplicity duplicates exposure or other totals Correct answer.
  3. Whether a larger row count automatically makes the analysis more accurate
  4. Only whether every output row has a policy number

Explanation

Joining different observational units can multiply values unintentionally. Reconcile keys, row counts, and aggregate quantities so repeated transaction rows do not duplicate policy-level exposure.

Question 27Data preparation, features, and visualization

A left join retains all policy rows but many claim fields are missing. What should be investigated?

  1. Only whether the total policy row count was preserved
  2. The assumption that every missing claim field proves no claim occurred
  3. Unmatched keys and the meaning of missing claim information Correct answer.
  4. The assumption that left joins can never have key problems

Explanation

Missing joined fields can result from unmatched identifiers or a genuinely absent record. Their meaning must be established before coding them as a no-claim outcome.

Question 28Data preparation, features, and visualization

For one key, table A has two rows and table B has three rows. A standard many-to-many join matches every pair. How many rows result for that key?

  1. 2
  2. 5
  3. 3
  4. 6 Correct answer.

Explanation

Each of the two rows in A matches each of the three rows in B. The Cartesian pairing produces six rows for that key.

Question 29Data preparation, features, and visualization

Multiple annual records from the same policyholder are randomly divided between training and test sets, but the intended evaluation is on new policyholders. What is the concern?

  1. The policyholder identifier becomes irrelevant after shuffling
  2. Related records can make the test less representative of genuinely new people Correct answer.
  3. Only the test-set file order affects generalization
  4. Random row splitting guarantees independence of policyholders

Explanation

The split should match the intended deployment population. Related observations across sets can share information and make performance on unseen policyholders look more favorable than it is.

Question 30Data preparation, features, and visualization

Training observations have values 4, 6, and 8 for a feature. A mean imputer is fitted on training data. What value fills a missing test observation?

  1. 18
  2. The mean of training and all test values combined
  3. 8
  4. 6 Correct answer.

Explanation

The fitted training mean is eighteen divided by three, or six. Apply that learned value to held-out data rather than refit using test observations.

Question 31Data preparation, features, and visualization

A standardizer fitted on training data has mean 100 and standard deviation 20. What transformed value corresponds to x = 140?

  1. 2 Correct answer.
  2. 0.5
  3. 7
  4. 40

Explanation

Standardization under the supplied parameters is x minus mean divided by standard deviation. Forty divided by twenty gives a standardized value of two.

Question 32Data preparation, features, and visualization

For the values 1, 2, 3, 4, and 100, what is the median?

  1. 4
  2. 100
  3. 3 Correct answer.
  4. 22

Explanation

The ordered list has five observations, so the middle, third value is three. The large final value affects the mean much more than the median.

Question 33Data preparation, features, and visualization

A missingness indicator improves validation performance. What interpretation is justified without further causal evidence?

  1. The indicator guarantees stability after collection processes change
  2. The missingness pattern carries predictive information in this setting Correct answer.
  3. Missingness itself must cause the insurance outcome
  4. Every missing value should therefore be imputed as zero

Explanation

Missingness can encode a collection process or other associated information. Predictive value alone does not establish a causal mechanism or future stability under changed data collection.

Question 34Data preparation, features, and visualization

A field uses zero for a genuine observed amount and blank for unavailable information. What is wrong with replacing every blank by zero without review?

  1. It erases a meaningful distinction between missing and observed zero Correct answer.
  2. The substitution always preserves the variable's meaning
  3. A numerical value cannot represent unavailable information
  4. Zero and missing are necessarily identical in every model

Explanation

Imputation should respect how the field was collected and used. An observed zero and an unavailable amount can have different implications for interpretation and prediction.

Question 35Data preparation, features, and visualization

A very large claim is verified as genuine. An analyst wants to delete it solely to reduce training error. What should be considered?

  1. That every extreme loss is necessarily a data entry error
  2. That lower training error alone proves deletion is appropriate
  3. That rare observations have no effect on expected insurance cost
  4. Whether removal changes the intended loss target and tail representation Correct answer.

Explanation

Insurance losses may be heavy-tailed, and genuine large claims can matter to the target. Exclusion needs a justified objective and validation, not merely a better training metric.

Question 36Data preparation, features, and visualization

A categorical target encoding uses each training row's outcome in its own category mean. What leakage control is especially relevant?

  1. Drop the category names after encoding to remove leakage
  2. Use the same target-inclusive value and rely on shuffling
  3. Encode using the final test labels to improve stability
  4. Generate training encodings out of fold within the appropriate training split Correct answer.

Explanation

A target-derived predictor can leak the response back into its own training representation. Appropriate out-of-fold construction limits that direct use while keeping validation separation intact.

Question 37Data preparation, features, and visualization

An unordered vehicle-type label is encoded as 1, 2, and 3 in a linear model with one numeric coefficient. What assumption may be introduced?

  1. A separate unrelated mean for every category automatically
  2. A guarantee that category 3 has the most exposure
  3. An ordered equally spaced relationship not inherent in the categories Correct answer.
  4. Complete removal of all category information

Explanation

A single numerical slope treats code increments as meaningful equal steps. That structure may be inappropriate for labels that have no natural order.

Question 38Data preparation, features, and visualization

A model has an intercept and a nominal factor with four levels. For ordinary reference coding without redundant columns, how many indicators are used?

  1. 3 Correct answer.
  2. 1
  3. 4
  4. 5

Explanation

With an intercept, one factor level serves as the reference and three indicators represent the others. Including all four indicators creates a linear dependency in this coding.

Question 39Data preparation, features, and visualization

A deployment record contains a category absent from training. What is the best preparation strategy?

  1. Change the input spelling until it matches a known category
  2. Assign the highest-risk category automatically without evaluation
  3. Refit the entire model using that record's unknown outcome
  4. Define and validate an explicit handling rule in the fitted pipeline Correct answer.

Explanation

Unseen levels are a foreseeable deployment condition. A documented compatible rule should preserve pipeline behavior without using unavailable outcomes or arbitrary risk assumptions.

Question 40Data preparation, features, and visualization

Training one-hot columns are ordered A, B, C, but a scoring table supplies C, A, B with no alignment. What can go wrong?

  1. The model automatically relearns the intended meaning
  2. Only the output file name can change
  3. Column order never matters to any model interface
  4. Values can be assigned to the wrong fitted coefficients Correct answer.

Explanation

A fitted model expects the same feature semantics and arrangement used in training. Misaligned columns can change predictions even when all category names appear somewhere.

Question 41Data preparation, features, and visualization

A renewal model is scored on January 1. A proposed feature counts claims reported through March 31 of that year. What is the issue?

  1. Using a later report date is harmless if accident dates are earlier
  2. All historical-looking aggregates are automatically safe
  3. The feature includes information unavailable at the scoring date Correct answer.
  4. A quarterly total cannot contain future information

Explanation

Prediction-time availability depends on when the information is known. A feature using later reports can leak future information even if events themselves occurred earlier.

Question 42Data preparation, features, and visualization

An event occurred before the prediction date but was entered into the source system afterward. What should feature construction respect?

  1. The information's actual availability at prediction time Correct answer.
  2. Only the event's occurrence date regardless of availability
  3. The rule that retrospective databases reproduce past availability automatically
  4. Only whether the eventual outcome is favorable

Explanation

Backtesting must reflect what could have been known then. An earlier event date does not establish that a delayed entry was available for the historical prediction.

Question 43Data preparation, features, and visualization

A feature is loss divided by exposure, and some rows have zero exposure. What must be addressed?

  1. Assuming division by zero equals zero by definition
  2. Replacing every undefined ratio by an arbitrarily large number
  3. Discarding all data without checking the observation process
  4. The meaning of those rows and a justified handling rule for the undefined ratio Correct answer.

Explanation

A zero denominator requires analysis of data definitions and intended use. An arbitrary numerical fill can misrepresent the quantity or hide a source-data problem.

Question 44Data preparation, features, and visualization

A territory plot shows a high mean loss for a group with three exposures. What additional information is essential to interpretation?

  1. A guarantee that it is the highest-risk territory in the population
  2. Its small volume and resulting uncertainty Correct answer.
  3. Only the portfolio-wide mean, without the group exposure count
  4. The assumption that group means have equal credibility regardless of volume

Explanation

A small cell can have an unstable mean. Exposure, outcome variability, and uncertainty should accompany a comparison rather than presenting all groups as equally well supported.

Question 45Data preparation, features, and visualization

A histogram uses unequal bin widths. What is important when the intended graphic represents a probability density?

  1. Equal raw counts always require equal heights regardless of width
  2. The widest bin must always be the tallest
  3. Heights must account for width so areas have the intended meaning Correct answer.
  4. Bin boundaries determine the response's causal effect

Explanation

In a density histogram, probability is represented by area. Unequal widths require appropriate height scaling so visual area does not misstate relative frequency.

Question 46Data preparation, features, and visualization

A positive-loss scatterplot uses logarithmic axes. What should the analyst explain?

  1. The transformed scale and its multiplicative interpretation, including treatment of nonpositive values Correct answer.
  2. That log axes remove every outlier automatically
  3. That equal visual gaps necessarily represent equal dollar differences
  4. That zero can be logged without any definition change

Explanation

A logarithmic scale changes how distances should be read and is not defined ordinarily at zero. Explain the scale and exclusions rather than imply unchanged linear-dollar interpretation.

Question 47Data preparation, features, and visualization

A boxplot flags a claim beyond its whisker rule. What is established?

  1. The claim must be fraudulent
  2. The claim is impossible under every loss distribution
  3. It can be removed without affecting the target definition
  4. It is unusual under that display rule, not necessarily erroneous Correct answer.

Explanation

A boxplot rule identifies relative extremeness in a dataset. Further validation and domain reasoning are needed before classifying the observation as error, fraud, or irrelevant.

Question 48Data preparation, features, and visualization

A sample deliberately overrepresents a small territory. What may be needed to estimate the original portfolio's overall mean?

  1. An assumption that territory proportions no longer matter
  2. Appropriate sampling weights and knowledge of the sampling design Correct answer.
  3. The unweighted sample mean as a guaranteed population mean
  4. The same unweighted mean after randomly reordering sample rows

Explanation

A disproportionate sample changes the observed composition. Population estimation may require weights consistent with the design rather than treating all sampled rows as equally representative.

Question 49Data preparation, features, and visualization

Training data are downsampled to increase the positive-class fraction. What issue matters when interpreting raw predicted probabilities for the original population?

  1. That class proportions affect only file size
  2. That precision cannot depend on prevalence
  3. The changed class prior and any required probability adjustment or calibration Correct answer.
  4. That downsampling guarantees original-population calibration

Explanation

Resampling can change the relationship between fitted probabilities and original-population event rates. Appropriate validation and correction may be needed for probability interpretation.

Question 50Data preparation, features, and visualization

Feature selection was performed using all labels before cross-validation. Why can the reported validation score be optimistic?

  1. Held-out outcomes influenced which predictors entered the model Correct answer.
  2. Cross-validation always repairs earlier full-data decisions
  3. Selecting fewer columns removes every selection bias
  4. Only feature scaling can leak information

Explanation

Selection is part of model fitting. If validation labels guide it, information has crossed the intended boundary even when the later estimator is refitted within folds.