Kostenlos und ohne Konto. Zu jeder Frage gibt es den vollständigen Lösungsweg, einschließlich der Begründung, warum jede falsche Option falsch ist. Die Fragen sind eigene Aufgaben von Preparaio auf Englisch, keine offiziellen Fragen des CFA Institute.
1. Mariela Etcheverry-Tan builds a probability tree for next year's dividend of Pukeko Ferries. In the first stage, fuel prices are either high (probability 0.40) or low (probability 0.60). If fuel prices are high, the board raises the dividend with probability 0.25. In her tree, what is the probability that fuel prices are high and the board raises the dividend?
A. 0.65, the sum of 0.40 and 0.25 B. 0.25, the probability of a raise given high fuel prices C. 0.10, the product of 0.40 and 0.25
Lösung und Erklärung Richtige Antwort: C 0.10, the product of 0.40 and 0.25
The second-stage probability of 0.25 is conditional: it applies only once fuel prices are high. The probability of reaching the end of the path 'high fuel, then raise' is therefore 0.40 × 0.25 = 0.10. Weighting each path's outcome by its joint probability and summing across all paths then gives the expected value (the total probability rule for expected value).
Warum die anderen Optionen falsch sind
A: Adding the branch probabilities is not a valid rule for a path through a tree; the probability of both events occurring cannot exceed the probability of either one. B: 0.25 is the conditional probability P(raise | high fuel); it ignores the 0.40 probability that fuel prices are high in the first place. Kernregel: In a probability tree, the joint probability of a path is the product of the probabilities along its branches: P(A and B) = P(B | A) × P(A).
Merksatz: Multiply along a branch for joint probabilities; add across branches for totals.
2. Teuila Fa'amausili uses annual data to regress the percentage change in regional sales of residential solar panels (Y) on the percentage change in retail electricity tariffs (X). The sample covariance between X and Y is 18.5, the sample variance of X is 12.5, the mean of X is 4.2% and the mean of Y is 9.6%. If electricity tariffs rise by 7.5% next year, what is the regression's forecast of the change in solar panel sales?
A. 11.10% B. 11.83% C. 14.48%
Lösung und Erklärung Richtige Antwort: C 14.48%
The slope is 18.5 / 12.5 = 1.48, so each 1 percentage point rise in tariffs is associated with a 1.48 percentage point rise in solar panel sales growth. Because the fitted line passes through the means, the intercept is 9.6 − 1.48 × 4.2 = 3.384. The forecast for a 7.5% tariff increase is 3.384 + 1.48 × 7.5 = 14.48%.
Rechenweg
Slope: b1 = Cov(X,Y) / Var(X) = 18.5 / 12.5 = 1.48 Intercept: b0 = Ybar − b1 × Xbar = 9.6 − 1.48 × 4.2 = 9.6 − 6.216 = 3.384 Forecast: Y = 3.384 + 1.48 × 7.5 = 3.384 + 11.10 = 14.484, or about 14.48% BA-II-Plus-Tastenfolge
18.5 ÷ 12.5 = STO 1; 7.5 − 4.2 = × RCL 1 + 9.6 = [14.48]
Warum die anderen Optionen falsch sind
A: This multiplies the slope by 7.5 but leaves out the intercept of 3.384, so it is not on the fitted regression line. B: This uses Var(X)/Cov(X,Y) = 0.676 as the slope; the slope is the covariance divided by the variance of X. Intermediate displayed workings may be rounded; retain full precision until the final answer.
Kernregel: The OLS slope is Cov(X,Y)/Var(X), the intercept is Ybar − b1 × Xbar, and the forecast is b0 + b1 × X.
Merksatz: Slope = Cov(X,Y)/Var(X); always add the intercept when forecasting.
3. Under a naming-rights agreement for a new indoor arena, Quercia Logistics will pay the arena's owner $23,750 at the end of each year for 16 years. Using a discount rate of 5.75% per year, what is the present value of the payments?
A. $234,481.03 B. $244,190.10 C. $258,231.03
Lösung und Erklärung Richtige Antwort: B $244,190.10
The payments are level and arrive at the end of each year, so this is an ordinary annuity with N = 16 and r = 5.75%. The annuity factor is (1 − 1.0575^−16) / 0.0575 = 10.2817, and the present value is 23,750 × 10.2817 = $244,190.10. On the calculator this is a standard TVM calculation in END mode.
Rechenweg
Ordinary annuity: N = 16, r = 0.0575, PMT = 23,750 paid at each year-end 1.0575^−16 = 0.40880 Annuity factor = (1 − 0.40880) / 0.0575 = 10.2817 PV = 23,750 × 10.2817 = 244,190.10 BA-II-Plus-Tastenfolge
Calculator setup first: 2ND [P/Y] 1 ENTER (C/Y = 1), 2ND [BGN] showing END, then 2ND [QUIT] to exit the settings. 2ND [CLR TVM] clears the TVM values but does not reset P/Y, C/Y or END/BGN. Periodic inputs below are entered manually.
2ND CLR TVM; 16 N; 5.75 I/Y; 23750 +|- PMT; 0 FV; CPT PV [244,190.10]
Warum die anderen Optionen falsch sind
A: This values only 15 payments; the agreement has 16 annual payments. C: This treats the payments as an annuity due, paid at the start of each year, which multiplies the correct value by 1.0575; the payments are made at year-end. Intermediate displayed workings may be rounded; retain full precision until the final answer.
Kernregel: The present value of an ordinary annuity is PMT × [1 − (1 + r)^−N] / r, with payments at the end of each period.
Merksatz: End-of-period payments = ordinary annuity; only payments starting today make it an annuity due.
4. Tarn Edge Wealth wants to predict which of its clients are likely to close their accounts within the next 12 months. It has ten years of client records, each marked with whether that client closed the account or stayed. The machine learning approach the firm should use is most likely:
A. unsupervised learning, because the model discovers groups of clients that behave in similar ways B. reinforcement learning, because the model is rewarded each time a prediction is correct C. supervised learning, because the model learns from past records labeled with the outcome
Lösung und Erklärung Richtige Antwort: C supervised learning, because the model learns from past records labeled with the outcome
Each historical record is labeled with the outcome the firm wants to predict (closed or stayed), so the task is a supervised classification problem. A model such as a classification tree, support vector machine or random forest can learn the link between client characteristics and the label and then score current clients. Unsupervised learning would be used if there were no labels and the goal were only to find structure.
Warum die anderen Optionen falsch sind
A: Unsupervised learning, such as clustering, finds groups in unlabeled data; it does not use the known closed-or-stayed outcome as a target. B: Reinforcement learning has an agent learn a sequence of actions through trial and reward; predicting a labeled outcome from historical records is not that kind of problem. Kernregel: Supervised learning trains a model on labeled data, inputs paired with a known target, to predict that target for new observations.
Merksatz: Labeled target = supervised; no labels = unsupervised; trial-and-reward actions = reinforcement.
5. Emeric Szalai estimates the probability that the real value of the Dunakanyar Conservatory endowment will fall over the next five years. He uses historical simulation, drawing scenarios from the endowment's monthly asset-class returns over the past 12 years, a period that contained no episode of rapid inflation. Which of his statements is least likely correct?
A. Switching to Monte Carlo with far more trials would remove the risk that my return assumptions are wrong B. My results cannot reflect an inflation shock worse than any that occurred during the look-back period C. Historical simulation does not require me to assume a particular probability distribution for asset returns
Lösung und Erklärung Richtige Antwort: A Switching to Monte Carlo with far more trials would remove the risk that my return assumptions are wrong
Monte Carlo simulation replaces the historical record with distributions and parameters that the analyst chooses. Running more trials makes the estimated outcome distribution more precise for that chosen model, but if the model or its inputs are wrong, the results remain wrong: model risk is not removed. Szalai's other two statements correctly describe historical simulation: it uses actual past returns, so no distribution has to be assumed, and it cannot produce an inflation shock more severe than any in the 12-year sample.
Warum die anderen Optionen falsch sind
B: This is correct: historical simulation can only reuse returns that actually occurred, so it misses events absent from the sample period. C: This is correct: historical simulation draws from actual past returns rather than from an assumed probability distribution. Kernregel: Historical simulation avoids distributional assumptions but cannot generate events absent from its look-back period, while Monte Carlo simulation depends on the assumed model, and adding trials reduces only sampling noise, not model risk.
Merksatz: More Monte Carlo trials cut sampling noise, not model risk; historical simulation is limited to its look-back period.
6. To estimate the average number of days that payments are overdue on its 48,000 small-business loans, the internal audit team of Kotmale Hills Savings Bank randomly selects 10 of the bank's 120 branches and reviews every loan booked at those branches. The team's sampling approach is most likely:
A. stratified random sampling, which ensures that every branch is represented in proportion to its size B. cluster sampling, which is usually cheaper but less precise than a simple random sample of equal size C. systematic sampling, because every loan at each selected branch is reviewed in turn
Lösung und Erklärung Richtige Antwort: B cluster sampling, which is usually cheaper but less precise than a simple random sample of equal size
The branches are natural groups (clusters) of loans, and the team randomly picks 10 whole branches and examines all of their loans. That is one-stage cluster sampling. It saves time and cost because auditors visit only 10 branches, but because loans within a branch tend to be similar, the estimate is usually less precise than a simple random sample of the same number of loans drawn from all 120 branches.
Warum die anderen Optionen falsch sind
A: Stratified sampling would draw a random sample from every subgroup; here 110 branches are not sampled at all. C: Systematic sampling selects every kth element from a list; reviewing all loans in randomly chosen groups is cluster sampling. Kernregel: Cluster sampling divides the population into groups that each resemble the population, randomly selects some whole groups and (in one-stage form) includes every member of the chosen groups.
Merksatz: Random whole groups = cluster sampling; random draws from every group = stratified sampling.
7. Kristoffer Mbanefo-Strand regresses the average bid-ask spread of 150 corporate bonds on the size of each bond issue. In a plot of the residuals against issue size, the residuals are centered on zero and show no curvature, but they are widely scattered for small issues and tightly clustered near zero for large issues. Which assumption of simple linear regression is most likely violated?
A. Homoskedasticity, because the variance of the errors changes with the size of the issue B. Linearity, because the residuals are spread unevenly across the range of issue sizes C. Independence of errors, because the residuals of small issues are related to one another
Lösung und Erklärung Richtige Antwort: A Homoskedasticity, because the variance of the errors changes with the size of the issue
The residual spread is large for small issues and narrows as issue size grows, which is a fan (or funnel) shape: the variance of the errors depends on the independent variable. That violates the homoskedasticity assumption, whether the fan opens toward small or large values of X. Because the residuals are centered on zero with no curvature, there is no sign of nonlinearity, and a cross-section of bonds plotted against size gives no evidence of correlated errors.
Warum die anderen Optionen falsch sind
B: A violation of linearity shows up as a curved pattern in the residuals; here they are centered on zero with no curvature, only uneven spread. C: Dependence of errors appears as runs of same-sign residuals, typically over time; a wider scatter at small issue sizes is a variance pattern, not correlation. Kernregel: Homoskedasticity requires the error variance to be the same for all values of the independent variable, so residuals whose spread changes with X signal heteroskedasticity.
Merksatz: Changing residual spread = heteroskedasticity; curvature = nonlinearity; same-sign runs = dependent errors.
8. A fund database reports that the monthly returns of the Kererū Orchards Fund over the past five years have a coefficient of variation of 1.5. Which interpretation of this figure is correct?
A. The standard deviation of monthly returns is 1.5 times the mean monthly return B. The mean monthly return is 1.5 times the standard deviation of the monthly returns C. The variance of monthly returns is 1.5 times the mean monthly return
Lösung und Erklärung Richtige Antwort: A The standard deviation of monthly returns is 1.5 times the mean monthly return
A CV of 1.5 means that s / Xbar = 1.5, so the standard deviation of the fund's monthly returns is one and a half times its mean monthly return. A higher CV indicates more risk for each unit of mean return, which is why the measure is useful for comparing assets with different average returns.
Warum die anderen Optionen falsch sind
B: This inverts the ratio: a mean equal to 1.5 times the standard deviation would correspond to a CV of about 0.67. C: The CV uses the standard deviation, not the variance; variance is in squared units and cannot be compared directly with the mean. Kernregel: The coefficient of variation is the standard deviation divided by the mean, CV = s / Xbar, so it measures risk per unit of mean return.
Merksatz: CV = standard deviation ÷ mean: risk per unit of mean return.
9. Ilona Haapala-Okoye, an analyst at Yindarra Resources Research, estimates that 22% of the junior gold miners she covers will announce an equity raise within the next six months. Among miners that raised equity in the past, 64% had reported cash below two quarters of planned spending at the preceding quarter-end; among miners that did not raise equity, 18% had reported such low cash. A miner she covers has just reported cash below two quarters of planned spending. Using Bayes' formula, what is the updated probability that this miner will announce an equity raise within six months?
A. 14.08% B. 50.07% C. 64.00%
Lösung und Erklärung Richtige Antwort: B 50.07%
The low-cash report is new information, so the 22% prior is updated with Bayes' formula. The joint probability of low cash and a raise is 0.64 × 0.22 = 0.1408. Low cash can also occur without a raise, with probability 0.18 × 0.78 = 0.1404, so the unconditional probability of a low-cash report is 0.2812. Dividing gives 0.1408 / 0.2812 = 50.07%, well above the prior because low cash is far more common among miners that go on to raise equity.
Rechenweg
Joint probability: P(low cash and raise) = 0.64 × 0.22 = 0.1408 Complement branch: P(low cash and no raise) = 0.18 × (1 − 0.22) = 0.18 × 0.78 = 0.1404 Unconditional probability: P(low cash) = 0.1408 + 0.1404 = 0.2812 Posterior: P(raise | low cash) = 0.1408 / 0.2812 = 0.5007, or 50.07% BA-II-Plus-Tastenfolge
0.64 × 0.22 = STO 1; 1 − 0.22 = × 0.18 = + RCL 1 = STO 2; RCL 1 ÷ RCL 2 = × 100 = [50.07]
Warum die anderen Optionen falsch sind
A: This is the joint probability of low cash and a raise (0.64 × 0.22); it still has to be divided by the probability of a low-cash report, 0.2812. C: This is P(low cash | raise), the likelihood; the question asks for the reverse conditional, P(raise | low cash). Intermediate displayed workings may be rounded; retain full precision until the final answer.
Kernregel: Bayes' formula gives the posterior as P(raise | low cash) = P(low cash | raise) × P(raise) / P(low cash), where P(low cash) sums both branches of the tree.
Merksatz: Posterior = joint probability ÷ total probability of the evidence, with both branches in the denominator.
10. Rukmini Joglekar studies the discount to net asset value at which listed closed-end infrastructure funds trade. In a random sample of 26 funds, the mean discount is 8.4% of net asset value and the sample standard deviation is 5.7%. The population of discounts is approximately normal, and its variance is unknown. Using the critical values in the exhibit, the lower limit of a 95% confidence interval for the population mean discount is closest to:
Selected critical values of Student's t (probability in the upper tail) Degrees of freedom p = 0.05 p = 0.025 24 1.711 2.064 25 1.708 2.060 26 1.706 2.056
A. 6.10% B. 7.95% C. 10.70%
Lösung und Erklärung Richtige Antwort: A 6.10%
Because the population variance is unknown, Joglekar uses the t-distribution with 26 − 1 = 25 degrees of freedom. A two-sided 95% interval places 2.5% in each tail, so the reliability factor is 2.060. The standard error is 5.7 / √26 = 1.1179, giving a half-width of 2.060 × 1.1179 = 2.3028. The lower limit is 8.4 − 2.3028 = 6.10% of net asset value.
Rechenweg
Reliability factor: t with df = 26 − 1 = 25 and 0.025 in each tail = 2.060 Standard error = s / √n = 5.7 / √26 = 5.7 / 5.0990 = 1.1179 Half-width = 2.060 × 1.1179 = 2.3028 Lower limit = 8.4 − 2.3028 = 6.0972, or about 6.10% BA-II-Plus-Tastenfolge
5.7 ÷ 26 √x × 2.06 = +|- + 8.4 = [6.10]
Warum die anderen Optionen falsch sind
B: This divides the standard deviation by n (26) instead of √26, which shrinks the standard error and gives an interval that is far too narrow. C: This is the upper limit of the interval, 8.4 + 2.3028; the question asks for the lower limit. Intermediate displayed workings may be rounded; retain full precision until the final answer.
Kernregel: With an unknown population variance and a normal population, the confidence interval is Xbar ± t(α/2, n − 1) × s / √n.
Merksatz: Confidence interval = point estimate ± t(α/2, n − 1) × s/√n; use n − 1 degrees of freedom and half of α in each tail.
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