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CFA Level I Quantitative Methods Practice Questions

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8
Updated
Nov 3, 2026
Share of the exam
6-9%
  1. Question 1, easy

    A manager's monthly return distribution is unknown and clearly non-normal. Using Chebyshev's inequality, at least what proportion of the observations lies within 2.5 standard deviations of the mean?

    • A.60%
    • B.75%
    • C.95%
    • D.84%
    Show answer and explanationHide answer

    Correct answer: D. 84%

    Chebyshev's inequality holds for any distribution and states that at least 1 - 1/k^2 of observations fall within k standard deviations of the mean. With k = 2.5, the bound is 1 - 1/6.25 = 0.84, or 84%.

    • A. This is 1 - 1/k with k = 2.5, which uses the wrong formula and omits the square.
    • B. This is the Chebyshev bound for k = 2, not 2.5.
    • C. This is the approximate share within two standard deviations of a normal distribution, which cannot be assumed here.
    • D. Correct: 1 - 1/(2.5)^2 = 84%.
  2. Question 2, easy

    A data set of annual advisor bonuses is strongly right-skewed (positively skewed), with a small number of very large bonuses pulling the tail. Compared to the median, what will typically be true of the mean?

    • A.The mode will be higher than both the mean and median because it captures the most frequent value in the distribution
    • B.The median will always be higher than the mean in a right-skewed distribution
    • C.The mean and median will always be exactly equal regardless of skew
    • D.The mean will typically be higher than the median, because the extreme high values pull the mean upward
    Show answer and explanationHide answer

    Correct answer: D. The mean will typically be higher than the median, because the extreme high values pull the mean upward

    In a right-skewed distribution, extreme values in the long right tail pull the arithmetic mean upward more than they affect the median. The typical ordering is mode < median < mean for right-skewed data.

    • A. Incorrect. The mode doesn't work this way; in a typical right-skewed unimodal distribution the ordering is mode < median < mean, not mode above both.
    • B. Incorrect. This states the opposite relationship; in a right-skewed distribution the mean typically exceeds the median, not the reverse.
    • C. Incorrect. Mean and median are only exactly equal in a symmetric distribution; skew by definition creates a gap between them.
    • D. Correct. In a right-skewed (positively skewed) distribution, a few extreme high values pull the mean above the median.
  3. Question 3, medium

    A sample of five monthly returns (in percent) is: 4, 8, 6, 10, 12. What is the sample standard deviation?

    • A.2.83
    • B.3.16
    • C.8.00
    • D.10.00
    Show answer and explanationHide answer

    Correct answer: B. 3.16

    Mean = 8. Squared deviations sum to 40. Sample variance = 40/(n−1) = 40/4 = 10, so sample standard deviation = √10 ≈ 3.16. Choice A mistakenly divides by n (population convention). Choice C reports the variance, not std dev.

    • A. 2.83 equals the square root of 8, which results from dividing the sum of squared deviations by n (the population convention) instead of n-1, then taking the square root.
    • B. Mean = 8; squared deviations sum to 40; sample variance = 40/(5-1) = 10; sample standard deviation = square root of 10, approximately 3.16.
    • C. 8.00 is the population variance (40/5), not a standard deviation. The square root step was never taken.
    • D. 10.00 is the correct sample variance itself (40/4), but the square root step needed to convert variance into standard deviation was omitted.
  4. Question 4, medium

    In hypothesis testing, a Type I error occurs when:

    • A.The null hypothesis is true, but the test fails to reject it at the 5% level.
    • B.The null hypothesis is false, and the test correctly rejects it.
    • C.The null hypothesis is true, but the test incorrectly rejects it.
    • D.The null hypothesis is false, but the test fails to reject it.
    Show answer and explanationHide answer

    Correct answer: C. The null hypothesis is true, but the test incorrectly rejects it.

    A Type I error is the rejection of a true null hypothesis. A 'false positive.' Its probability is the test's significance level, α. A Type II error, by contrast, is the failure to reject a false null hypothesis. A 'false negative,' whose probability is denoted β.

    • A. Incorrect. This describes a correct decision (true null, correctly not rejected) at the 5% level, not an error.
    • B. Incorrect. This describes a correct decision (false null, correctly rejected), related to the test's power, not an error.
    • C. Correct. Rejecting a true null hypothesis is precisely the definition of a Type I error, with probability α.
    • D. Incorrect. This describes a Type II error, not a Type I error.
  5. Question 5, medium

    An analyst has a random sample of 20 monthly returns and wants to construct a 95% confidence interval for the mean return. The population variance is unknown. Which distribution should the analyst use to obtain the appropriate reliability factor?

    • A.Student's t-distribution with 20 degrees of freedom
    • B.The standard normal (z) distribution
    • C.Student's t-distribution with 19 degrees of freedom
    • D.The chi-square distribution with 19 degrees of freedom
    Show answer and explanationHide answer

    Correct answer: C. Student's t-distribution with 19 degrees of freedom

    When the population variance is unknown and the sample is small, the t-distribution with n − 1 degrees of freedom is the appropriate basis for the reliability factor. With n = 20, degrees of freedom = 19.

    • A. Incorrect. This uses the correct distribution but the wrong degrees of freedom (should be n − 1, not n).
    • B. Incorrect. The z-distribution is appropriate when the population variance is known or the sample size is very large, neither of which applies here.
    • C. Correct. Unknown population variance calls for the t-distribution, and degrees of freedom equal n − 1 = 19.
    • D. Incorrect. The chi-square distribution is used for tests concerning a population variance, not for confidence intervals around a mean.
  6. Question 6, easy

    A retiree wants to withdraw $5,000 at the end of each year for the next 4 years from an account earning 8% annually. Assuming the first withdrawal occurs one year from today, what amount must she deposit today to fund these withdrawals exactly?

    • A.$20,000
    • B.$16,561
    • C.$17,885
    • D.$19,964
    Show answer and explanationHide answer

    Correct answer: B. $16,561

    This is an ordinary annuity. PV = PMT × [1 − (1+r)^−n] / r = 5,000 × [1 − 1.08^−4] / 0.08 = 5,000 × 3.31213 = $16,561 (rounded). Choice A ignores time value of money entirely. Choice C mistakenly treats the cash flows as an annuity due, and D mistakenly uses n=5.

    • A. Incorrect. $20,000 is simply 4 × $5,000 with no discounting at all; it ignores the time value of money implied by the 8% rate.
    • B. Correct. PV = $5,000 × [1−(1.08)^−4]/0.08 ≈ $5,000 × 3.3121 ≈ $16,561.
    • C. Incorrect. $17,885 doesn't match the ordinary annuity formula at 8% for 4 years; recheck whether payments were mistakenly treated as occurring at the start of each year (annuity due) instead of the end.
    • D. Incorrect. $19,964 is too close to the undiscounted total and doesn't reflect proper discounting of all four cash flows at 8%.
  7. Question 7, easy

    An investor deposits $10,000 today into an account earning 6% annual interest, compounded annually. What is the value of the account at the end of 5 years?

    • A.$13,000
    • B.$13,382
    • C.$12,625
    • D.$14,185
    Show answer and explanationHide answer

    Correct answer: B. $13,382

    FV = PV(1+r)^n = 10,000 × (1.06)^5 = 10,000 × 1.338226 = $13,382 (rounded). Choice A results from mistakenly applying simple interest. Choice C uses n=4 instead of 5, and Choice D uses n=6 instead of 5.

    • A. Incorrect. $13,000 reflects simple interest ($10,000 × 6% × 5 years added on top), not annual compounding.
    • B. Correct. FV = $10,000 × 1.06^5 = $13,382.
    • C. Incorrect. $12,625 corresponds to compounding over only 4 years (1.06^4), one period short of the required 5.
    • D. Incorrect. $14,185 corresponds to compounding over 6 years (1.06^6), one period too many.
  8. Question 8, medium

    A bank advertises a savings account with a stated (nominal) annual interest rate of 6%, compounded monthly. What is the effective annual rate (EAR)?

    • A.6.00%
    • B.6.09%
    • C.6.17%
    • D.6.30%
    Show answer and explanationHide answer

    Correct answer: C. 6.17%

    EAR = (1 + periodic rate)^m − 1 = (1 + 0.06/12)^12 − 1 ≈ 1.0617 − 1 = 6.17%. Monthly compounding produces an EAR above the stated annual rate because interest earned within the year itself earns additional interest.

    • A. Wrong. This is just the stated nominal annual rate, ignoring the effect of intra-year compounding.
    • B. Wrong. Resembles a value from assuming an incorrect (e.g., quarterly) compounding frequency.
    • C. Correct.
    • D. Wrong. Overstates the compounding effect, consistent with assuming a higher compounding frequency than monthly.

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