Free Online CFA Level 1 Mock Exam 1

Quantitative Methods

4. Cassandra Turnbull is planning a college fund for her daughter, who will be attending college in six years. College costs are expected to be $55,000 per year, payable at the beginning of each of the four years. Turnbull plans to invest in a fund that yields 6%. Turnbull currently has $100,000 in savings.

Must she invest equal amounts at the end of Years 1, 2, 3, 4, and 5 to fund her daughter’s four years of college?

A. USD10,069.

B. USD18,116.

C. USD24,000.

Correct Answer: A

Answer Explanation:

N=4, I/Y=6, PMT=-55,000, FV=0, CPT PV.

PV=190,580.81.

Note: This is the amount needed for the year before the first payment of $55,000.Cassandra needs $190,580.81 at the end of year 5.

Now we need to calculate the equivalent amount that needs to be paid at the end of each year.

N=5, I/Y=6, PV=100,000, FV=-190,580.81, CPT PMT.

PMT=10,068.7. she needs to save an additional $10,068.7.

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5. An equity analyst has categorized a number of stocks according to the industry sector to which they belong. Which measure would this analyst most likely use?

A. Interval.

B. Nominal.

C. Ordinal.

Correct Answer: B

Answer Explanation:

Nominal scales categorize data but do not rank them. In ordinal scales, data are graded, whereas in interval scales, data are graded and separated by equal intervals.

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6. Steve Rowling is analyzing the historical performance of a portfolio manager who claims to have outperformed a benchmark 70% of the time over a one-year period. Steve has eight months of monthly performance history for the portfolio manager. The portfolio manager has outperformed the benchmark for five months.

If all of the outperformance results against the benchmark are independent and the portfolio manager performs as he claims, what is the probability of observing two or fewer profitable recommendations out of a total of eight recommendations?

A. 7.5 %

B. 5.76 %

C. 1.13 %

Correct Answer: C

Answer Explanation:

The observed success rate is 0.7 or 70%.

According to the binomial distribution with n = 8 and p = 0.70, the probability of two or fewer successes is F(2) = p(2) + p(1) + p(0), where p(2), p(1), and p(0) are the probabilities of 2, 1, and 0 successes, respectively.

We have P(x) = nCx px(1-p)n-x = [n!/(n -x)!x!] px(1 – p)n-x

p(2) = (8!/2!6!)(0.70)2(0.30)6 = 28(0.000357) = 0.010002

p(1) = (8!/1!7!)(0.70)1(0.30)7 = 8(0.000153) = 0.001224

p(0) = (8!/0!8!)(0.70)0(0.30)8 = 1(0.000066) = 0.000066

Adding up all these probabilities gives F(2) = 0.010002 + 0.001224 + 0.000066 = 0.011292, or 1.13%. The probability of observing 2 or fewer profitable recommendations out of a total of 8 profitable recommendations is 1.13%.

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