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CFA® Level I Chartered Financial Analyst Questions 1031–1035 (page: 208)

Updated On: 9-Feb-2026

What is the Net Present Value of this series of annual cash flows using an interest rate of 15% per year: Year
0: <$25,000>, Year 1: $5,000, Year 2: $0, Year 3: $30,000? (Note that the <> are used to indicate a negative number).

  1. $14.37
  2. <$926.69>
  3. $258.11
  4. <$291.21>
  5. <$204.58>

Answer(s): B

Explanation:

On the BAII Plus, press CF 2nd CLRWork 25000 +/- ENTER DownArrow 5000 ENTER DownArrow DownArrow 0 ENTER DownArrow DownArrow 30000 ENTER DownArrow DownArrow 2nd Quit. Then press NPV 15 ENTER DownArrow CPT. On the HP12C, press these keys: 25000 CHS BlueShift CFo 5000 BlueShift CFj 0 BlueShift CFj 30000 BlueShift CFj Then press 15 i, YellowShift NPV. The answer is negative, indicating that the PV of the cash flows to be received is less than the original cash outflow. The "DownArrow" represents the downward-pointing arrow on the top row of the BAII Plus keyboard. Make sure the BAII Plus has the P/Y value set to 1.



A researcher has a sample of 600 observations from a population whose standard deviation is known to be 51.44. The mean of the sample is calculated to be 3.6. The null hypothesis is stated as Ho: mean > 8.4. The researcher should

  1. accept the null at 95% significance.
  2. reject the null at 99% significance.
  3. fail to reject the null at 95% significance.
  4. none of these answers.

Answer(s): B

Explanation:

To test the hypothesis, you need to calculate the largest z-statistic since the null hypothesis is unidirectional and to the right. This makes it the hardest to reject the null and you should always use the most stringent criterion for rejecting the null. After all, the null is the hypothesis maintained to be true by default and only a sufficient weight of evidence should be used to reject that view. The largest z-statistic calculated under the null hypothesis is equal to (3.6 - 8.4)/2.1 = -2.4. Now, the alternative hypothesis is H1: mean <= 8.4. Since the alternative is directional, we use a one-tailed test. For this, the critical value at 95% level is -1.645 and for 99%, it is -2.32. -2.32 is the value below which only 1% of the probability mass of the standard normal distribution lies. Similarly, only 5% of probability lies below -1.645). Since even the calculated z-statistic that is most favorable to the null lies in the rejection region, we can reject the null at the 99% level in a one-sided test.



If you make 22 deposits of $33 with the first deposit occurring next month, how much do you have immediately after the last deposit, if the account earns 4.44% per year, compounded annually?

  1. $666.00
  2. $1,207.81
  3. $$457.44
  4. $1,313.51
  5. $1,189.62

Answer(s): E

Explanation:

On the BAII Plus, press 22 N, 4.44 I/Y, 0 PV, 33 PMT, CPT FV. On the HP12C, press 22 n, 4.44 i, 0 PV, 33 PMT, FV. Note that the answer will be displayed as a negative number. Make sure the BAII Plus has the P/Y value set to 1.



As the alternative mean approaches the hypothesized mean, what can we say about the risk?

  1. Smaller risk of a Type II error
  2. Greater risk of a Type II error
  3. Smaller risk of a Type I error
  4. None of these answers
  5. Greater risk of a Type I error

Answer(s): B

Explanation:

When the alternative mean approaches the hypothesized mean, there is a greater possibility of accepting the null when it is actually false since they are close together.



Chances are 50-50 that a newborn baby will be a girl. For families with five children, what is the probability that all the children are girls?

  1. 0.100
  2. None of these answers
  3. 0.250
  4. 0.001
  5. 0.031

Answer(s): E

Explanation:

This is a binomial probability. The probability of getting r successes out of n trials where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p) is given by: n!(p^r)[q^(n-r)]/r!(n-r)!.
Here n = 5, r = 5,p = 0.5 and q = 0.5. Therefore we have 5!(0.5^5)(0.5^0)/5!0! = 0.03125.



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