You need €25 000 in 6 years. An account pays 4% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €25 0004% / 12n = 6
PV = €
2.Time value of moneyHard
You save €500 at the end of every month for 25 years. The account pays 4.2% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €5004.2% / 12N = 300
FV = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 4: €8 000, €5 000, €9 000, €8 000. The cost of capital is 6%. Find the net present value NPV.
Project A costs €20 000 and returns €11 000, €8 000, €4 500 at the end of years 1 to 3. Project B costs €25 000 and returns €6 500, €9 500, €14 000. Find the crossover rate r* at which both projects have the same NPV, in percent.
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 4% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 4%n = 7
P = €
6.Bond pricing & yieldHard
A bond with 12 years to maturity (face value €1 000, annual coupon 7%) sells for €1 228. The issuer can call it in 4 years at a call price of €1 070. Find the yield to call YTC in percent.
F = €1 000coupon = 7%P = €1 228call price = €1 070call in 4 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €6 000 000 and debt worth €3 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 5% and the tax rate is 21%. Find the WACC.
A company's business has an unlevered (asset) beta of 0.8. It plans a debt-to-equity ratio of 1.5. The tax rate is 28%, the risk-free rate is 1.5% and the market risk premium is 7%. Find the cost of equity re at the new capital structure.
βU = 0.8D/E = 1.5T = 28%rf = 1.5%rm − rf = 7%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 30, 43, 50, 18, 25, 50.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is above 78.75?
μ = 60σ = 15x = 78.75
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 10, 20, 30, 40p(x): 0.35, 0.3, 0.25, 0.1
σ =
12.ProbabilityHard
A game costs €3 to play. You win €15 with probability 0.25 and €30 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€15: p = 0.25€30: p = 0.01
E(G) = €
13.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 108.2 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 108.2 s
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 500σ = 40P = 97.5%
x = h
15.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 1.9. What is the margin of error E₂ if the same study uses 450 observations instead?
90%n₁ = 50E₁ = 1.9n₂ = 450
E₂ =
16.Confidence intervalsHard
A pilot study suggests that about 0.7 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.01?
p ≈ 0.7E = 0.0190%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 80; the population standard deviation is known to be 15. A random sample of 100 observations has a mean of 83. Test H₀: μ = 80 against H₁: μ > 80. Find the p-value p.
μ₀ = 80σ = 15n = 100x̄ = 83H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 101.5. The population standard deviation is 5. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 101.5σ = 5n = 8199%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.
n = 5x = 2.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 19674 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Rate and number of periods
i = 0.04/12 = 0.003333
N = 12·6 = 72
Present value
PV = FV/(1 + i)N = 25 000/1.271 = €19 673.55
Answer
PV = €19 673.55
2.Time value of money
Answer: FV = 264631 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
You paid in €150 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 5890 €
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
Present value of each cash flow
CF0 = −20 000
CF1/(1 + 0.06)1 = 8 000/1.06 = 7 547.17
CF2/(1 + 0.06)2 = 5 000/1.124 = 4 449.98
CF3/(1 + 0.06)3 = 9 000/1.191 = 7 556.57
CF4/(1 + 0.06)4 = 8 000/1.262 = 6 336.75
Add them up
NPV = €5 890.47
Answer
NPV = €5 890.47
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: r* = 6.315 %
Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.
Difference in cash flows
ΔCF0 = −25 000 − (−20 000) = −5 000
ΔCF1 = 6 500 − 11 000 = −4 500
ΔCF2 = 9 500 − 8 000 = 1 500
ΔCF3 = 14 000 − 4 500 = 9 500
IRR of the difference
NPVΔ(6%) = 66.09
NPVΔ(7%) = −140.62
r* = 6.315%
Answer
r* = 6.315%
Check: at 6.315% both NPVs are €1 169.12.
5.Bond pricing & yield
Answer: P = 1121 €
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Per period
C = €30.00, i = 0.02, N = 14
Present value of the coupons
€30.00·(1 − 0.7579)/0.02 = €363.19
Present value of the face value
€1 000·0.7579 = €757.88
Price
P = €1 121.06
Answer
P = €1 121.06
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTC = 2.607 %
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Equation for the yield
1228 = 70·(1 − (1 + y)−4)/y + 1070/(1 + y)4
Solve numerically
y = 2%: €1 255.06
y = 3%: €1 210.88
YTC = 2.607%
Answer
YTC = 2.607%
The yield to maturity is 4.5%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.
7.CAPM & WACC
Answer: WACC = 9.983 %
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Weights
V = €9 000 000, E/V = 0.6667, D/V = 0.3333
WACC
WACC = 0.6667·13% + 0.3333·5%·(1 − 0.21) = 9.983%
Answer
WACC = 9.983%
8.CAPM & WACC
Answer: re = 13.15 %
Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.
Levered beta
βL = 0.8·(1 + 0.72·1.5) = 1.664
Cost of equity (CAPM)
re = 1.5% + 1.664·7% = 13.15%
Answer
re = 13.15%
9.Descriptive statistics
Answer: s = 13.58
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
11.Probability
Answer: σ = 9.95
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Net outcomes
G: 12 (p = 0.25), 27 (p = 0.01), −3 (p = 0.74)
Expected value
E(G) = 12·0.25 + 27·0.01 − 3·0.74 = 1.05 €
Answer
E(G) = 1.05 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.3085
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (108.2 − 105)/6.403 = 0.5
Upper tail
P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085
Answer
P = 0.3085
14.Normal distribution
Answer: x = 578.4 h
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z for the percentile
Φ(z) = 0.975 → z = 1.96
Back to x
x = 500 + 1.96·40 = 578.4 h
Answer
x = 578.4 h
15.Confidence intervals
Answer: E₂ = 0.6333
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Ratio of margins
E₂/E₁ = (1.645/1.645)·√(50/450) = 0.3333
New margin
E₂ = 1.9·0.3333 = 0.6333
Answer
E₂ = 0.6333
16.Confidence intervals
Answer: n = 5683
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.645²·0.7·0.3/0.01² = 5683
Round up
n = 5683
Answer
n = 5683
17.Hypothesis tests & confidence intervals
Answer: p = 0.02275
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Test statistic
SE = σ/√n = 15/√100 = 1.5
z = (83 − 80)/1.5 = 2
Tail area
Φ(2) = 0.9772
1 − Φ(2) = 0.0228
p-value
p = 0.02275
Answer
p = 0.02275
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: U = 102.9
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Margin of error
E = 2.576·5/√81 = 1.431
Interval limit
U = 101.5 + 1.431 = 102.9
Answer
U = 102.9
19.Linear regression & correlation
Answer: r = 0.9931
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 48/6 = 8
ȳ = 508/6 = 84.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 354
Syy = Σ(y − ȳ)² = 1815
Correlation coefficient
r = 354/√(70·1815) = 0.9931
Answer
r = 0.9931
20.Linear regression & correlation
Answer: ŷ = 17.75
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 15/5 = 3
ȳ = 84/5 = 16.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −19
Regression line
b = −19/10 = −1.9
a = 16.8 − (−1.9)·3 = 22.5
ŷ = 22.5 − 1.9·x
Prediction
ŷ = 22.5 + (−1.9)·2.5 = 17.75
Answer
ŷ = 17.75
This is independent practice material. ExamProblems is not affiliated with or endorsed by the organisation that runs this exam, and the exam name is used only to describe the content. CFA® and Chartered Financial Analyst® are trademarks of CFA Institute, which is not affiliated with ExamProblems and does not endorse it or vouch for its accuracy.