A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 3% per year forever. The required return is 7% per year. Find the value PV of the property today.
C1 = €8 000g = 3%r = 7%
PV = €
2.Time value of moneyHard
A loan of €250 000 is repaid with equal monthly payments over 20 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.
P = €250 0003% / 12N = 240
A = €
3.NPV & IRRMedium
A machine costs €30 000 today, lasts 6 years and costs €2 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.
C0 = €30 000M = €2 000n = 6r = 12%
EAC = €
4.NPV & IRRHard
A project costs €25 000 and returns €16 500, €11 000, €13 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 16 500CF2 = 11 000CF3 = 13 500
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 3 years to maturity (face value €1 000, annual coupon 8%) when its yield to maturity is 2.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 3%. Find your holding period return HPR in percent.
F = €1 000coupon = 8%n = 3y0 = 2.5%y1 = 3%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 4 years and sells for €1 055. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 4P = €1 055
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €8 000 000, preferred shares worth €2 000 000 and debt worth €2 000 000. The cost of equity is 11%, the cost of preferred stock 8% and the pre-tax cost of debt 4%. The tax rate is 20%. Find the WACC.
A company has 2 000 000 shares at €44.50 each and 10 000 bonds trading at €980 each. The pre-tax cost of debt is 5% and the tax rate is 28%. The share’s beta is 1.3, the risk-free rate 2% and the market return 7%. Find the WACC.
β = 1.3rf = 2%rm = 7%rd = 5%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 48, 13, 59, 16, 44, 55, 12.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 62.
μ = 70σ = 8x = 62
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.25, 0.25, 0.3, 0.05, 0.15
Var(X) =
12.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 1. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 0.5X + 1
σY =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.8029. Find z to two decimals.
A = 0.8029
z =
14.Normal distributionHard
Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 2.6. What is the margin of error E₂ of a 99% interval from the same data?
95%n₁ = 25E₁ = 2.699%
E₂ =
16.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.05?
p = 0.5E = 0.0595%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 60 parts has mean length 140 mm, and the line's known standard deviation is 18 mm. Line 2: a sample of 50 parts has mean 145.3 mm, known standard deviation 4 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
In a trial, 98 of 150 patients on treatment A recover, and 85 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 98n₁ = 150x₂ = 85n₂ = 120
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.
n = 6x = 4.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 200000 €
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.
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.
Monthly rate and number of payments
i = 0.03/12 = 0.0025
N = 12·20 = 240
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.5492
A = 250 000·0.0025/(1 − 0.5492) = €1 386.49
Answer
A = €1 386.49
3.NPV & IRR
Answer: EAC = 9297 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.12−6)/0.12 = 4.111
Annualised purchase price
30 000/4.111 = 7 296.77
EAC
EAC = 7 296.77 + 2 000 = €9 296.77
Answer
EAC = €9 296.77
4.NPV & IRR
Answer: IRR = 31.03 %
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.
Bracket the IRR
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
NPV(30%) = 345.93
NPV(35%) = −1 255.14
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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.
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
€1 055 = €50·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴
Solve numerically
P(3%) = €1 074.34, P(4%) = €1 036.30
YTM = 3.503%
Answer
YTM = 3.503%
7.CAPM & WACC
Answer: WACC = 9.2 %
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).
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).
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 = (62 − 70)/8 = −1
Answer
z = −1
11.Probability
Answer: Var(X) = 1.74
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.
Spread of X
μ = 1, Var(X) = 0.7, σX = 0.8367
Linear change
σY = |0.5|·0.8367 = 0.4183
Answer
σY = 0.4183
The mean does move: E(Y) = 0.5·1 + 1 = 1.5.
13.Normal distribution
Answer: z = 1.29
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left of z
Φ(z) = (1 + 0.8029)/2 = 0.9015
Read the z-table backwards
Find 0.9015 inside the table and read off z.
z = 1.29
Answer
z = 1.29
14.Normal distribution
Answer: P = 0.9772
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 difference
μ = 25.05 − 25 = 0.05 mm, σ = 0.025 mm
Standardise
z = (0 − 0.05)/0.025 = −2
Probability
P = 1 − Φ(−2) = 0.9772
Answer
P = 0.9772
15.Confidence intervals
Answer: E₂ = 3.417
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 = (2.576/1.96)·√(25/25) = 1.314
New margin
E₂ = 2.6·1.314 = 3.417
Answer
E₂ = 3.417
16.Confidence intervals
Answer: n = 385
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 = 1.96²·0.5·0.5/0.05² = 384.2
Round up
n = 385
Answer
n = 385
17.Hypothesis tests & confidence intervals
Answer: z = −2.216
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.
Standard error of the difference
SE = √(18²/60 + 4²/50) = √5.72 = 2.392
Test statistic
z = (140 − 145.3)/2.392 = −2.216
Answer
z = −2.216
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −0.9609
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.
Sample proportions
p̂₁ = 98/150 = 0.6533
p̂₂ = 85/120 = 0.7083
p̂ = 183/270 = 0.6778
Standard error under H₀
SE = √(0.6778·0.3222·(1/150 + 1/120)) = 0.05724
Test statistic
z = (0.6533 − 0.7083)/0.05724 = −0.9609
Answer
z = −0.9609
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.992
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 = 30/5 = 6
ȳ = 91/5 = 18.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 104
Syy = Σ(y − ȳ)² = 274.8
Correlation coefficient
r = 104/√(40·274.8) = 0.992
Answer
r = 0.992
20.Linear regression & correlation
Answer: ŷ = 64
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 = 27/6 = 4.5
ȳ = 384/6 = 64
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 92
Regression line
b = 92/17.5 = 5.257
a = 64 − 5.257·4.5 = 40.34
ŷ = 40.34 + 5.257·x
Prediction
ŷ = 40.34 + 5.257·4.5 = 64
Answer
ŷ = 64
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.