You invest €8 000 at a nominal rate of 3% per year, compounded monthly. Find the value FV of the investment after 2 years.
PV = €8 0003% / 12n = 2
FV = €
2.Time value of moneyHard
A mortgage of €250 000 is repaid with equal monthly payments over 20 years at 3.6% per year, compounded monthly. How much is still owed just after the payment at the end of year 9? Find the outstanding balance B.
P = €250 0003.6% / 12N = 240k = 108
B = €
3.NPV & IRRMedium
A project costs €20 000 today and returns €5 000, €5 500, €6 000, €7 000, €4 500, €6 500 at the end of years 1 to 6. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €40 000 and returns €22 500, €27 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −40 000CF1 = 22 500CF2 = 27 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 7% per year. Find the price P of the bond.
F = €1 000coupon = 4.5%YTM = 7%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 5 years and sells for €935. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 5P = €935
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €5 000 000, preferred shares worth €2 000 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 6.5% and the pre-tax cost of debt 3%. The tax rate is 28%. Find the WACC.
A company has 2 000 000 shares at €29.50 each and 20 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 25.8%. The share’s beta is 1.5, the risk-free rate 2% and the market return 9%. Find the WACC.
β = 1.5rf = 2%rm = 9%rd = 3%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 41, 29, 19, 48, 27, 33.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 470.
μ = 500σ = 60x = 470
z =
11.ProbabilityMedium
In how many ways N can a committee of 5 people be chosen from a group of 7? The order does not matter.
n = 7k = 5
N =
12.ProbabilityHard
A game costs €5 to play. You win €10 with probability 0.2 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€10: p = 0.2€100: p = 0.05
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9780. Find z to two decimals.
A = 0.9780
z =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h90%
x = h
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 81 h, known standard deviation 19 h. Model 2: 100 phones, mean 70.5 h, known standard deviation 13 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.
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 0.025?
p = 0.5E = 0.02595%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 43.4 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 43.4s = 12n = 16
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 12. Find the smallest sample size n.
σ = 12E = 195%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (3, 13) is made. Find its residual e = y − ŷ.
n = 5(3, 13)
e =
3Solutions
1.Time value of money
Answer: FV = 8494 €
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.
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 240 − 108 = 132
B = 1463·(1 − (1 + 0.003)−132)/0.003 = €159 245.02
Answer
B = €159 245.02
3.NPV & IRR
Answer: DPB = 4.244 years
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 5 000/1.08 = 4 629.63 → Σ = 4 629.63
year 2: 5 500/1.166 = 4 715.36 → Σ = 9 344.99
year 3: 6 000/1.26 = 4 762.99 → Σ = 14 107.99
year 4: 7 000/1.36 = 5 145.21 → Σ = 19 253.20
year 5: 4 500/1.469 = 3 062.62 → Σ = 22 315.82
Fraction of the last year
DPB = 4 + 746.80/3 062.62 = 4.244
Answer
DPB = 4.244 years
The discounted payback is always longer than the simple payback (3.5 years).
4.NPV & IRR
Answer: IRR = 15.68 %
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(15%) = 359.17
NPV(20%) = −2 152.78
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.
Per period
C = €22.50, i = 0.035, N = 24
Present value of the coupons
€22.50·(1 − 0.438)/0.035 = €361.31
Present value of the face value
€1 000·0.438 = €437.96
Price
P = €799.27
Answer
P = €799.27
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 5.523 %
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
€935 = €40·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(5%) = €956.71, P(6%) = €915.75
YTM = 5.523%
Answer
YTM = 5.523%
7.CAPM & WACC
Answer: WACC = 7.4 %
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 = (470 − 500)/60 = −0.5
Answer
z = −0.5
11.Probability
Answer: N = 21
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Combinations
C(7, 5) = (7·6·5·4·3)/(5·4·3·2·1) = 21
Answer
N = 21
12.Probability
Answer: E(G) = 2 €
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: 5 (p = 0.2), 95 (p = 0.05), −5 (p = 0.75)
Expected value
E(G) = 5·0.2 + 95·0.05 − 5·0.75 = 2 €
Answer
E(G) = 2 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = 2.29
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 + 0.9780)/2 = 0.9890
Read the z-table backwards
Find 0.9890 inside the table and read off z.
z = 2.29
Answer
z = 2.29
14.Normal distribution
Answer: x = 551 h
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
Φ(z) = 0.9
z from the table
z ≈ 1.28
Back to the original units
x = 500 + 1.28·40 = 551 h
Answer
x = 551 h
15.Confidence intervals
Answer: E = 5.851 h
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Point estimate
x̄₁ − x̄₂ = 81 − 70.5 = 10.5
Standard error
SE = √(19²/50 + 13²/100) = 2.985
Margin of error
E = 1.96·2.985 = 5.851
Answer
E = 5.851 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 1537
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 = 1.96²·0.5·0.5/0.025² = 1537
Round up
n = 1537
Answer
n = 1537
17.Hypothesis tests & confidence intervals
Answer: t = −2.2
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).
Standard error
SE = 12/√16 = 12/4 = 3
Test statistic
t = (43.4 − 50)/3 = −2.2
Answer
t = −2.2
t0.025; df = 15 = 2.131
|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 554
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).
Solve for n
n = (1.96·12/1)² = 553.2
Round up
n = 554
Answer
n = 554
19.Linear regression & correlation
Answer: r = 0.7531
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 = 25/5 = 5
ȳ = 218/5 = 43.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 26
Syy = Σ(y − ȳ)² = 119.2
Correlation coefficient
r = 26/√(10·119.2) = 0.7531
Answer
r = 0.7531
20.Linear regression & correlation
Answer: e = 1.65
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 = 107/5 = 21.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 134
Regression line
b = 134/40 = 3.35
a = 21.4 − 3.35·6 = 1.3
ŷ = 1.3 + 3.35·x
Predicted value
ŷ = 1.3 + 3.35·3 = 11.35
Residual
e = 13 − 11.35 = 1.65
Answer
e = 1.65
The point lies above the line.
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