You need €20 000 in 2 years. An account pays 5% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €20 0005% / 12n = 2
PV = €
2.Time value of moneyHard
A mortgage of €150 000 is repaid with equal monthly payments over 20 years at 4.8% per year, compounded monthly. How much is still owed just after the payment at the end of year 15? Find the outstanding balance B.
P = €150 0004.8% / 12N = 240k = 180
B = €
3.NPV & IRRMedium
A machine costs €60 000 today. It generates €13 500 at the end of each year for 6 years and can then be sold for €6 000. The cost of capital is 12%. Find the net present value NPV.
C0 = €60 000CF = €13 500S = €6 000n = 6r = 12%
NPV = €
4.NPV & IRRHard
A project costs €80 000 today. Its cash flow is €11 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 7. The cost of capital is 12%. Find the net present value NPV.
C0 = €80 000CF1 = €11 500g = 3%n = 7r = 12%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 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 = 5%YTM = 7%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 7 years. It sells for €642. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 2% (2×)n = 7P = €642
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €12 000 000, preferred shares worth €1 500 000 and debt worth €6 000 000. The cost of equity is 10%, the cost of preferred stock 6.5% and the pre-tax cost of debt 4%. The tax rate is 28%. Find the WACC.
A company has 4 000 000 shares at €44.00 each and 5 000 bonds trading at €1 050 each. The pre-tax cost of debt is 6% and the tax rate is 21%. The share’s beta is 0.8, the risk-free rate 2% and the market return 6.5%. Find the WACC.
β = 0.8rf = 2%rm = 6.5%rd = 6%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 49, 30, 58, 44, 30.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 20. Find the z-score z of a score of 26.
μ = 50σ = 20x = 26
z =
11.ProbabilityMedium
A bag holds 4 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
4 / 6
P =
12.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 2 correct answers?
n = 12p = 0.3k = 2
P =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9692. Find z to two decimals.
A = 0.9692
z =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 80% of them lie (the 80th percentile).
μ = 500σ = 40P = 80%
x = h
15.Confidence intervalsMedium
The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.5 and standard deviation 5.9. Find the lower limit of the 99% confidence interval for the mean difference.
n = 10d̄ = 6.5sd = 5.999%t(9) = 3.25
lower limit = mmHg
16.Confidence intervalsHard
The population standard deviation is 30. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1?
σ = 30E = 195%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 250; the population standard deviation is known to be 25. A random sample of 100 observations has a mean of 252.25. Test H₀: μ = 250 against H₁: μ > 250. Find the p-value p.
μ₀ = 250σ = 25n = 100x̄ = 252.25H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.10 with a sample of 25; the standard deviation is known to be 15. In reality the mean is 84.5. Find the probability of a Type II error β in percent.
μ₀ = 80μ = 84.5σ = 15n = 25α = 0.10
β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
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. For which value x does the line predict ŷ = 20?
n = 5ŷ = 20
x =
3Solutions
1.Time value of money
Answer: PV = 18101 €
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 = 0.05/12 = 0.004167
N = 12·2 = 24
Present value
PV = FV/(1 + i)N = 20 000/1.105 = €18 100.51
Answer
PV = €18 100.51
2.Time value of money
Answer: B = 51834 €
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 payment
i = 0.004, N = 240
A = 150 000·0.004/(1 − (1 + 0.004)−240) = €973.44
Present value of the remaining payments
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 240 − 180 = 60
B = 973.4·(1 − (1 + 0.004)−60)/0.004 = €51 834.38
Answer
B = €51 834.38
3.NPV & IRR
Answer: NPV = −1456 €
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 the yearly cash flows
13 500·4.111 = 55 504.00
Present value of the salvage value
6 000/1.974 = 3 039.79
NPV
NPV = −60 000 + 55 504.00 + 3 039.79 = −€1 456.21
Answer
NPV = −€1 456.21
4.NPV & IRR
Answer: NPV = −23309 €
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.
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: P = 839.4 €
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 = €25.00, i = 0.035, N = 24
Present value of the coupons
€25.00·(1 − 0.438)/0.035 = €401.46
Present value of the face value
€1 000·0.438 = €437.96
Price
P = €839.42
Answer
P = €839.42
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 9.005 %
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 = €10.00, N = 14
Solve numerically
i = 4.503% per half-year
YTM
YTM = 2·4.503% = 9.005%
Answer
YTM = 9.005%
7.CAPM & WACC
Answer: WACC = 7.54 %
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 = (26 − 50)/20 = −1.2
Answer
z = −1.2
11.Probability
Answer: P = 0.1333
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.
First and second draw
P = 4/10 · 3/9 = 12/90 = 0.1333
Answer
P = 0.1333
12.Probability
Answer: P = 0.1678
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.
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 + 0.9692)/2 = 0.9846
Read the z-table backwards
Find 0.9846 inside the table and read off z.
z = 2.16
Answer
z = 2.16
14.Normal distribution
Answer: x = 533.7 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) = 0.8 → z = 0.8416
Back to x
x = 500 + 0.8416·40 = 533.66 h
Answer
x = 533.66 h
15.Confidence intervals
Answer: lower limit = 0.4363 mmHg
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.
Critical value
df = 9, t = 3.25
Margin of error
E = 3.25·5.9/√10 = 6.064
Interval
6.5 ± 6.064 → [0.4363, 12.56]
Answer
lower limit = 0.4363 mmHg
16.Confidence intervals
Answer: n = 3458
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.
Solve for n
n = (1.96·30/1)² = 3457
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 3458
Answer
n = 3458
17.Hypothesis tests & confidence intervals
Answer: p = 0.1841
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 = 25/√100 = 2.5
z = (252.25 − 250)/2.5 = 0.9
Tail area
Φ(0.9) = 0.8159
1 − Φ(0.9) = 0.1841
p-value
p = 0.1841
Answer
p = 0.1841
p > 0.05, so H₀ is not rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: β = 41.37 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 15/√25 = 3
x̄c = 80 + 1.282·3 = 83.85
z under the true mean
z = (83.85 − 84.5)/3 = −0.218
Type II error
power = P(Z > −0.218) = 58.63%
β = 100% − 58.63% = 41.37%
Answer
β = 41.37%
19.Linear regression & correlation
Answer: r = 0.9173
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.
Centred sums
Sxy = 8706 − 138·372/6 = 150
Sxx = 3244 − 138²/6 = 70
Syy = 23446 − 372²/6 = 382
Correlation coefficient
r = 150/√(70·382) = 0.9173
Answer
r = 0.9173
20.Linear regression & correlation
Answer: x = 3.824
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 = 107/5 = 21.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −17
Regression line
b = −17/10 = −1.7
a = 21.4 − (−1.7)·3 = 26.5
ŷ = 26.5 − 1.7·x
Solve for x
x = (20 − 26.5)/(−1.7) = 3.824
Answer
x = 3.824
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