You invest €5 000 at a nominal rate of 5% per year, compounded quarterly. Find the value FV of the investment after 8 years.
PV = €5 0005% / 4n = 8
FV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 20 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.
P = €200 0004.2% / 12N = 240
A = €
3.NPV & IRRMedium
A machine costs €100 000 today. It generates €20 000 at the end of each year for 5 years and can then be sold for €15 000. The cost of capital is 10%. Find the net present value NPV.
C0 = €100 000CF = €20 000S = €15 000n = 5r = 10%
NPV = €
4.NPV & IRRHard
A project costs €10 000 and returns €4 500, €7 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 4 500CF2 = 7 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €1 435 when the yield to maturity is 3.5%. Find the coupon rate c in percent.
F = €1 000n = 12P = €1 435YTM = 3.5%
c = %
6.Bond pricing & yieldHard
A bond with 12 years to maturity (face value €1 000, annual coupon 8%) sells for €1 215. The issuer can call it in 5 years at a call price of €1 080. Find the yield to call YTC in percent.
F = €1 000coupon = 8%P = €1 215call price = €1 080call in 5 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €5 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the WACC.
A company's shares have a beta of 1.4 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 21%, the risk-free rate 3% and the market risk premium 6%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 23, 44, 36, 54, 34, 14.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 48?
μ = 60σ = 10x = 48
P =
11.ProbabilityMedium
A fair die is rolled 5 times. What is the probability P of getting at least one six?
n = 5
P =
12.ProbabilityHard
9 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 9p = 0.6
P =
13.Normal distributionMedium
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 10 of them exceeds 248.4 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 10> 248.4 kg
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded bottom 90% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm90%
x = cm
15.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 36 observations and a known σ, has a margin of error of 10.8. What is the margin of error E₂ of a 95% interval based on 324 observations?
90%n₁ = 36E₁ = 10.895%n₂ = 324
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 99% confidence interval for the proportion has a margin of error of at most 0.025?
p = 0.5E = 0.02599%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 123.6 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 123.6s = 12n = 49
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 12. Find the smallest sample size n.
σ = 12E = 290%
n =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 118Σx² = 364Σy² = 2810Σxy = 1010
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 79?
n = 5ŷ = 79
x =
3Solutions
1.Time value of money
Answer: FV = 7441 €
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 = 0.042/12 = 0.0035
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 = 0.4323
A = 200 000·0.0035/(1 − 0.4323) = €1 233.14
Answer
A = €1 233.14
3.NPV & IRR
Answer: NPV = −14870 €
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.
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(5%) = 634.92
NPV(10%) = −123.97
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 5 + 5·634.92/(634.92 − (−123.97)) = 9.183%
exact: NPV(9.139%) = 0 → IRR = 9.139%
Answer
IRR = 9.139%
5.Bond pricing & yield
Answer: c = 8.002 %
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.
Present value of the face value
1000·(1 + 0.035)−12 = €661.78
Coupon
C = (1435 − 661.78)/9.663 = €80.02
Coupon rate
c = 80.02/1000 = 8.002%
Answer
c = 8.002%
6.Bond pricing & yield
Answer: YTC = 4.556 %
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
1215 = 80·(1 − (1 + y)−5)/y + 1080/(1 + y)5
Solve numerically
y = 4%: €1 243.83
y = 5%: €1 192.57
YTC = 4.556%
Answer
YTC = 4.556%
The yield to maturity is 5.505%. 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 = 6 %
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 = €8 000 000, E/V = 0.375, D/V = 0.625
WACC
WACC = 0.375·11% + 0.625·4%·(1 − 0.25) = 6%
Answer
WACC = 6%
8.CAPM & WACC
Answer: re = 16.16 %
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.
Unlever the beta
βU = 1.4/(1 + 0.79·0.5) = 1.004
Relever at the new D/E
βL = 1.004·(1 + 0.79·1.5) = 2.193
Cost of equity (CAPM)
re = 3% + 2.193·6% = 16.16%
Answer
re = 16.16%
9.Descriptive statistics
Answer: s = 14.32
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849
Answer
P = 0.8849
11.Probability
Answer: P = 0.5981
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.
Complement
P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981
Answer
P = 0.5981
12.Probability
Answer: P = 0.003801
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.
Zero and one failure
P(X = 0) = 0.49 = 0.0002621
P(X = 1) = 9·0.6·0.48 = 0.003539
Add them
P = 0.0002621 + 0.003539 = 0.003801
Answer
P = 0.003801
13.Normal distribution
Answer: P = 0.8438
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 total
μT = 250 kg, σT = 0.5·√10 = 1.581 kg
Standardise
z = (248.4 − 250)/1.581 = −1.01
Probability
P = 1 − Φ(−1.01) = 0.8438
Answer
P = 0.8438
14.Normal distribution
Answer: x = 187 cm
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 = 178 + 1.28·7 = 187 cm
Answer
x = 187 cm
15.Confidence intervals
Answer: E₂ = 4.289
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 = (1.96/1.645)·√(36/324) = 0.3972
New margin
E₂ = 10.8·0.3972 = 4.289
Answer
E₂ = 4.289
16.Confidence intervals
Answer: n = 2655
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 = 2.576²·0.5·0.5/0.025² = 2654
Round up
n = 2655
Answer
n = 2655
17.Hypothesis tests & confidence intervals
Answer: t = 2.1
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/√49 = 12/7 = 1.714
Test statistic
t = (123.6 − 120)/1.714 = 2.1
Answer
t = 2.1
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 98
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 = (1.645·12/2)² = 97.42
Round up
n = 98
Answer
n = 98
19.Linear regression & correlation
Answer: r = 0.9942
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 = 1010 − 42·118/6 = 184
Sxx = 364 − 42²/6 = 70
Syy = 2810 − 118²/6 = 489.3
Correlation coefficient
r = 184/√(70·489.3) = 0.9942
Answer
r = 0.9942
20.Linear regression & correlation
Answer: x = 24.63
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 = 100/5 = 20
ȳ = 317/5 = 63.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 303
Regression line
b = 303/90 = 3.367
a = 63.4 − 3.367·20 = −3.933
ŷ = −3.933 + 3.367·x
Solve for x
x = (79 − (−3.933))/3.367 = 24.63
Answer
x = 24.63
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