You need €20 000 in 6 years. An account pays 3% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €20 0003% / 4n = 6
PV = €
2.Time value of moneyHard
You want €10 000 in 17 years. You save the same amount at the end of every month in an account paying 4.2% per year, compounded monthly. Find the monthly deposit A.
FV = €10 0004.2% / 12N = 204
A = €
3.NPV & IRRMedium
A machine costs €80 000 today, lasts 10 years and costs €4 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.
C0 = €80 000M = €4 000n = 10r = 12%
EAC = €
4.NPV & IRRHard
An investment of €20 000 returns €4 500 at the end of each year for 6 years. Find its internal rate of return IRR in percent.
C0 = €20 000CF = €4 500n = 6
IRR = %
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 8 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 5%YTM = 3.5%n = 8
P = €
6.Bond pricing & yieldHard
A bond with 10 years to maturity (face value €1 000, annual coupon 8%) sells for €1 188. The issuer can call it in 4 years at a call price of €1 080. Find the yield to call YTC in percent.
F = €1 000coupon = 8%P = €1 188call price = €1 080call in 4 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €2 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 30%. Find the WACC.
A company's shares have a beta of 1.5 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 1. The tax rate is 25%, the risk-free rate 3% and the market risk premium 4.5%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 52, 18, 58, 27, 38.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 75.
μ = 70σ = 5x = 75
z =
11.ProbabilityMedium
A bag holds 3 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 9
P =
12.ProbabilityHard
8 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 8p = 0.6
P =
13.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 16 of them is below 198.4 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 16< 198.4 kg
P =
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 0.02 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 = 0.02 mm
P =
15.Confidence intervalsMedium
The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 9.2 and standard deviation 4.1. Find the margin of error E of the 95% confidence interval for the mean difference.
n = 15d̄ = 9.2sd = 4.195%t(14) = 2.145
E = mmHg
16.Confidence intervalsHard
The population standard deviation is 19. 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 1.5?
σ = 19E = 1.595%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 50; the population standard deviation is known to be 20. A random sample of 64 observations has a mean of 46.2. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.
μ₀ = 50σ = 20n = 64x̄ = 46.2H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.05 with a sample of 36; the standard deviation is known to be 12. In reality the mean is 94.6. Find the power 1 − β of the test in percent.
μ₀ = 100μ = 94.6σ = 12n = 36α = 0.05
1 − β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 6 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 = 18?
n = 6ŷ = 18
x =
3Solutions
1.Time value of money
Answer: PV = 16717 €
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 = 0.03/4 = 0.0075
N = 4·6 = 24
Present value
PV = FV/(1 + i)N = 20 000/1.196 = €16 716.63
Answer
PV = €16 716.63
2.Time value of money
Answer: A = 33.67 €
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.
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 − 1.12−10)/0.12 = 5.65
Annualised purchase price
80 000/5.65 = 14 158.73
EAC
EAC = 14 158.73 + 4 000 = €18 158.73
Answer
EAC = €18 158.73
4.NPV & IRR
Answer: IRR = 9.312 %
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.
Required annuity factor
(1 − (1 + i)−6)/i = 20 000/4 500 = 4.444
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 9%: 4.486
i = 10%: 4.355
IRR = 9.312%
Answer
IRR = 9.312%
5.Bond pricing & yield
Answer: P = 1104 €
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 = 0.0175, N = 16
Present value of the coupons
€25.00·(1 − 0.7576)/0.0175 = €346.26
Present value of the face value
€1 000·0.7576 = €757.62
Price
P = €1 103.88
Answer
P = €1 103.88
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTC = 4.613 %
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
1188 = 80·(1 − (1 + y)−4)/y + 1080/(1 + y)4
Solve numerically
y = 4%: €1 213.58
y = 5%: €1 172.19
YTC = 4.613%
Answer
YTC = 4.613%
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 = 9.171 %
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 = €7 000 000, E/V = 0.7143, D/V = 0.2857
WACC
WACC = 0.7143·12% + 0.2857·3%·(1 − 0.3) = 9.171%
Answer
WACC = 9.171%
8.CAPM & WACC
Answer: re = 13.27 %
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 = 1.5/(1 + 0.75·0.2) = 1.304
Relever at the new D/E
βL = 1.304·(1 + 0.75·1) = 2.283
Cost of equity (CAPM)
re = 3% + 2.283·4.5% = 13.27%
Answer
re = 13.27%
9.Descriptive statistics
Answer: s = 16.7
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 = (75 − 70)/5 = 1
Answer
z = 1
11.Probability
Answer: P = 0.04545
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 = 3/12 · 2/11 = 6/132 = 0.04545
Answer
P = 0.04545
12.Probability
Answer: P = 0.00852
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) = 0.48 = 0.0006554
P(X = 1) = 8·0.6·0.47 = 0.007864
Add them
P = 0.0006554 + 0.007864 = 0.00852
Answer
P = 0.00852
13.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 total
μT = 192 kg, σT = 0.8·√16 = 3.2 kg
Standardise
z = (198.4 − 192)/3.2 = 2
Probability
P = Φ(2) = 0.9772
Answer
P = 0.9772
14.Normal distribution
Answer: P = 0.9616
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 = 0.02828 mm
Standardise
z = (0 − 0.05)/0.02828 = −1.77
Probability
P = 1 − Φ(−1.77) = 0.9616
Answer
P = 0.9616
15.Confidence intervals
Answer: E = 2.271 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 = 14, t = 2.145
Margin of error
E = 2.145·4.1/√15 = 2.271
Answer
E = 2.271 mmHg
16.Confidence intervals
Answer: n = 617
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 = (1.96·19/1.5)² = 616.4
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 617
Answer
n = 617
17.Hypothesis tests & confidence intervals
Answer: p = 0.06426
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 = 20/√64 = 2.5
z = (46.2 − 50)/2.5 = −1.52
Tail area
Φ(1.52) = 0.9357
1 − Φ(1.52) = 0.0643
p-value
p = 0.06426
Answer
p = 0.06426
p > 0.05, so H₀ is not rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 85.43 %
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 = 12/√36 = 2
x̄c = 100 − 1.645·2 = 96.71
z under the true mean
z = (96.71 − 94.6)/2 = 1.055
Power
power = P(Z < 1.055) = 85.43%
1 − β = 85.43%
Answer
1 − β = 85.43%
19.Linear regression & correlation
Answer: r = 0.9728
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 = 42/6 = 7
ȳ = 73/6 = 12.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 107
Syy = Σ(y − ȳ)² = 172.8
Correlation coefficient
r = 107/√(70·172.8) = 0.9728
Answer
r = 0.9728
20.Linear regression & correlation
Answer: x = 3.978
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 = 21/6 = 3.5
ȳ = 113/6 = 18.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −30.5
Regression line
b = −30.5/17.5 = −1.743
a = 18.83 − (−1.743)·3.5 = 24.93
ŷ = 24.93 − 1.743·x
Solve for x
x = (18 − 24.93)/(−1.743) = 3.978
Answer
x = 3.978
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