You invest €5 000 at a nominal rate of 5% per year, compounded monthly. Find the value FV of the investment after 6 years.
PV = €5 0005% / 12n = 6
FV = €
2.Time value of moneyHard
A loan of €25 000 is repaid with equal monthly payments over 4 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.
P = €25 0002.4% / 12N = 48
A = €
3.NPV & IRRMedium
An investment of €30 000 gives these cash flows at the end of years 1 to 4: €7 000, €7 000, €6 000, €12 000. The cost of capital is 9%. Find the net present value NPV.
A project costs €20 000 and returns €8 000, €11 500, €11 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 8 000CF2 = 11 500CF3 = 11 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 6% and matures in 4 years. Its yield to maturity changes from 8% to 8.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 6%n = 4y1 = 8%y2 = 8.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 10 years and sells for €1 180. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 10P = €1 180
YTM = %
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €3 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 6% and the tax rate is 25%. Find the WACC.
A company has 2 000 000 shares trading at €20.00; the book value of its equity is €24 000 000. It just paid a dividend of €1.90 per share, and dividends grow by 3% per year. Its debt has a market value of €20 000 000 and a pre-tax cost of 4.5%. The tax rate is 25.8%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 35, 32, 32, 41, 55.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 8. Find the z-score z of a score of 53.2.
μ = 50σ = 8x = 53.2
z =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4p(x): 0.3, 0.2, 0.35, 0.15
σ =
12.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 4 correct answers?
n = 9p = 0.3k = 4
P =
13.Normal distributionMedium
X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.6 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.6 mm
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 99% of them lie (the 99th percentile).
μ = 250σ = 5P = 99%
x = g
15.Confidence intervalsMedium
A 95% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 12. What is the margin of error E₂ if the same study uses 16 observations instead?
95%n₁ = 64E₁ = 12n₂ = 16
E₂ =
16.Confidence intervalsHard
The population standard deviation is 10. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2.5?
σ = 10E = 2.599%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 122.4 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 122.4s = 12n = 36
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.01 with a sample of 25; the standard deviation is known to be 15. In reality the mean is 62. Find the probability of a Type II error β in percent.
μ₀ = 50μ = 62σ = 15n = 25α = 0.01
β = %
19.Linear regression & correlationMedium
The table gives 5 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 = 5Σx = 30Σy = 70Σx² = 220Σy² = 1162Σxy = 502
r =
20.Linear regression & correlationHard
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. Predict ŷ for x = 3.5.
n = 5x = 3.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 6745 €
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 = 0.024/12 = 0.002
N = 12·4 = 48
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 = 0.9086
A = 25 000·0.002/(1 − 0.9086) = €546.75
Answer
A = €546.75
3.NPV & IRR
Answer: NPV = −4552 €
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 each cash flow
CF0 = −30 000
CF1/(1 + 0.09)1 = 7 000/1.09 = 6 422.02
CF2/(1 + 0.09)2 = 7 000/1.188 = 5 891.76
CF3/(1 + 0.09)3 = 6 000/1.295 = 4 633.10
CF4/(1 + 0.09)4 = 12 000/1.412 = 8 501.10
Add them up
NPV = €−4 552.02
Answer
NPV = €−4 552.02
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 23.88 %
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(20%) = 1 307.87
NPV(25%) = −352.00
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.
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 = €11 000 000, E/V = 0.7273, D/V = 0.2727
WACC
WACC = 0.7273·11% + 0.2727·6%·(1 − 0.25) = 9.227%
Answer
WACC = 9.227%
8.CAPM & WACC
Answer: WACC = 9.636 %
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 = (53.2 − 50)/8 = 0.4
Answer
z = 0.4
11.Probability
Answer: σ = 1.062
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.
Mean
μ = 1·0.3 + 2·0.2 + 3·0.35 + 4·0.15 = 2.35
Mean of the squares
E(X²) = 1·0.3 + 4·0.2 + 9·0.35 + 16·0.15 = 6.65
Variance
Var(X) = 6.65 − 2.35² = 1.128
σ = √1.128 = 1.062
Answer
σ = 1.062
12.Probability
Answer: P = 0.1715
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.
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 sum
μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm
Standardise
z = (200.6 − 200)/0.5 = 1.2
Upper tail
P = 1 − Φ(1.2) = 1 − 0.8849 = 0.1151
Answer
P = 0.1151
14.Normal distribution
Answer: x = 261.6 g
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) = 0.99 → z = 2.326
Back to x
x = 250 + 2.326·5 = 261.63 g
Answer
x = 261.63 g
15.Confidence intervals
Answer: E₂ = 24
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 = (1.96/1.96)·√(64/16) = 2
New margin
E₂ = 12·2 = 24
Answer
E₂ = 24
16.Confidence intervals
Answer: n = 107
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 = (2.576·10/2.5)² = 106.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 107
Answer
n = 107
17.Hypothesis tests & confidence intervals
Answer: t = 1.2
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/√36 = 12/6 = 2
Test statistic
t = (122.4 − 120)/2 = 1.2
Answer
t = 1.2
t0.025; df = 35 = 2.03
|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: β = 4.707 %
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 = 50 + 2.326·3 = 56.98
z under the true mean
z = (56.98 − 62)/3 = −1.674
Type II error
power = P(Z > −1.674) = 95.29%
β = 100% − 95.29% = 4.707%
Answer
β = 4.707%
19.Linear regression & correlation
Answer: r = 0.9611
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 = 502 − 30·70/5 = 82
Sxx = 220 − 30²/5 = 40
Syy = 1162 − 70²/5 = 182
Correlation coefficient
r = 82/√(40·182) = 0.9611
Answer
r = 0.9611
20.Linear regression & correlation
Answer: ŷ = 16.6
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 = 90/5 = 18
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −28
Regression line
b = −28/10 = −2.8
a = 18 − (−2.8)·3 = 26.4
ŷ = 26.4 − 2.8·x
Prediction
ŷ = 26.4 + (−2.8)·3.5 = 16.6
Answer
ŷ = 16.6
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