You invest €1 000 at a nominal rate of 5% per year, compounded quarterly. Find the value FV of the investment after 2 years.
PV = €1 0005% / 4n = 2
FV = €
2.Time value of moneyHard
You want €75 000 in 16 years. You save the same amount at the end of every month in an account paying 7.2% per year, compounded monthly. Find the monthly deposit A.
FV = €75 0007.2% / 12N = 192
A = €
3.NPV & IRRMedium
A machine costs €30 000 today. It generates €5 500 at the end of each year for 8 years and can then be sold for €6 000. The cost of capital is 10%. Find the net present value NPV.
C0 = €30 000CF = €5 500S = €6 000n = 8r = 10%
NPV = €
4.NPV & IRRHard
An investment of €80 000 returns €28 000 at the end of each year for 4 years. Find its internal rate of return IRR in percent.
C0 = €80 000CF = €28 000n = 4
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 3 years. The yield to maturity is 4% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 4%n = 3
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 6 years and sells for €1 075. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 6P = €1 075
YTM = %
7.CAPM & WACCMedium
A company has equity worth €6 000 000 and debt worth €2 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 3.5% and the tax rate is 25.8%. Find the WACC.
A company's business has an unlevered (asset) beta of 1. It plans a debt-to-equity ratio of 0.6. The tax rate is 21%, the risk-free rate is 2% and the market risk premium is 6.5%. Find the cost of equity re at the new capital structure.
βU = 1D/E = 0.6T = 21%rf = 2%rm − rf = 6.5%
re = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 58, 37, 37, 10, 32.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is below 66?
μ = 50σ = 8x = 66
P =
11.ProbabilityMedium
A bag holds 7 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 7
P =
12.ProbabilityHard
X has the distribution shown, and Y = 10X − 3. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.35, 0.35, 0.3Y = 10X − 3
σY =
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 10 of them is below 118.7 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 10< 118.7 kg
P =
14.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 95 h, known standard deviation 16 h. Model 2: 100 phones, mean 96.5 h, known standard deviation 17 h. Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.
The diameters of 10 machined shafts have a sample standard deviation of 7.8 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.
n = 10s = 7.8 µmχ² = 2.7, 19.023
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 123.9 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 123.9s = 12n = 81
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 195 of 300 patients on treatment A recover, and 92 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 195n₁ = 300x₂ = 92n₂ = 120
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (6, 90) is made. Find its residual e = y − ŷ.
n = 5(6, 90)
e =
3Solutions
1.Time value of money
Answer: FV = 1104 €
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.
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
5 500·5.335 = 29 342.09
Present value of the salvage value
6 000/2.144 = 2 799.04
NPV
NPV = −30 000 + 29 342.09 + 2 799.04 = €2 141.14
Answer
NPV = €2 141.14
4.NPV & IRR
Answer: IRR = 14.96 %
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)−4)/i = 80 000/28 000 = 2.857
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 14%: 2.914
i = 15%: 2.855
IRR = 14.96%
Answer
IRR = 14.96%
5.Bond pricing & yield
Answer: P = 1084 €
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 = €35.00, i = 0.02, N = 6
Present value of the coupons
€35.00·(1 − 0.888)/0.02 = €196.05
Present value of the face value
€1 000·0.888 = €887.97
Price
P = €1 084.02
Answer
P = €1 084.02
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 4.544 %
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
€1 075 = €60·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶
Solve numerically
P(4%) = €1 104.84, P(5%) = €1 050.76
YTM = 4.544%
Answer
YTM = 4.544%
7.CAPM & WACC
Answer: WACC = 6.649 %
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 = 0.75, D/V = 0.25
WACC
WACC = 0.75·8% + 0.25·3.5%·(1 − 0.258) = 6.649%
Answer
WACC = 6.649%
8.CAPM & WACC
Answer: re = 11.58 %
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.
Levered beta
βL = 1·(1 + 0.79·0.6) = 1.474
Cost of equity (CAPM)
re = 2% + 1.474·6.5% = 11.58%
Answer
re = 11.58%
9.Descriptive statistics
Answer: s = 17.11
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 0.9772
Answer
P = 0.9772
11.Probability
Answer: P = 0.2308
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 = 7/14 · 6/13 = 42/182 = 0.2308
Answer
P = 0.2308
12.Probability
Answer: σY = 8.047
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.
Spread of X
μ = 0.95, Var(X) = 0.6475, σX = 0.8047
Linear change
σY = |10|·0.8047 = 8.047
Answer
σY = 8.047
The mean does move: E(Y) = 10·0.95 − 3 = 6.5.
13.Normal distribution
Answer: P = 0.305
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 = 120 kg, σT = 0.8·√10 = 2.53 kg
Standardise
z = (118.7 − 120)/2.53 = −0.51
Probability
P = Φ(−0.51) = 0.305
Answer
P = 0.305
14.Normal distribution
Answer: P = 0.7389
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 = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
15.Confidence intervals
Answer: E = 5.243 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 = 95 − 96.5 = −1.5
Standard error
SE = √(16²/60 + 17²/100) = 2.675
Margin of error
E = 1.96·2.675 = 5.243
Answer
E = 5.243 h
The interval contains 0, so the data do not show a significant difference.
16.Confidence intervals
Answer: σ = 5.365 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 9·7.8² = 547.6
Limits for σ²
547.6/19.023 = 28.78
547.6/2.7 = 202.8
Limit for σ
σ = √28.78 = 5.365 µm
Answer
σ = 5.365 µm
95% interval for σ: [5.365, 14.24] µm.
17.Hypothesis tests & confidence intervals
Answer: t = 2.925
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/√81 = 12/9 = 1.333
Test statistic
t = (123.9 − 120)/1.333 = 2.925
Answer
t = 2.925
t0.025; df = 80 = 1.99
|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −2.322
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.
Sample proportions
p̂₁ = 195/300 = 0.65
p̂₂ = 92/120 = 0.7667
p̂ = 287/420 = 0.6833
Standard error under H₀
SE = √(0.6833·0.3167·(1/300 + 1/120)) = 0.05024
Test statistic
z = (0.65 − 0.7667)/0.05024 = −2.322
Answer
z = −2.322
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9674
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 = 126/6 = 21
ȳ = 431/6 = 71.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 191
Syy = Σ(y − ȳ)² = 556.8
Correlation coefficient
r = 191/√(70·556.8) = 0.9674
Answer
r = 0.9674
20.Linear regression & correlation
Answer: e = 5.45
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 = 391/5 = 78.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 254
Regression line
b = 254/40 = 6.35
a = 78.2 − 6.35·5 = 46.45
ŷ = 46.45 + 6.35·x
Predicted value
ŷ = 46.45 + 6.35·6 = 84.55
Residual
e = 90 − 84.55 = 5.45
Answer
e = 5.45
The point lies above the line.
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