You invest €1 000 at 5% per year with continuous compounding. Find the value FV after 10 years.
PV = €1 000r = 5%t = 10
FV = €
2.Time value of moneyHard
You borrow €10 000 at 6% per year, compounded monthly, and repay €325 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.
P = €10 000A = €3256% / 12
N = payments
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 3: €9 000, €5 000, €9 000. The cost of capital is 6%. Find the net present value NPV.
A project costs €20 000 and returns €12 500, €12 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 12 500CF2 = 12 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 7% per year. Find the price P of the bond.
F = €1 000coupon = 2%YTM = 7%n = 7
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 3% matures in 8 years and sells for €900. Find its yield to maturity YTM in percent.
F = €1 000coupon = 3%n = 8P = €900
YTM = %
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €3 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.
A company has 1 000 000 shares trading at €25.00; the book value of its equity is €12 500 000. It just paid a dividend of €2.20 per share, and dividends grow by 5% per year. Its debt has a market value of €20 000 000 and a pre-tax cost of 5%. The tax rate is 25%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 22, 51, 44, 57, 28, 59, 12.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is above 116?
μ = 100σ = 8x = 116
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.05, 0.25, 0.3, 0.1
Var(X) =
12.ProbabilityHard
A game costs €3 to play. You win €15 with probability 0.25 and €30 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€15: p = 0.25€30: p = 0.02
E(G) = €
13.Normal distributionMedium
X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 754.9 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 754.9 g
P =
14.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
A sample of 10 items has mean 101.5 and standard deviation s = 6. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 10x̄ = 101.5s = 695%t(9) = 2.262
lower limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 8 plots, mean yield 68.5 kg, standard deviation 6.5 kg. Plot group 2: 6 plots, mean 75.5 kg, standard deviation 6.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 99% confidence interval for μ₁ − μ₂.
Two teaching methods are compared. Group 1: 9 students, mean score 60.5, standard deviation 3. Group 2: 17 students, mean score 55.6, standard deviation 4.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 9x̄₁ = 60.5s₁ = 3n₂ = 17x̄₂ = 55.6s₂ = 4.5
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 48 of 80 patients on treatment A recover, and 86 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 48n₁ = 80x₂ = 86n₂ = 120
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 35) is made. Find its residual e = y − ŷ.
n = 5(9, 35)
e =
3Solutions
1.Time value of money
Answer: FV = 1649 €
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.
Exponent
r·t = 0.05·10 = 0.5
e0.5 = 1.649
Future value
FV = 1 000·1.649 = €1 648.72
Answer
FV = €1 648.72
2.Time value of money
Answer: N = 34 payments
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.
Rearrange the annuity formula
i = 0.005
(1 + i)−N = 1 − 10 000·0.005/325 = 0.8462
Solve with logarithms
N = −ln(0.8462)/ln(1.005) = 33.49
Round up to whole payments
33 full payments leave a small balance, which the last, smaller payment clears.
N = 34
Answer
N = 34 payments
3.NPV & IRR
Answer: NPV = 497.1 €
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 = −20 000
CF1/(1 + 0.06)1 = 9 000/1.06 = 8 490.57
CF2/(1 + 0.06)2 = 5 000/1.124 = 4 449.98
CF3/(1 + 0.06)3 = 9 000/1.191 = 7 556.57
Add them up
NPV = €497.12
Answer
NPV = €497.12
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 16.26 %
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(15%) = 321.36
NPV(20%) = −902.78
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 15 + 5·321.36/(321.36 − (−902.78)) = 16.31%
exact: NPV(16.26%) = 0 → IRR = 16.26%
Answer
IRR = 16.26%
5.Bond pricing & yield
Answer: P = 727 €
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, i = 0.035, N = 14
Present value of the coupons
€10.00·(1 − 0.6178)/0.035 = €109.21
Present value of the face value
€1 000·0.6178 = €617.78
Price
P = €726.99
Answer
P = €726.99
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 4.517 %
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
€900 = €30·(1 − (1 + y)⁻⁸)/y + €1 000/(1 + y)⁸
Solve numerically
P(4%) = €932.67, P(5%) = €870.74
YTM = 4.517%
Answer
YTM = 4.517%
7.CAPM & WACC
Answer: WACC = 7.437 %
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 = €13 000 000, E/V = 0.7692, D/V = 0.2308
WACC
WACC = 0.7692·9% + 0.2308·3%·(1 − 0.258) = 7.437%
Answer
WACC = 7.437%
8.CAPM & WACC
Answer: WACC = 9.578 %
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
11.Probability
Answer: Var(X) = 1.928
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.
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.
Net outcomes
G: 12 (p = 0.25), 27 (p = 0.02), −3 (p = 0.73)
Expected value
E(G) = 12·0.25 + 27·0.02 − 3·0.73 = 1.35 €
Answer
E(G) = 1.35 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.3085
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 = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (754.9 − 750)/9.849 = 0.5
Upper tail
P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085
Answer
P = 0.3085
14.Normal distribution
Answer: P = 0.8413
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 = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
15.Confidence intervals
Answer: lower limit = 97.21
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.
Critical value
df = n − 1 = 9, t = 2.262
Margin of error
E = 2.262·6/√10 = 4.292
Interval
101.5 ± 4.292 → [97.21, 105.8]
Answer
lower limit = 97.21
16.Confidence intervals
Answer: lower limit = −17.72 kg
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.
Pooled standard deviation
sp² = (7·6.5² + 5·6.5²)/12 = 42.25
sp = 6.5
Margin of error
E = 3.055·6.5·√(1/8 + 1/6) = 10.72
Interval
−7 ± 10.72 → [−17.72, 3.724]
Answer
lower limit = −17.72 kg
17.Hypothesis tests & confidence intervals
Answer: t = 2.926
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.
Pooled variance
sp² = (8·3² + 16·4.5²)/24 = 16.5
sp = 4.062
Standard error
SE = 4.062·√(1/9 + 1/17) = 1.674
Test statistic
t = (60.5 − 55.6)/1.674 = 2.926
Answer
t = 2.926
With df = 24, the two-sided 5% critical value is 2.064; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: z = −1.719
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 = 48/80 = 0.6
p̂₂ = 86/120 = 0.7167
p̂ = 134/200 = 0.67
Standard error under H₀
SE = √(0.67·0.33·(1/80 + 1/120)) = 0.06787
Test statistic
z = (0.6 − 0.7167)/0.06787 = −1.719
Answer
z = −1.719
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r² = 0.9713
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 = 135/6 = 22.5
ȳ = 575/6 = 95.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 757.5
Syy = Σ(y − ȳ)² = 3751
Coefficient of determination
r² = 757.5²/(157.5·3751) = 0.9713
Answer
r² = 0.9713
About 97.13% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = 4.55
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 = 30/5 = 6
ȳ = 99/5 = 19.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 142
Regression line
b = 142/40 = 3.55
a = 19.8 − 3.55·6 = −1.5
ŷ = −1.5 + 3.55·x
Predicted value
ŷ = −1.5 + 3.55·9 = 30.45
Residual
e = 35 − 30.45 = 4.55
Answer
e = 4.55
The point lies above the line.
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