You invest €1 000 at 6% per year with continuous compounding. Find the value FV after 19 years.
PV = €1 000r = 6%t = 19
FV = €
2.Time value of moneyHard
You deposit €2 500 at the start of every year for 5 years, the first deposit today. The account pays 7% per year. How much is in the account at the end of year 5? Find FV.
A = €2 500i = 7%n = 5
FV = €
3.NPV & IRRMedium
A project costs €50 000 today and returns €15 000 at the end of each year for 5 years. The cost of capital is 10%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €20 000 and returns €9 000, €13 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 9 000CF2 = 13 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 8 years. The yield to maturity is 3% per year. Find the price P of the bond.
F = €1 000coupon = 4.5%YTM = 3%n = 8
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 8 years. It sells for €1 173. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 6% (2×)n = 8P = €1 173
YTM = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 0.4. Its cost of equity is 14%, its pre-tax cost of debt is 4% and the tax rate is 20%. Find the WACC.
D/E = 0.4re = 14%rd = 4%T = 20%
WACC = %
8.CAPM & WACCHard
A company has equity with a market value of €6 000 000 and a cost of equity of 10%. Its debt consists of 5 000 bonds (face value €1 000, annual coupon 4%, 6 years to maturity) that trade at €1 054 each. The tax rate is 21%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.
Find the sample standard deviation s of the data: 19, 31, 18, 18, 20, 47.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 12. Find the z-score z of a score of 44.
μ = 50σ = 12x = 44
z =
11.ProbabilityMedium
A fair die is rolled 7 times. What is the probability P of getting at least one six?
n = 7
P =
12.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 3 correct answers?
n = 6p = 0.6k = 3
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 199.8 mm. Round z to two decimals.
μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 199.8 mm
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.03 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.03 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
A sample of 6 items has mean 88.5 and standard deviation s = 5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 6x̄ = 88.5s = 590%t(5) = 2.015
E =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 6 plots, mean yield 14.5 kg, standard deviation 4 kg. Plot group 2: 6 plots, mean 4 kg, standard deviation 10 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
A sample of 49 students has a mean score of 501.5 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 501.5s = 15n = 49
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 2.5. In reality the mean is 78.9. Find the power 1 − β of the test in percent.
μ₀ = 80μ = 78.9σ = 2.5n = 25α = 0.05
1 − β = %
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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. Predict ŷ for x = 17.
n = 5x = 17
ŷ =
3Solutions
1.Time value of money
Answer: FV = 3127 €
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 = 0.06·19 = 1.14
e1.14 = 3.127
Future value
FV = 1 000·3.127 = €3 126.77
Answer
FV = €3 126.77
2.Time value of money
Answer: FV = 15383 €
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.
Ordinary annuity factor
((1 + 0.07)5 − 1)/0.07 = 5.751
Annuity due
Each deposit is invested one year longer.
FV = 2 500·5.751·1.07 = €15 383.23
Answer
FV = €15 383.23
3.NPV & IRR
Answer: DPB = 4.263 years
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 15 000/1.1 = 13 636.36 → Σ = 13 636.36
year 2: 15 000/1.21 = 12 396.69 → Σ = 26 033.06
year 3: 15 000/1.331 = 11 269.72 → Σ = 37 302.78
year 4: 15 000/1.464 = 10 245.20 → Σ = 47 547.98
year 5: 15 000/1.611 = 9 313.82 → Σ = 56 861.80
Fraction of the last year
DPB = 4 + 2 452.02/9 313.82 = 4.263
Answer
DPB = 4.263 years
The discounted payback is always longer than the simple payback (3.333 years).
4.NPV & IRR
Answer: IRR = 7.684 %
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%) = 816.33
NPV(10%) = −661.16
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 5 + 5·816.33/(816.33 − (−661.16)) = 7.763%
exact: NPV(7.684%) = 0 → IRR = 7.684%
Answer
IRR = 7.684%
5.Bond pricing & yield
Answer: P = 1106 €
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 = €22.50, i = 0.015, N = 16
Present value of the coupons
€22.50·(1 − 0.788)/0.015 = €317.95
Present value of the face value
€1 000·0.788 = €788.03
Price
P = €1 105.98
Answer
P = €1 105.98
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 3.502 %
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 = €30.00, N = 16
Solve numerically
i = 1.751% per half-year
YTM
YTM = 2·1.751% = 3.502%
Answer
YTM = 3.502%
7.CAPM & WACC
Answer: WACC = 10.91 %
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 from D/E
E/V = 1/(1 + 0.4) = 0.7143
D/V = 0.4/(1 + 0.4) = 0.2857
WACC
WACC = 0.7143·14% + 0.2857·4%·(1 − 0.2) = 10.91%
Answer
WACC = 10.91%
8.CAPM & WACC
Answer: WACC = 6.433 %
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 = (44 − 50)/12 = −0.5
Answer
z = −0.5
11.Probability
Answer: P = 0.7209
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)7 = 0.2791
P = 1 − 0.2791 = 0.7209
Answer
P = 0.7209
12.Probability
Answer: P = 0.2765
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.
Binomial formula
P(X = 3) = 20·0.63·0.43 = 20·0.216·0.064 = 0.2765
Answer
P = 0.2765
13.Normal distribution
Answer: P = 0.6554
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 = (199.8 − 200)/0.5 = −0.4
Upper tail
P = 1 − Φ(−0.4) = 1 − 0.3446 = 0.6554
Answer
P = 0.6554
14.Normal distribution
Answer: P = 0.9207
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 = 25.03 − 25 = 0.03 mm, σ = 0.02121 mm
Standardise
z = (0 − 0.03)/0.02121 = −1.41
Probability
P = 1 − Φ(−1.41) = 0.9207
Answer
P = 0.9207
15.Confidence intervals
Answer: E = 4.113
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 = 5, t = 2.015
Margin of error
E = 2.015·5/√6 = 4.113
Answer
E = 4.113
16.Confidence intervals
Answer: lower limit = 0.7036 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 = (5·4² + 5·10²)/10 = 58
sp = 7.616
Margin of error
E = 2.228·7.616·√(1/6 + 1/6) = 9.796
Interval
10.5 ± 9.796 → [0.7036, 20.3]
Answer
lower limit = 0.7036 kg
17.Hypothesis tests & confidence intervals
Answer: t = 0.7
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 = 15/√49 = 15/7 = 2.143
Test statistic
t = (501.5 − 500)/2.143 = 0.7
Answer
t = 0.7
t0.025; df = 48 = 2.011
|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 71.06 %
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 = 2.5/√25 = 0.5
x̄c = 80 − 1.645·0.5 = 79.18
z under the true mean
z = (79.18 − 78.9)/0.5 = 0.555
Power
power = P(Z < 0.555) = 71.06%
1 − β = 71.06%
Answer
1 − β = 71.06%
19.Linear regression & correlation
Answer: r² = 0.6923
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 = 106/5 = 21.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −12
Syy = Σ(y − ȳ)² = 20.8
Coefficient of determination
r² = −12²/(10·20.8) = 0.6923
Answer
r² = 0.6923
About 69.23% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 59.45
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 = 90/5 = 18
ȳ = 317/5 = 63.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 158
Regression line
b = 158/40 = 3.95
a = 63.4 − 3.95·18 = −7.7
ŷ = −7.7 + 3.95·x
Prediction
ŷ = −7.7 + 3.95·17 = 59.45
Answer
ŷ = 59.45
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