You invest €1 000 at 7% per year with continuous compounding. Find the value FV after 6 years.
PV = €1 000r = 7%t = 6
FV = €
2.Time value of moneyHard
A loan of €25 000 is repaid with equal monthly payments over 3 years at 4.2% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €25 0004.2% / 12N = 36
I = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 3: €18 000, €18 000, €13 000. The cost of capital is 9%. Find the net present value NPV.
A project costs €50 000 and returns €29 500, €29 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 29 500CF2 = 29 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 5% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 5%n = 5
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 4 years. It sells for €1 170. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 7% (2×)n = 4P = €1 170
YTM = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €2 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.
A company has 2 000 000 shares at €16.50 each and 10 000 bonds trading at €1 000 each. The pre-tax cost of debt is 3% and the tax rate is 25.8%. The share’s beta is 1.3, the risk-free rate 3% and the market return 9.5%. Find the WACC.
β = 1.3rf = 3%rm = 9.5%rd = 3%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 58, 20, 11, 32, 46.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 15. What proportion P of values is below 79?
μ = 70σ = 15x = 79
P =
11.ProbabilityMedium
A fair die is rolled 3 times. What is the probability P of getting at least one six?
n = 3
P =
12.ProbabilityHard
6 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 6p = 0.3
P =
13.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 53 and 70 points? Give P as a decimal.
μ = 65σ = 10x1 = 53x2 = 70
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g70%
x = g
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 80 phones, mean 130 h, known standard deviation 10 h. Model 2: 80 phones, mean 128.5 h, known standard deviation 19 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
A pilot study suggests that about 0.6 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.025?
p ≈ 0.6E = 0.02595%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 502.8 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 502.8s = 15n = 81
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 11. Find the minimum sample size n.
The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
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 = 4.5.
n = 5x = 4.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 1522 €
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 = 0.07·6 = 0.42
e0.42 = 1.522
Future value
FV = 1 000·1.522 = €1 521.96
Answer
FV = €1 521.96
2.Time value of money
Answer: I = 1652 €
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 payment
i = 0.0035, N = 36
A = 25 000·0.0035/(1 − (1 + 0.0035)−36) = €740.33
Total interest
I = N·A − P = 36·740.3 − 25 000 = €1 651.73
Answer
I = €1 651.73
3.NPV & IRR
Answer: NPV = −3298 €
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 = −45 000
CF1/(1 + 0.09)1 = 18 000/1.09 = 16 513.76
CF2/(1 + 0.09)2 = 18 000/1.188 = 15 150.24
CF3/(1 + 0.09)3 = 13 000/1.295 = 10 038.39
Add them up
NPV = €−3 297.61
Answer
NPV = €−3 297.61
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 11.78 %
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(10%) = 1 198.35
NPV(15%) = −2 041.59
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.
Per period
C = €40.00, i = 0.025, N = 10
Present value of the coupons
€40.00·(1 − 0.7812)/0.025 = €350.08
Present value of the face value
€1 000·0.7812 = €781.20
Price
P = €1 131.28
Answer
P = €1 131.28
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 2.507 %
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, N = 8
Solve numerically
i = 1.253% per half-year
YTM
YTM = 2·1.253% = 2.507%
Answer
YTM = 2.507%
7.CAPM & WACC
Answer: WACC = 8.493 %
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).
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 0.7257
Answer
P = 0.7257
11.Probability
Answer: P = 0.4213
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)3 = 0.5787
P = 1 − 0.5787 = 0.4213
Answer
P = 0.4213
12.Probability
Answer: P = 0.4202
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) = 0.76 = 0.1176
P(X = 1) = 6·0.3·0.75 = 0.3025
Add them
P = 0.1176 + 0.3025 = 0.4202
Answer
P = 0.4202
13.Normal distribution
Answer: P = 0.5764
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-scores
z₁ = (53 − 65)/10 = −1.2
z₂ = (70 − 65)/10 = 0.5
Area between
P = Φ(0.5) − Φ(−1.2) = 0.6915 − 0.1151 = 0.5764
Answer
P = 0.5764
14.Normal distribution
Answer: x = 252.6 g
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left
Φ(z) = 0.7
z from the table
z ≈ 0.52
Back to the original units
x = 250 + 0.52·5 = 252.6 g
Answer
x = 252.6 g
15.Confidence intervals
Answer: E = 3.949 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 = 130 − 128.5 = 1.5
Standard error
SE = √(10²/80 + 19²/80) = 2.401
Margin of error
E = 1.645·2.401 = 3.949
Answer
E = 3.949 h
The interval contains 0, so the data do not show a significant difference.
16.Confidence intervals
Answer: n = 1476
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.96²·0.6·0.4/0.025² = 1475
Round up
n = 1476
Answer
n = 1476
17.Hypothesis tests & confidence intervals
Answer: t = 1.68
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/√81 = 15/9 = 1.667
Test statistic
t = (502.8 − 500)/1.667 = 1.68
Answer
t = 1.68
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 47
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((1.645 + 0.842)·11/4)² = 46.78
Round up
n = 47
Answer
n = 47
19.Linear regression & correlation
Answer: r = 0.983
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 = 1841 − 25·358/5 = 51
Sxx = 135 − 25²/5 = 10
Syy = 25902 − 358²/5 = 269.2
Correlation coefficient
r = 51/√(10·269.2) = 0.983
Answer
r = 0.983
20.Linear regression & correlation
Answer: ŷ = 18.95
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 = 103/5 = 20.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −11
Regression line
b = −11/10 = −1.1
a = 20.6 − (−1.1)·3 = 23.9
ŷ = 23.9 − 1.1·x
Prediction
ŷ = 23.9 + (−1.1)·4.5 = 18.95
Answer
ŷ = 18.95
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