You invest €5 000 at 6% per year with continuous compounding. Find the value FV after 7 years.
PV = €5 000r = 6%t = 7
FV = €
2.Time value of moneyHard
You borrow €20 000 at 3% per year, compounded monthly, and repay €250 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 = €20 000A = €2503% / 12
N = payments
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 4: €9 000, €6 000, €8 000, €6 000. The cost of capital is 7.000000000000001%. Find the net present value NPV.
A project costs €200 000 today. Its cash flow is €55 000 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 4. The cost of capital is 12%. Find the net present value NPV.
C0 = €200 000CF1 = €55 000g = 3%n = 4r = 12%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 6 years. The yield to maturity is 5% per year. Find the price P of the bond.
F = €1 000coupon = 4%YTM = 5%n = 6
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 7 years and sells for €1 060. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 7P = €1 060
YTM = %
7.CAPM & WACCMedium
A share with a beta of 1.6 has a required return of 10.6%. The risk-free rate is 1%. What expected market return rm does CAPM imply?
β = 1.6re = 10.6%rf = 1%
rm = %
8.CAPM & WACCHard
A company has equity with a market value of €6 000 000 and a cost of equity of 11%. Its debt consists of 2 000 bonds (face value €1 000, annual coupon 4%, 10 years to maturity) that trade at €1 042 each. The tax rate is 30%. 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: 54, 24, 42, 53, 51.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 182?
μ = 170σ = 12x = 182
P =
11.ProbabilityMedium
A bag holds 3 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 8
P =
12.ProbabilityHard
10 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?
n = 10p = 0.7
P =
13.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65.5 and 67 points? Give P as a decimal.
μ = 65σ = 10x1 = 65.5x2 = 67
P =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 95% is to the right of it. Find the cut-off x.
μ = 100 pointsσ = 15 points95%
x = points
15.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 7.6. What is the margin of error E₂ of a 95% interval based on 200 observations?
99%n₁ = 100E₁ = 7.695%n₂ = 200
E₂ =
16.Confidence intervalsHard
The population standard deviation is 21. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 5?
σ = 21E = 590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 96.88 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 96.88s = 12n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 50.2. The population standard deviation is 12. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 50.2σ = 12n = 2590%
U =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales 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 = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (3.5, 31) is made. Find its residual e = y − ŷ.
n = 5(3.5, 31)
e =
3Solutions
1.Time value of money
Answer: FV = 7610 €
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 = 0.06·7 = 0.42
e0.42 = 1.522
Future value
FV = 5 000·1.522 = €7 609.81
Answer
FV = €7 609.81
2.Time value of money
Answer: N = 90 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 = 0.0025
(1 + i)−N = 1 − 20 000·0.0025/250 = 0.8
Solve with logarithms
N = −ln(0.8)/ln(1.002) = 89.37
Round up to whole payments
89 full payments leave a small balance, which the last, smaller payment clears.
N = 90
Answer
N = 90 payments
3.NPV & IRR
Answer: NPV = 4760 €
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 + 0.07)1 = 9 000/1.07 = 8 411.21
CF2/(1 + 0.07)2 = 6 000/1.145 = 5 240.63
CF3/(1 + 0.07)3 = 8 000/1.225 = 6 530.38
CF4/(1 + 0.07)4 = 6 000/1.311 = 4 577.37
Add them up
NPV = €4 759.60
Answer
NPV = €4 759.60
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: NPV = −26005 €
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.
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: P = 948.7 €
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 = €20.00, i = 0.025, N = 12
Present value of the coupons
€20.00·(1 − 0.7436)/0.025 = €205.16
Present value of the face value
€1 000·0.7436 = €743.56
Price
P = €948.71
Answer
P = €948.71
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 4.964 %
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
€1 060 = €60·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷
Solve numerically
P(4%) = €1 120.04, P(5%) = €1 057.86
YTM = 4.964%
Answer
YTM = 4.964%
7.CAPM & WACC
Answer: rm = 7 %
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).
Market risk premium
rm − rf = (10.6% − 1%)/1.6 = 6%
Market return
rm = 1% + 6% = 7%
Answer
rm = 7%
8.CAPM & WACC
Answer: WACC = 8.795 %
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 = 1
Standard normal table
Φ(1) = 0.8413
P = 0.8413
Answer
P = 0.8413
11.Probability
Answer: P = 0.05455
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 = 3/11 · 2/10 = 6/110 = 0.05455
Answer
P = 0.05455
12.Probability
Answer: P = 0.0001437
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) = 0.310 = 5.905 × 10⁻⁶
P(X = 1) = 10·0.7·0.39 = 0.0001378
Add them
P = 5.905 × 10⁻⁶ + 0.0001378 = 0.0001437
Answer
P = 0.0001437
13.Normal distribution
Answer: P = 0.05932
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 = (65.5 − 65)/10 = 0.05
z₂ = (67 − 65)/10 = 0.2
Area between
P = Φ(0.2) − Φ(0.05) = 0.5793 − 0.5199 = 0.05932
Answer
P = 0.05932
14.Normal distribution
Answer: x = 75.4 points
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) = 0.05
z from the table
z ≈ −1.64
Back to the original units
x = 100 − 1.64·15 = 75.4 points
Answer
x = 75.4 points
15.Confidence intervals
Answer: E₂ = 4.089
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 = (1.96/2.576)·√(100/200) = 0.538
New margin
E₂ = 7.6·0.538 = 4.089
Answer
E₂ = 4.089
16.Confidence intervals
Answer: n = 48
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 = (1.645·21/5)² = 47.73
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 48
Answer
n = 48
17.Hypothesis tests & confidence intervals
Answer: t = −2.6
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/√100 = 12/10 = 1.2
Test statistic
t = (96.88 − 100)/1.2 = −2.6
Answer
t = −2.6
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 54.15
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).
Margin of error
E = 1.645·12/√25 = 3.948
Interval limit
U = 50.2 + 3.948 = 54.15
Answer
U = 54.15
19.Linear regression & correlation
Answer: r² = 0.7994
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 = 35/5 = 7
ȳ = 202/5 = 40.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 58
Syy = Σ(y − ȳ)² = 105.2
Coefficient of determination
r² = 58²/(40·105.2) = 0.7994
Answer
r² = 0.7994
About 79.94% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = −1.75
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.