You need €5 000 in 8 years. An account pays 3.6% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €5 0003.6% / 12n = 8
PV = €
2.Time value of moneyHard
You want €10 000 in 15 years. You save the same amount at the end of every month in an account paying 4.8% per year, compounded monthly. Find the monthly deposit A.
FV = €10 0004.8% / 12N = 180
A = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 3: €13 000, €20 000, €21 000. The cost of capital is 10%. Find the net present value NPV.
An investment of €40 000 returns €9 000 at the end of each year for 7 years. Find its internal rate of return IRR in percent.
C0 = €40 000CF = €9 000n = 7
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 9% per year. Find the price P of the bond.
F = €1 000coupon = 5%YTM = 9%n = 5
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 3% matures in 4 years and sells for €1 080. Find its yield to maturity YTM in percent.
F = €1 000coupon = 3%n = 4P = €1 080
YTM = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 0.25. Its cost of equity is 13%, its pre-tax cost of debt is 5% and the tax rate is 21%. Find the WACC.
D/E = 0.25re = 13%rd = 5%T = 21%
WACC = %
8.CAPM & WACCHard
A company has 5 000 000 shares at €39.00 each and 10 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 20%. The share’s beta is 0.6, the risk-free rate 2% and the market return 9%. Find the WACC.
β = 0.6rf = 2%rm = 9%rd = 3%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 19, 25, 16, 44, 29.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 67.5.
μ = 70σ = 5x = 67.5
z =
11.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
12.ProbabilityHard
A game costs €3 to play. You win €6 with probability 0.2 and €150 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€6: p = 0.2€150: p = 0.01
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9651. Find z to two decimals.
A = 0.9651
z =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 90% of them lie (the 90th percentile).
μ = 250σ = 5P = 90%
x = g
15.Confidence intervalsMedium
A sample of 30 items has mean 110 and standard deviation s = 9.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 30x̄ = 110s = 9.599%t(29) = 2.756
upper limit =
16.Confidence intervalsHard
The diameters of 10 machined shafts have a sample standard deviation of 4.9 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 = 4.9 µmχ² = 2.7, 19.023
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 101.5 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 101.5s = 5n = 16
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.10 with a sample of 36; the standard deviation is known to be 3. In reality the mean is 98.85. Find the power 1 − β of the test in percent.
μ₀ = 100μ = 98.85σ = 3n = 36α = 0.10
1 − β = %
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 15Σy = 103Σx² = 55Σy² = 2163Σxy = 292
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. For which value x does the line predict ŷ = 111?
n = 5ŷ = 111
x =
3Solutions
1.Time value of money
Answer: PV = 3750 €
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.
Rate and number of periods
i = 0.036000000000000004/12 = 0.003
N = 12·8 = 96
Present value
PV = FV/(1 + i)N = 5 000/1.333 = €3 750.42
Answer
PV = €3 750.42
2.Time value of money
Answer: A = 38.04 €
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 each cash flow
CF0 = −45 000
CF1/(1 + 0.1)1 = 13 000/1.1 = 11 818.18
CF2/(1 + 0.1)2 = 20 000/1.21 = 16 528.93
CF3/(1 + 0.1)3 = 21 000/1.331 = 15 777.61
Add them up
NPV = €−875.28
Answer
NPV = €−875.28
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 12.84 %
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)−7)/i = 40 000/9 000 = 4.444
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 12%: 4.564
i = 13%: 4.423
IRR = 12.84%
Answer
IRR = 12.84%
5.Bond pricing & yield
Answer: P = 841.7 €
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 = €25.00, i = 0.045, N = 10
Present value of the coupons
€25.00·(1 − 0.6439)/0.045 = €197.82
Present value of the face value
€1 000·0.6439 = €643.93
Price
P = €841.75
Answer
P = €841.75
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 0.9522 %
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
€1 080 = €30·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴
Solve numerically
P(1%) = €1 078.04, P(1%) = €1 078.04
YTM = 0.9522%
Answer
YTM = 0.9522%
7.CAPM & WACC
Answer: WACC = 11.19 %
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 + 0.25) = 0.8
D/V = 0.25/(1 + 0.25) = 0.2
WACC
WACC = 0.8·13% + 0.2·5%·(1 − 0.21) = 11.19%
Answer
WACC = 11.19%
8.CAPM & WACC
Answer: WACC = 6.006 %
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 = (67.5 − 70)/5 = −0.5
Answer
z = −0.5
11.Probability
Answer: P = 0.3056
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)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
12.Probability
Answer: E(G) = −0.3 €
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: 3 (p = 0.2), 147 (p = 0.01), −3 (p = 0.79)
Expected value
E(G) = 3·0.2 + 147·0.01 − 3·0.79 = −0.3 €
Answer
E(G) = −0.3 €
A negative value means you lose money on average.
13.Normal distribution
Answer: z = 2.11
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 of z
Φ(z) = (1 + 0.9651)/2 = 0.9826
Read the z-table backwards
Find 0.9826 inside the table and read off z.
z = 2.11
Answer
z = 2.11
14.Normal distribution
Answer: x = 256.4 g
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 for the percentile
Φ(z) = 0.9 → z = 1.282
Back to x
x = 250 + 1.282·5 = 256.41 g
Answer
x = 256.41 g
15.Confidence intervals
Answer: upper limit = 114.8
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 = 29, t = 2.756
Margin of error
E = 2.756·9.5/√30 = 4.78
Interval
110 ± 4.78 → [105.2, 114.8]
Answer
upper limit = 114.8
16.Confidence intervals
Answer: σ = 3.37 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 = 9·4.9² = 216.1
Limits for σ²
216.1/19.023 = 11.36
216.1/2.7 = 80.03
Limit for σ
σ = √11.36 = 3.37 µm
Answer
σ = 3.37 µm
95% interval for σ: [3.37, 8.946] µm.
17.Hypothesis tests & confidence intervals
Answer: t = 1.2
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 = 5/√16 = 5/4 = 1.25
Test statistic
t = (101.5 − 100)/1.25 = 1.2
Answer
t = 1.2
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 84.57 %
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 = 3/√36 = 0.5
x̄c = 100 − 1.282·0.5 = 99.36
z under the true mean
z = (99.36 − 98.85)/0.5 = 1.018
Power
power = P(Z < 1.018) = 84.57%
1 − β = 84.57%
Answer
1 − β = 84.57%
19.Linear regression & correlation
Answer: r = −0.8375
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 = 292 − 15·103/5 = −17
Sxx = 55 − 15²/5 = 10
Syy = 2163 − 103²/5 = 41.2
Correlation coefficient
r = −17/√(10·41.2) = −0.8375
Answer
r = −0.8375
20.Linear regression & correlation
Answer: x = 9.062
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 = 444/5 = 88.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 290
Regression line
b = 290/40 = 7.25
a = 88.8 − 7.25·6 = 45.3
ŷ = 45.3 + 7.25·x
Solve for x
x = (111 − 45.3)/7.25 = 9.062
Answer
x = 9.062
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