You invest €12 000 at a nominal rate of 4.8% per year, compounded quarterly. Find the value FV of the investment after 7 years.
PV = €12 0004.8% / 4n = 7
FV = €
2.Time value of moneyHard
You borrow €10 000 at 7.2% per year, compounded monthly, and repay €150 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 = €1507.2% / 12
N = payments
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 4: €20 000, €34 000, €26 000, €16 000. The cost of capital is 10%. Find the net present value NPV.
A project costs €25 000 and returns €12 000, €13 000, €14 000 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 12 years. Its yield to maturity changes from 3.5% to 4%. By how much ΔP does the bond price change?
F = €1 000coupon = 4%n = 12y1 = 3.5%y2 = 4%
ΔP = €
6.Bond pricing & yieldHard
A bond with 13 years to maturity (face value €1 000, annual coupon 8%) sells for €1 228. The issuer can call it in 5 years at a call price of €1 080. Find the yield to call YTC in percent.
F = €1 000coupon = 8%P = €1 228call price = €1 080call in 5 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €6 000 000 and debt worth €1 000 000. Its WACC is 7.5%, its pre-tax cost of debt is 6% and the tax rate is 25%. Find the cost of equity re.
A company has 2 500 000 shares at €35.00 each and 5 000 bonds trading at €1 020 each. The pre-tax cost of debt is 3% and the tax rate is 30%. The share’s beta is 1.5, the risk-free rate 2.5% and the market return 9.5%. Find the WACC.
β = 1.5rf = 2.5%rm = 9.5%rd = 3%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 32, 60, 28, 45, 10, 59, 59.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 462.5.
μ = 500σ = 25x = 462.5
z =
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
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 2 correct answers?
n = 7p = 0.3k = 2
P =
13.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 29.3 and 33.2 min? Give P as a decimal.
μ = 32σ = 6x1 = 29.3x2 = 33.2
P =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 10% of them lie (the 10th percentile).
μ = 65σ = 10P = 10%
x = points
15.Confidence intervalsMedium
The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 3.3 and standard deviation 4.2. Find the upper limit of the 99% confidence interval for the mean difference.
n = 10d̄ = 3.3sd = 4.299%t(9) = 3.25
upper limit = mmHg
16.Confidence intervalsHard
The population standard deviation is 18. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 5?
σ = 18E = 599%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 21 students, mean score 22.5, standard deviation 5.5. Group 2: 10 students, mean score 27.1, standard deviation 9.5. 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).
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 199%
n =
19.Linear regression & correlationMedium
The table gives 6 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 = 6
r² =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 29?
n = 6ŷ = 29
x =
3Solutions
1.Time value of money
Answer: FV = 16759 €
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.
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 = 0.006
(1 + i)−N = 1 − 10 000·0.006/150 = 0.6
Solve with logarithms
N = −ln(0.6)/ln(1.006) = 85.39
Round up to whole payments
85 full payments leave a small balance, which the last, smaller payment clears.
N = 86
Answer
N = 86 payments
3.NPV & IRR
Answer: NPV = 1743 €
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 = −75 000
CF1/(1 + 0.1)1 = 20 000/1.1 = 18 181.82
CF2/(1 + 0.1)2 = 34 000/1.21 = 28 099.17
CF3/(1 + 0.1)3 = 26 000/1.331 = 19 534.18
CF4/(1 + 0.1)4 = 16 000/1.464 = 10 928.22
Add them up
NPV = €1 743.39
Answer
NPV = €1 743.39
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: MIRR = 18.18 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
12 000·1.062 = 13 483.20
13 000·1.061 = 13 780.00
14 000·1.060 = 14 000.00
TV = 41 263.20
MIRR
MIRR = (41 263.20/25 000)1/3 − 1 = 18.18%
Answer
MIRR = 18.18%
5.Bond pricing & yield
Answer: ΔP = −48.32 €
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.
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
1228 = 80·(1 − (1 + y)−5)/y + 1080/(1 + y)5
Solve numerically
y = 4%: €1 243.83
y = 5%: €1 192.57
YTC = 4.303%
Answer
YTC = 4.303%
The yield to maturity is 5.499%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.
7.CAPM & WACC
Answer: re = 8 %
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 = €7 000 000, E/V = 0.8571, D/V = 0.1429
Debt part of the WACC
0.1429·6%·(1 − 0.25) = 0.6429%
Cost of equity
re = (7.5% − 0.6429%)/0.8571 = 8%
Answer
re = 8%
8.CAPM & WACC
Answer: WACC = 12.4 %
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 values
E = 2 500 000·€35.00 = €87 500 000
D = 5 000·€1 020 = €5 100 000
Cost of equity (CAPM)
re = 2.5% + 1.5·7% = 13%
Weights
V = €92 600 000, E/V = 0.9449, D/V = 0.05508
WACC
WACC = 0.9449·13% + 0.05508·3%·(1 − 0.3) = 12.4%
Answer
WACC = 12.4%
9.Descriptive statistics
Answer: s = 19.28
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 = (462.5 − 500)/25 = −1.5
Answer
z = −1.5
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 = 0.3177
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 = 2) = 21·0.32·0.75 = 21·0.09·0.1681 = 0.3177
Answer
P = 0.3177
13.Normal distribution
Answer: P = 0.2529
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 = (29.3 − 32)/6 = −0.45
z₂ = (33.2 − 32)/6 = 0.2
Area between
P = Φ(0.2) − Φ(−0.45) = 0.5793 − 0.3264 = 0.2529
Answer
P = 0.2529
14.Normal distribution
Answer: x = 52.18 points
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) = 0.1 → z = −1.282
Back to x
x = 65 + −1.282·10 = 52.18 points
Answer
x = 52.18 points
15.Confidence intervals
Answer: upper limit = 7.617 mmHg
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.
Critical value
df = 9, t = 3.25
Margin of error
E = 3.25·4.2/√10 = 4.317
Interval
3.3 ± 4.317 → [−1.017, 7.617]
Answer
upper limit = 7.617 mmHg
16.Confidence intervals
Answer: n = 86
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 = (2.576·18/5)² = 86
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 86
Answer
n = 86
17.Hypothesis tests & confidence intervals
Answer: t = −1.713
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 = (20·5.5² + 9·9.5²)/29 = 48.87
sp = 6.991
Standard error
SE = 6.991·√(1/21 + 1/10) = 2.686
Test statistic
t = (22.5 − 27.1)/2.686 = −1.713
Answer
t = −1.713
With df = 29, the two-sided 5% critical value is 2.045; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: n = 2655
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).
Solve for n
n = (2.576·20/1)² = 2654
Round up
n = 2655
Answer
n = 2655
19.Linear regression & correlation
Answer: r² = 0.9884
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 = 42/6 = 7
ȳ = 358/6 = 59.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 384
Syy = Σ(y − ȳ)² = 2131
Coefficient of determination
r² = 384²/(70·2131) = 0.9884
Answer
r² = 0.9884
About 98.84% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: x = 7.447
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 = 42/6 = 7
ȳ = 165/6 = 27.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 235
Regression line
b = 235/70 = 3.357
a = 27.5 − 3.357·7 = 4
ŷ = 4 + 3.357·x
Solve for x
x = (29 − 4)/3.357 = 7.447
Answer
x = 7.447
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