You invest €2 000 at a nominal rate of 4% per year, compounded quarterly. Find the value FV of the investment after 2 years.
PV = €2 0004% / 4n = 2
FV = €
2.Time value of moneyHard
A loan of €10 000 is repaid with equal monthly payments over 6 years at 7.2% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €10 0007.2% / 12N = 72
I = €
3.NPV & IRRMedium
A machine costs €40 000 today. It generates €6 000 at the end of each year for 8 years and can then be sold for €8 000. The cost of capital is 8%. Find the net present value NPV.
C0 = €40 000CF = €6 000S = €8 000n = 8r = 8%
NPV = €
4.NPV & IRRHard
An investment of €20 000 returns €9 500 at the end of each year for 3 years. Find its internal rate of return IRR in percent.
C0 = €20 000CF = €9 500n = 3
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 4 years. The yield to maturity is 6% per year. Find the price P of the bond.
F = €1 000coupon = 5%YTM = 6%n = 4
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 6 years. It sells for €924. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 4% (2×)n = 6P = €924
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 6% and the tax rate is 25.8%. Find the WACC.
A company has 2 000 000 shares trading at €44.50; the book value of its equity is €53 400 000. It just paid a dividend of €2.70 per share, and dividends grow by 3% per year. Its debt has a market value of €44 500 000 and a pre-tax cost of 4.5%. The tax rate is 25%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 10, 19, 50, 39, 19, 43, 32.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is below 180?
μ = 170σ = 10x = 180
P =
11.ProbabilityMedium
A fair die is rolled 4 times. What is the probability P of getting at least one six?
n = 4
P =
12.ProbabilityHard
A game costs €5 to play. You win €20 with probability 0.3 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€20: p = 0.3€50: p = 0.05
E(G) = €
13.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 250.25 and 260 g? Give P as a decimal.
μ = 250σ = 5x1 = 250.25x2 = 260
P =
14.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 95% of them lie (the 95th percentile).
μ = 32σ = 6P = 95%
x = min
15.Confidence intervalsMedium
The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 3.7 and standard deviation 6.8. Find the lower limit of the 99% confidence interval for the mean difference.
n = 10d̄ = 3.7sd = 6.899%t(9) = 3.25
lower limit = mmHg
16.Confidence intervalsHard
The population standard deviation is 17. 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 4?
σ = 17E = 490%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 122.9 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 122.9s = 5n = 25
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 120.8. The population standard deviation is 8. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 120.8σ = 8n = 6499%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 17?
n = 6ŷ = 17
x =
3Solutions
1.Time value of money
Answer: FV = 2166 €
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 = r/m = 0.04/4 = 0.01
N = m·n = 4·2 = 8
Future value
FV = 2 000·(1 + 0.01)8 = 2 000·1.083 = €2 165.71
Answer
FV = €2 165.71
2.Time value of money
Answer: I = 2345 €
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 = 0.006, N = 72
A = 10 000·0.006/(1 − (1 + 0.006)−72) = €171.45
Total interest
I = N·A − P = 72·171.5 − 10 000 = €2 344.55
Answer
I = €2 344.55
3.NPV & IRR
Answer: NPV = −1198 €
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 the yearly cash flows
6 000·5.747 = 34 479.83
Present value of the salvage value
8 000/1.851 = 4 322.15
NPV
NPV = −40 000 + 34 479.83 + 4 322.15 = −€1 198.02
Answer
NPV = −€1 198.02
4.NPV & IRR
Answer: IRR = 20.04 %
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)−3)/i = 20 000/9 500 = 2.105
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 20%: 2.106
i = 21%: 2.074
IRR = 20.04%
Answer
IRR = 20.04%
5.Bond pricing & yield
Answer: P = 964.9 €
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 = 0.03, N = 8
Present value of the coupons
€25.00·(1 − 0.7894)/0.03 = €175.49
Present value of the face value
€1 000·0.7894 = €789.41
Price
P = €964.90
Answer
P = €964.90
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 5.505 %
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, N = 12
Solve numerically
i = 2.752% per half-year
YTM
YTM = 2·2.752% = 5.505%
Answer
YTM = 5.505%
7.CAPM & WACC
Answer: WACC = 9.129 %
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 = 1
Standard normal table
Φ(1) = 0.8413
P = 0.8413
Answer
P = 0.8413
11.Probability
Answer: P = 0.5177
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)4 = 0.4823
P = 1 − 0.4823 = 0.5177
Answer
P = 0.5177
12.Probability
Answer: E(G) = 3.5 €
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: 15 (p = 0.3), 45 (p = 0.05), −5 (p = 0.65)
Expected value
E(G) = 15·0.3 + 45·0.05 − 5·0.65 = 3.5 €
Answer
E(G) = 3.5 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.4573
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 = (250.25 − 250)/5 = 0.05
z₂ = (260 − 250)/5 = 2
Area between
P = Φ(2) − Φ(0.05) = 0.9772 − 0.5199 = 0.4573
Answer
P = 0.4573
14.Normal distribution
Answer: x = 41.87 min
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) = 0.95 → z = 1.645
Back to x
x = 32 + 1.645·6 = 41.87 min
Answer
x = 41.87 min
15.Confidence intervals
Answer: lower limit = −3.289 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 = 3.25·6.8/√10 = 6.989
Interval
3.7 ± 6.989 → [−3.289, 10.69]
Answer
lower limit = −3.289 mmHg
16.Confidence intervals
Answer: n = 49
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 = (1.645·17/4)² = 48.88
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 49
Answer
n = 49
17.Hypothesis tests & confidence intervals
Answer: t = 2.9
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/√25 = 5/5 = 1
Test statistic
t = (122.9 − 120)/1 = 2.9
Answer
t = 2.9
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: L = 118.2
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).
Margin of error
E = 2.576·8/√64 = 2.576
Interval limit
L = 120.8 − 2.576 = 118.2
Answer
L = 118.2
19.Linear regression & correlation
Answer: r = 0.9443
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 = 6670 − 100·326/5 = 150
Sxx = 2040 − 100²/5 = 40
Syy = 21886 − 326²/5 = 630.8
Correlation coefficient
r = 150/√(40·630.8) = 0.9443
Answer
r = 0.9443
20.Linear regression & correlation
Answer: x = 3.976
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 = 21/6 = 3.5
ȳ = 110/6 = 18.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −49
Regression line
b = −49/17.5 = −2.8
a = 18.33 − (−2.8)·3.5 = 28.13
ŷ = 28.13 − 2.8·x
Solve for x
x = (17 − 28.13)/(−2.8) = 3.976
Answer
x = 3.976
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