You need €5 000 in 8 years. An account pays 4% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €5 0004% / 12n = 8
PV = €
2.Time value of moneyHard
You save €200 at the end of every month for 5 years. The account pays 3.6% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €2003.6% / 12N = 60
FV = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 4: €7 000, €6 000, €5 000, €8 000. The cost of capital is 10%. Find the net present value NPV.
An investment of €40 000 returns €10 500 at the end of each year for 8 years. Find its internal rate of return IRR in percent.
C0 = €40 000CF = €10 500n = 8
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 11 years and pays annual coupons. It sells for €1 166 when the yield to maturity is 5%. Find the coupon rate c in percent.
F = €1 000n = 11P = €1 166YTM = 5%
c = %
6.Bond pricing & yieldHard
A bond with 13 years to maturity (face value €1 000, annual coupon 6%) sells for €1 258. The issuer can call it in 6 years at a call price of €1 060. Find the yield to call YTC in percent.
F = €1 000coupon = 6%P = €1 258call price = €1 060call in 6 years
YTC = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €12 000 000, preferred shares worth €1 000 000 and debt worth €4 000 000. The cost of equity is 14%, the cost of preferred stock 7% and the pre-tax cost of debt 6%. The tax rate is 20%. Find the WACC.
A company has 2 000 000 shares at €51.00 each and 10 000 bonds trading at €1 000 each. The pre-tax cost of debt is 6% and the tax rate is 20%. The share’s beta is 0.8, the risk-free rate 2.5% and the market return 6.5%. Find the WACC.
β = 0.8rf = 2.5%rm = 6.5%rd = 6%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 45, 58, 40, 27, 25.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 10. Find the z-score z of a score of 174.
μ = 170σ = 10x = 174
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.35, 0.25, 0.3, 0.1
Var(X) =
12.ProbabilityHard
A game costs €1 to play. You win €3 with probability 0.25 and €20 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€3: p = 0.25€20: p = 0.02
E(G) = €
13.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 10 of them is below 869 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 10< 869 kg
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
A sample of 30 items has mean 95 and standard deviation s = 8. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 30x̄ = 95s = 899%t(29) = 2.756
upper limit =
16.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.02?
p = 0.5E = 0.0290%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 78.64 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 78.64s = 15n = 49
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 88 of 250 patients on treatment A recover, and 50 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 88n₁ = 250x₂ = 50n₂ = 100
z =
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 = 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: PV = 3633 €
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 = 0.04/12 = 0.003333
N = 12·8 = 96
Present value
PV = FV/(1 + i)N = 5 000/1.376 = €3 632.68
Answer
PV = €3 632.68
2.Time value of money
Answer: FV = 13126 €
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.
You paid in €12 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 543 €
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 + 0.1)1 = 7 000/1.1 = 6 363.64
CF2/(1 + 0.1)2 = 6 000/1.21 = 4 958.68
CF3/(1 + 0.1)3 = 5 000/1.331 = 3 756.57
CF4/(1 + 0.1)4 = 8 000/1.464 = 5 464.11
Add them up
NPV = €543.00
Answer
NPV = €543.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 20.24 %
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)−8)/i = 40 000/10 500 = 3.81
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 20%: 3.837
i = 21%: 3.726
IRR = 20.24%
Answer
IRR = 20.24%
5.Bond pricing & yield
Answer: c = 6.998 %
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.
Present value of the face value
1000·(1 + 0.05)−11 = €584.68
Coupon
C = (1166 − 584.68)/8.306 = €69.98
Coupon rate
c = 69.98/1000 = 6.998%
Answer
c = 6.998%
6.Bond pricing & yield
Answer: YTC = 2.293 %
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
1258 = 60·(1 − (1 + y)−6)/y + 1060/(1 + y)6
Solve numerically
y = 2%: €1 277.34
y = 3%: €1 212.76
YTC = 2.293%
Answer
YTC = 2.293%
The yield to maturity is 3.496%. 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: WACC = 11.42 %
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).
Market values
E = 2 000 000·€51.00 = €102 000 000
D = 10 000·€1 000 = €10 000 000
Cost of equity (CAPM)
re = 2.5% + 0.8·4% = 5.7%
Weights
V = €112 000 000, E/V = 0.9107, D/V = 0.08929
WACC
WACC = 0.9107·5.7% + 0.08929·6%·(1 − 0.2) = 5.62%
Answer
WACC = 5.62%
9.Descriptive statistics
Answer: s = 13.58
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 = (174 − 170)/10 = 0.4
Answer
z = 0.4
11.Probability
Answer: Var(X) = 1.028
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.
Mean
μ = 1·0.35 + 2·0.25 + 3·0.3 + 4·0.1 = 2.15
Mean of the squares
E(X²) = 1·0.35 + 4·0.25 + 9·0.3 + 16·0.1 = 5.65
Variance
Var(X) = 5.65 − 2.15² = 1.028
Answer
Var(X) = 1.028
12.Probability
Answer: E(G) = 0.15 €
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: 2 (p = 0.25), 19 (p = 0.02), −1 (p = 0.73)
Expected value
E(G) = 2·0.25 + 19·0.02 − 1·0.73 = 0.15 €
Answer
E(G) = 0.15 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.6915
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the total
μT = 850 kg, σT = 12·√10 = 37.95 kg
Standardise
z = (869 − 850)/37.95 = 0.5
Probability
P = Φ(0.5) = 0.6915
Answer
P = 0.6915
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 = 99.03
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 = 2.756·8/√30 = 4.025
Interval
95 ± 4.025 → [90.97, 99.03]
Answer
upper limit = 99.03
16.Confidence intervals
Answer: n = 1692
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 = 1.645²·0.5·0.5/0.02² = 1691
Round up
n = 1692
Answer
n = 1692
17.Hypothesis tests & confidence intervals
Answer: t = 1.699
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/√49 = 15/7 = 2.143
Test statistic
t = (78.64 − 75)/2.143 = 1.699
Answer
t = 1.699
t0.025; df = 48 = 2.011
|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −2.56
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.
Sample proportions
p̂₁ = 88/250 = 0.352
p̂₂ = 50/100 = 0.5
p̂ = 138/350 = 0.3943
Standard error under H₀
SE = √(0.3943·0.6057·(1/250 + 1/100)) = 0.05782
Test statistic
z = (0.352 − 0.5)/0.05782 = −2.56
Answer
z = −2.56
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r² = 0.9929
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 = 48/6 = 8
ȳ = 420/6 = 70
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 458
Syy = Σ(y − ȳ)² = 3018
Coefficient of determination
r² = 458²/(70·3018) = 0.9929
Answer
r² = 0.9929
About 99.29% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: x = 3.917
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 = 106/6 = 17.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −28
Regression line
b = −28/17.5 = −1.6
a = 17.67 − (−1.6)·3.5 = 23.27
ŷ = 23.27 − 1.6·x
Solve for x
x = (17 − 23.27)/(−1.6) = 3.917
Answer
x = 3.917
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