You invest €1 000 at a nominal rate of 3.6% per year, compounded quarterly. Find the value FV of the investment after 4 years.
PV = €1 0003.6% / 4n = 4
FV = €
2.Time value of moneyHard
You borrow €10 000 at 3.6% per year, compounded monthly, and repay €200 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 = €2003.6% / 12
N = payments
3.NPV & IRRMedium
An investment of €30 000 gives these cash flows at the end of years 1 to 3: €12 000, €8 000, €13 000. The cost of capital is 15%. Find the net present value NPV.
A project costs €40 000 and returns €17 000, €24 500, €24 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −40 000CF1 = 17 000CF2 = 24 500CF3 = 24 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 4% per year. Find the price P of the bond.
F = €1 000coupon = 4.5%YTM = 4%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 8 years and sells for €1 110. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 8P = €1 110
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €6 000 000, preferred shares worth €1 500 000 and debt worth €4 000 000. The cost of equity is 10%, the cost of preferred stock 6% and the pre-tax cost of debt 3%. The tax rate is 28%. Find the WACC.
A company has 2 000 000 shares at €29.50 each and 5 000 bonds trading at €1 020 each. The pre-tax cost of debt is 7% and the tax rate is 30%. The share’s beta is 1.3, the risk-free rate 1% and the market return 7%. Find the WACC.
β = 1.3rf = 1%rm = 7%rd = 7%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 21, 12, 51, 27, 56, 50, 34.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 8. What proportion P of values is below 63.2?
μ = 60σ = 8x = 63.2
P =
11.ProbabilityMedium
A fair die is rolled 6 times. What is the probability P of getting at least one six?
n = 6
P =
12.ProbabilityHard
8 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 8p = 0.6
P =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.3688. Find z to two decimals.
A = 0.3688
z =
14.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 80 phones, mean 76.5 h, known standard deviation 4 h. Model 2: 50 phones, mean 68.5 h, known standard deviation 15 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
A pilot study suggests that about 0.15 of products are defective. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.04?
p ≈ 0.15E = 0.0490%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 204.8 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 204.8s = 15n = 81
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 1.599%
n =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. 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. A new observation (5.5, 10) is made. Find its residual e = y − ŷ.
n = 6(5.5, 10)
e =
3Solutions
1.Time value of money
Answer: FV = 1154 €
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 = 0.003
(1 + i)−N = 1 − 10 000·0.003/200 = 0.85
Solve with logarithms
N = −ln(0.85)/ln(1.003) = 54.25
Round up to whole payments
54 full payments leave a small balance, which the last, smaller payment clears.
N = 55
Answer
N = 55 payments
3.NPV & IRR
Answer: NPV = −4968 €
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 = −30 000
CF1/(1 + 0.15)1 = 12 000/1.15 = 10 434.78
CF2/(1 + 0.15)2 = 8 000/1.322 = 6 049.15
CF3/(1 + 0.15)3 = 13 000/1.521 = 8 547.71
Add them up
NPV = €−4 968.36
Answer
NPV = €−4 968.36
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 27.87 %
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.
Bracket the IRR
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
NPV(25%) = 1 824.00
NPV(30%) = −1 274.47
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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 = €22.50, i = 0.02, N = 24
Present value of the coupons
€22.50·(1 − 0.6217)/0.02 = €425.56
Present value of the face value
€1 000·0.6217 = €621.72
Price
P = €1 047.28
Answer
P = €1 047.28
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 2.468 %
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
€1 110 = €40·(1 − (1 + y)⁻⁸)/y + €1 000/(1 + y)⁸
Solve numerically
P(2%) = €1 146.51, P(3%) = €1 070.20
YTM = 2.468%
Answer
YTM = 2.468%
7.CAPM & WACC
Answer: WACC = 6.751 %
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·€29.50 = €59 000 000
D = 5 000·€1 020 = €5 100 000
Cost of equity (CAPM)
re = 1% + 1.3·6% = 8.8%
Weights
V = €64 100 000, E/V = 0.9204, D/V = 0.07956
WACC
WACC = 0.9204·8.8% + 0.07956·7%·(1 − 0.3) = 8.49%
Answer
WACC = 8.49%
9.Descriptive statistics
Answer: s = 16.87
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 0.6554
Answer
P = 0.6554
11.Probability
Answer: P = 0.6651
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)6 = 0.3349
P = 1 − 0.3349 = 0.6651
Answer
P = 0.6651
12.Probability
Answer: P = 0.00852
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) = 0.48 = 0.0006554
P(X = 1) = 8·0.6·0.47 = 0.007864
Add them
P = 0.0006554 + 0.007864 = 0.00852
Answer
P = 0.00852
13.Normal distribution
Answer: z = 0.48
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 + 0.3688)/2 = 0.6844
Read the z-table backwards
Find 0.6844 inside the table and read off z.
z = 0.48
Answer
z = 0.48
14.Normal distribution
Answer: P = 0.7389
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 difference
μ = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
15.Confidence intervals
Answer: lower limit = 4.434 h
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.
Point estimate
x̄₁ − x̄₂ = 76.5 − 68.5 = 8
Standard error
SE = √(4²/80 + 15²/50) = 2.168
Margin of error
E = 1.645·2.168 = 3.566
Interval
8 ± 3.566 → [4.434, 11.57]
Answer
lower limit = 4.434 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 216
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 = 1.645²·0.15·0.85/0.04² = 215.6
Round up
n = 216
Answer
n = 216
17.Hypothesis tests & confidence intervals
Answer: t = 2.88
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/√81 = 15/9 = 1.667
Test statistic
t = (204.8 − 200)/1.667 = 2.88
Answer
t = 2.88
t0.025; df = 80 = 1.99
|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 295
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 = (2.576·10/1.5)² = 294.9
Round up
n = 295
Answer
n = 295
19.Linear regression & correlation
Answer: r² = 0.9876
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 = 141/6 = 23.5
ȳ = 715/6 = 119.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 844.5
Syy = Σ(y − ȳ)² = 4585
Coefficient of determination
r² = 844.5²/(157.5·4585) = 0.9876
Answer
r² = 0.9876
About 98.76% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = −2.724
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 = 112/6 = 18.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −52
Regression line
b = −52/17.5 = −2.971
a = 18.67 − (−2.971)·3.5 = 29.07
ŷ = 29.07 − 2.971·x
Predicted value
ŷ = 29.07 + (−2.971)·5.5 = 12.72
Residual
e = 10 − 12.72 = −2.724
Answer
e = −2.724
The point lies below the line.
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