A credit card charges a nominal interest rate of 17.5% per year, compounded quarterly. Find the effective annual rate EAR in percent.
r = 17.5%m = 4
EAR = %
2.Time value of moneyHard
A mortgage of €250 000 is repaid with equal monthly payments over 30 years at 6% per year, compounded monthly. How much is still owed just after the payment at the end of year 5? Find the outstanding balance B.
P = €250 0006% / 12N = 360k = 60
B = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 4: €28 000, €21 000, €22 000, €37 000. The cost of capital is 8%. Find the net present value NPV.
A project costs €20 000 and returns €8 000, €14 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 8 000CF2 = 14 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 8 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 3.5%n = 8
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 5 years and sells for €1 090. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 5P = €1 090
YTM = %
7.CAPM & WACCMedium
A company has equity worth €12 000 000 and debt worth €2 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 25%. Find the WACC.
A company has 1 000 000 shares at €36.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 6% and the tax rate is 28%. The share’s beta is 1.1, the risk-free rate 1.5% and the market return 5.5%. Find the WACC.
β = 1.1rf = 1.5%rm = 5.5%rd = 6%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 54, 36, 49, 51, 60.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 15. Find the z-score z of a score of 152.
μ = 170σ = 15x = 152
z =
11.ProbabilityMedium
A fair die is rolled 8 times. What is the probability P of getting at least one six?
n = 8
P =
12.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 2 correct answers?
n = 7p = 0.2k = 2
P =
13.Normal distributionMedium
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 33.3 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 33.3 min
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h75%
x = h
15.Confidence intervalsMedium
A sample of 6 items has mean 110.5 and standard deviation s = 11. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 6x̄ = 110.5s = 1190%t(5) = 2.015
lower limit =
16.Confidence intervalsHard
The diameters of 21 machined shafts have a sample standard deviation of 7.4 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 20, 2.5% of the χ² distribution lies below 9.591 and 2.5% lies above 34.17.
n = 21s = 7.4 µmχ² = 9.591, 34.17
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 82.2 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 82.2s = 15n = 25
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 28 of 80 patients on treatment A recover, and 100 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 28n₁ = 80x₂ = 100n₂ = 200
z =
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. Predict ŷ for x = 4.5.
n = 6x = 4.5
ŷ =
3Solutions
1.Time value of money
Answer: EAR = 18.68 %
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 per period
r/m = 0.175/4 = 0.04375
Compound over one year
EAR = (1 + 0.04375)4 − 1 = 1.187 − 1 = 18.68%
Answer
EAR = 18.68%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: B = 232636 €
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.
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 360 − 60 = 300
B = 1499·(1 − (1 + 0.005)−300)/0.005 = €232 635.89
Answer
B = €232 635.89
3.NPV & IRR
Answer: NPV = 13590 €
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 + 0.08)1 = 28 000/1.08 = 25 925.93
CF2/(1 + 0.08)2 = 21 000/1.166 = 18 004.12
CF3/(1 + 0.08)3 = 22 000/1.26 = 17 464.31
CF4/(1 + 0.08)4 = 37 000/1.36 = 27 196.10
Add them up
NPV = €13 590.46
Answer
NPV = €13 590.46
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 7.464 %
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(5%) = 770.98
NPV(10%) = −743.80
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 5 + 5·770.98/(770.98 − (−743.80)) = 7.545%
exact: NPV(7.464%) = 0 → IRR = 7.464%
Answer
IRR = 7.464%
5.Bond pricing & yield
Answer: P = 1242 €
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 = €35.00, i = 0.0175, N = 16
Present value of the coupons
€35.00·(1 − 0.7576)/0.0175 = €484.77
Present value of the face value
€1 000·0.7576 = €757.62
Price
P = €1 242.38
Answer
P = €1 242.38
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 3.033 %
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
€1 090 = €50·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(3%) = €1 091.59, P(4%) = €1 044.52
YTM = 3.033%
Answer
YTM = 3.033%
7.CAPM & WACC
Answer: WACC = 10.61 %
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 = €14 000 000, E/V = 0.8571, D/V = 0.1429
WACC
WACC = 0.8571·12% + 0.1429·3%·(1 − 0.25) = 10.61%
Answer
WACC = 10.61%
8.CAPM & WACC
Answer: WACC = 5.085 %
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.
Mean
x̄ = 250/5 = 50
Squared deviations
Σ(x − x̄)² = 4² + (−14)² + (−1)² + 1² + 10² = 314
Standard deviation
s = √(314/4) = 8.86
Answer
s = 8.86
10.Descriptive statistics
Answer: z = −1.2
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 = (152 − 170)/15 = −1.2
Answer
z = −1.2
11.Probability
Answer: P = 0.7674
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)8 = 0.2326
P = 1 − 0.2326 = 0.7674
Answer
P = 0.7674
12.Probability
Answer: P = 0.2753
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) = 21·0.22·0.85 = 21·0.04·0.3277 = 0.2753
Answer
P = 0.2753
13.Normal distribution
Answer: P = 0.3015
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 sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (33.3 − 32)/2.5 = 0.52
Upper tail
P = 1 − Φ(0.52) = 1 − 0.6985 = 0.3015
Answer
P = 0.3015
14.Normal distribution
Answer: x = 527 h
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
Φ(z) = 0.75
z from the table
z ≈ 0.67
Back to the original units
x = 500 + 0.67·40 = 527 h
Answer
x = 527 h
15.Confidence intervals
Answer: lower limit = 101.5
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 = 5, t = 2.015
Margin of error
E = 2.015·11/√6 = 9.049
Interval
110.5 ± 9.049 → [101.5, 119.5]
Answer
lower limit = 101.5
16.Confidence intervals
Answer: σ = 10.69 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 = 20·7.4² = 1095
Limits for σ²
1095/34.17 = 32.05
1095/9.591 = 114.2
Limit for σ
σ = √114.2 = 10.69 µm
Answer
σ = 10.69 µm
95% interval for σ: [5.661, 10.69] µm.
17.Hypothesis tests & confidence intervals
Answer: t = 2.4
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/√25 = 15/5 = 3
Test statistic
t = (82.2 − 75)/3 = 2.4
Answer
t = 2.4
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: z = −2.276
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 = 28/80 = 0.35
p̂₂ = 100/200 = 0.5
p̂ = 128/280 = 0.4571
Standard error under H₀
SE = √(0.4571·0.5429·(1/80 + 1/200)) = 0.0659
Test statistic
z = (0.35 − 0.5)/0.0659 = −2.276
Answer
z = −2.276
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9707
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 = 7181 − 110·314/5 = 273
Sxx = 2510 − 110²/5 = 90
Syy = 20598 − 314²/5 = 878.8
Correlation coefficient
r = 273/√(90·878.8) = 0.9707
Answer
r = 0.9707
20.Linear regression & correlation
Answer: ŷ = 12.72
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 = 95/6 = 15.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −54.5
Regression line
b = −54.5/17.5 = −3.114
a = 15.83 − (−3.114)·3.5 = 26.73
ŷ = 26.73 − 3.114·x
Prediction
ŷ = 26.73 + (−3.114)·4.5 = 12.72
Answer
ŷ = 12.72
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