You need €25 000 in 7 years. An account pays 3.6% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €25 0003.6% / 12n = 7
PV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 25 years. The interest rate is 3.6% per year, compounded monthly. Find the monthly payment A.
P = €200 0003.6% / 12N = 300
A = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 3: €16 000, €37 000, €35 000. The cost of capital is 5%. Find the net present value NPV.
A project costs €50 000 and returns €26 000, €35 000, €29 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 26 000CF2 = 35 000CF3 = 29 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 3.5%n = 10
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 8 years. It sells for €929. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 2% (2×)n = 8P = €929
YTM = %
7.CAPM & WACCMedium
A company has equity worth €6 000 000 and debt worth €4 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the WACC.
A company has equity with a market value of €15 000 000 and a cost of equity of 9%. Its debt consists of 5 000 bonds (face value €1 000, annual coupon 6%, 10 years to maturity) that trade at €1 119 each. The tax rate is 20%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.
Find the sample standard deviation s of the data: 22, 16, 25, 53, 35, 45.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 5. Find the z-score z of a score of 180.
μ = 170σ = 5x = 180
z =
11.ProbabilityMedium
A bag holds 8 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 5
P =
12.ProbabilityHard
5 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 5p = 0.6
P =
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 12 of them exceeds 1041 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 12> 1041 kg
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm60%
x = cm
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 40 phones, mean 88 h, known standard deviation 9 h. Model 2: 30 phones, mean 88.5 h, known standard deviation 15 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
The population standard deviation is 22. 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 3?
σ = 22E = 399%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 101.3 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 101.3s = 10n = 81
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 81 measurements has mean 199.6. The population standard deviation is 8. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 199.6σ = 8n = 8199%
U =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 57?
n = 6ŷ = 57
x =
3Solutions
1.Time value of money
Answer: PV = 19438 €
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 = 0.036000000000000004/12 = 0.003
N = 12·7 = 84
Present value
PV = FV/(1 + i)N = 25 000/1.286 = €19 438.45
Answer
PV = €19 438.45
2.Time value of money
Answer: A = 1012 €
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 rate and number of payments
i = 0.036000000000000004/12 = 0.003
N = 12·25 = 300
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.4071
A = 200 000·0.003/(1 − 0.4071) = €1 012.01
Answer
A = €1 012.01
3.NPV & IRR
Answer: NPV = 4033 €
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 + 0.05)1 = 16 000/1.05 = 15 238.10
CF2/(1 + 0.05)2 = 37 000/1.103 = 33 560.09
CF3/(1 + 0.05)3 = 35 000/1.158 = 30 234.32
Add them up
NPV = €4 032.50
Answer
NPV = €4 032.50
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 35.66 %
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(35%) = 453.69
NPV(40%) = −2 820.70
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 = €30.00, i = 0.0175, N = 20
Present value of the coupons
€30.00·(1 − 0.7068)/0.0175 = €502.59
Present value of the face value
€1 000·0.7068 = €706.82
Price
P = €1 209.41
Answer
P = €1 209.41
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 3.005 %
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 = €10.00, N = 16
Solve numerically
i = 1.503% per half-year
YTM
YTM = 2·1.503% = 3.005%
Answer
YTM = 3.005%
7.CAPM & WACC
Answer: WACC = 7.2 %
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 = €10 000 000, E/V = 0.6, D/V = 0.4
WACC
WACC = 0.6·10% + 0.4·4%·(1 − 0.25) = 7.2%
Answer
WACC = 7.2%
8.CAPM & WACC
Answer: WACC = 7.532 %
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 = (180 − 170)/5 = 2
Answer
z = 2
11.Probability
Answer: P = 0.359
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 = 8/13 · 7/12 = 56/156 = 0.359
Answer
P = 0.359
12.Probability
Answer: P = 0.08704
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) = 0.45 = 0.01024
P(X = 1) = 5·0.6·0.44 = 0.0768
Add them
P = 0.01024 + 0.0768 = 0.08704
Answer
P = 0.08704
13.Normal distribution
Answer: P = 0.3085
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 = 1020 kg, σT = 12·√12 = 41.57 kg
Standardise
z = (1041 − 1020)/41.57 = 0.5
Probability
P = 1 − Φ(0.5) = 0.3085
Answer
P = 0.3085
14.Normal distribution
Answer: x = 176.3 cm
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) = 0.4
z from the table
z ≈ −0.25
Back to the original units
x = 178 − 0.25·7 = 176.3 cm
Answer
x = 176.3 cm
15.Confidence intervals
Answer: lower limit = −6.549 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 = 88 − 88.5 = −0.5
Standard error
SE = √(9²/40 + 15²/30) = 3.086
Margin of error
E = 1.96·3.086 = 6.049
Interval
−0.5 ± 6.049 → [−6.549, 5.549]
Answer
lower limit = −6.549 h
The interval contains 0, so the data do not show a significant difference.
16.Confidence intervals
Answer: n = 357
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 = (2.576·22/3)² = 356.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 357
Answer
n = 357
17.Hypothesis tests & confidence intervals
Answer: t = 1.17
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 = 10/√81 = 10/9 = 1.111
Test statistic
t = (101.3 − 100)/1.111 = 1.17
Answer
t = 1.17
t0.025; df = 80 = 1.99
|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 201.9
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).
Margin of error
E = 2.576·8/√81 = 2.29
Interval limit
U = 199.6 + 2.29 = 201.9
Answer
U = 201.9
19.Linear regression & correlation
Answer: r = 0.9957
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 = 30/5 = 6
ȳ = 135/5 = 27
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 154
Syy = Σ(y − ȳ)² = 598
Correlation coefficient
r = 154/√(40·598) = 0.9957
Answer
r = 0.9957
20.Linear regression & correlation
Answer: x = 19.02
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 = 126/6 = 21
ȳ = 394/6 = 65.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 306
Regression line
b = 306/70 = 4.371
a = 65.67 − 4.371·21 = −26.13
ŷ = −26.13 + 4.371·x
Solve for x
x = (57 − (−26.13))/4.371 = 19.02
Answer
x = 19.02
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