You need €10 000 in 6 years. An account pays 5% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €10 0005% / 4n = 6
PV = €
2.Time value of moneyHard
You deposit €2 000 at the start of every year for 18 years, the first deposit today. The account pays 6% per year. How much is in the account at the end of year 18? Find FV.
A = €2 000i = 6%n = 18
FV = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 3: €37 000, €29 000, €32 000. The cost of capital is 8%. Find the net present value NPV.
A project costs €25 000 and returns €7 500, €13 500, €14 000 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 12%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 4 years. The yield to maturity is 3% per year. Find the price P of the bond.
F = €1 000coupon = 4.5%YTM = 3%n = 4
P = €
6.Bond pricing & yieldHard
A bond trades at €920 with a yield to maturity of 4% (annual compounding) and a Macaulay duration of 7.2 years. Use the modified duration to estimate the new price P if the yield rises by 25 basis points.
P0 = €920y = 4%D = 7.2Δy = +25 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €1 000 000. Its WACC is 8.09%, its pre-tax cost of debt is 3% and the tax rate is 21%. Find the cost of equity re.
A company has 1 000 000 shares trading at €40.50; the book value of its equity is €12 200 000. It just paid a dividend of €1.40 per share, and dividends grow by 5% per year. Its debt has a market value of €40 500 000 and a pre-tax cost of 3%. 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: 36, 44, 12, 26, 30, 37, 60.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 170 and standard deviation 8. Find the z-score z of a score of 173.2.
μ = 170σ = 8x = 173.2
z =
11.ProbabilityMedium
A bag holds 6 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
6 / 8
P =
12.ProbabilityHard
A game costs €10 to play. You win €50 with probability 0.15 and €200 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €10€50: p = 0.15€200: p = 0.02
E(G) = €
13.Normal distributionMedium
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 16 of them exceeds 632.8 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 16> 632.8 g
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 5% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g5%
x = g
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 50 phones, mean 41 h, known standard deviation 3 h. Model 2: 50 phones, mean 47 h, known standard deviation 18 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
A pilot study suggests that about 0.3 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 0.02?
p ≈ 0.3E = 0.0290%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 48.5 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 48.5s = 10n = 36
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 49 measurements has mean 51.3. The population standard deviation is 15. Find the upper limit U of the 90% confidence interval for the mean.
x̄ = 51.3σ = 15n = 4990%
U =
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 5 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 (1.5, 34) is made. Find its residual e = y − ŷ.
n = 5(1.5, 34)
e =
3Solutions
1.Time value of money
Answer: PV = 7422 €
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 = 0.05/4 = 0.0125
N = 4·6 = 24
Present value
PV = FV/(1 + i)N = 10 000/1.347 = €7 421.97
Answer
PV = €7 421.97
2.Time value of money
Answer: FV = 65520 €
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.
Ordinary annuity factor
((1 + 0.06)18 − 1)/0.06 = 30.91
Annuity due
Each deposit is invested one year longer.
FV = 2 000·30.91·1.06 = €65 519.98
Answer
FV = €65 519.98
3.NPV & IRR
Answer: NPV = 9525 €
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 = 37 000/1.08 = 34 259.26
CF2/(1 + 0.08)2 = 29 000/1.166 = 24 862.83
CF3/(1 + 0.08)3 = 32 000/1.26 = 25 402.63
Add them up
NPV = €9 524.72
Answer
NPV = €9 524.72
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: MIRR = 15.51 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
7 500·1.122 = 9 408.00
13 500·1.121 = 15 120.00
14 000·1.120 = 14 000.00
TV = 38 528.00
MIRR
MIRR = (38 528.00/25 000)1/3 − 1 = 15.51%
Answer
MIRR = 15.51%
5.Bond pricing & yield
Answer: P = 1056 €
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 = 0.015, N = 8
Present value of the coupons
€22.50·(1 − 0.8877)/0.015 = €168.43
Present value of the face value
€1 000·0.8877 = €887.71
Price
P = €1 056.14
Answer
P = €1 056.14
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: P = 904.1 €
Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.
Modified duration
Dmod = 7.2/1.04 = 6.923
Estimated price change
%ΔP ≈ −6.923·(0.0025) = −1.731%
P ≈ 920·(1 − 0.01731) = €904.08
Answer
P = €904.08
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: re = 9.997 %
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 = €4 000 000, E/V = 0.75, D/V = 0.25
Debt part of the WACC
0.25·3%·(1 − 0.21) = 0.5925%
Cost of equity
re = (8.09% − 0.5925%)/0.75 = 9.997%
Answer
re = 9.997%
8.CAPM & WACC
Answer: WACC = 5.44 %
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 equity
D1 = 1.4·1.05 = €1.47
re = 1.47/40.5 + 0.05 = 8.63%
Market values
E = 1 000 000·€40.50 = €40 500 000
V = €81 000 000, E/V = 0.5, D/V = 0.5
WACC
WACC = 0.5·8.63% + 0.5·3%·(1 − 0.25) = 5.44%
Answer
WACC = 5.44%
9.Descriptive statistics
Answer: s = 14.98
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 = (173.2 − 170)/8 = 0.4
Answer
z = 0.4
11.Probability
Answer: P = 0.1648
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 = 6/14 · 5/13 = 30/182 = 0.1648
Answer
P = 0.1648
12.Probability
Answer: E(G) = 1.5 €
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: 40 (p = 0.15), 190 (p = 0.02), −10 (p = 0.83)
Expected value
E(G) = 40·0.15 + 190·0.02 − 10·0.83 = 1.5 €
Answer
E(G) = 1.5 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.9332
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 = 640 g, σT = 1.2·√16 = 4.8 g
Standardise
z = (632.8 − 640)/4.8 = −1.5
Probability
P = 1 − Φ(−1.5) = 0.9332
Answer
P = 0.9332
14.Normal distribution
Answer: x = 241.8 g
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) = 0.05
z from the table
z ≈ −1.64
Back to the original units
x = 250 − 1.64·5 = 241.8 g
Answer
x = 241.8 g
15.Confidence intervals
Answer: E = 4.245 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 = 41 − 47 = −6
Standard error
SE = √(3²/50 + 18²/50) = 2.581
Margin of error
E = 1.645·2.581 = 4.245
Answer
E = 4.245 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 1421
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 = 1.645²·0.3·0.7/0.02² = 1421
Round up
n = 1421
Answer
n = 1421
17.Hypothesis tests & confidence intervals
Answer: t = −0.9
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 = 10/√36 = 10/6 = 1.667
Test statistic
t = (48.5 − 50)/1.667 = −0.9
Answer
t = −0.9
t0.025; df = 35 = 2.03
|t| < 2.03, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 54.82
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 = 1.645·15/√49 = 3.525
Interval limit
U = 51.3 + 3.525 = 54.82
Answer
U = 54.82
19.Linear regression & correlation
Answer: r = 0.9845
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 = 8750 − 110·389/5 = 192
Sxx = 2460 − 110²/5 = 40
Syy = 31215 − 389²/5 = 950.8
Correlation coefficient
r = 192/√(40·950.8) = 0.9845
Answer
r = 0.9845
20.Linear regression & correlation
Answer: e = 5.4
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 = 15/5 = 3
ȳ = 131/5 = 26.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −16
Regression line
b = −16/10 = −1.6
a = 26.2 − (−1.6)·3 = 31
ŷ = 31 − 1.6·x
Predicted value
ŷ = 31 + (−1.6)·1.5 = 28.6
Residual
e = 34 − 28.6 = 5.4
Answer
e = 5.4
The point lies above the line.
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