An annuity pays €500 at the end of each year for 15 years. The interest rate is 6% per year. Find the present value PV of the annuity.
A = €500i = 6%n = 15
PV = €
2.Time value of moneyHard
You save €100 at the end of every month for 30 years. The account pays 4.2% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €1004.2% / 12N = 360
FV = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 3: €7 000, €6 000, €9 000. The cost of capital is 8%. Find the net present value NPV.
A project costs €20 000 and returns €8 500, €12 500, €12 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 8 500CF2 = 12 500CF3 = 12 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 10 years. Its yield to maturity changes from 3.5% to 4%. By how much ΔP does the bond price change?
F = €1 000coupon = 5%n = 10y1 = 3.5%y2 = 4%
ΔP = €
6.Bond pricing & yieldHard
A bond with 16 years to maturity (face value €1 000, annual coupon 7%) sells for €1 281. The issuer can call it in 3 years at a call price of €1 070. Find the yield to call YTC in percent.
F = €1 000coupon = 7%P = €1 281call price = €1 070call in 3 years
YTC = %
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €2 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 5% and the tax rate is 21%. Find the WACC.
A company has 2 000 000 shares at €47.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 4.5% and the tax rate is 25%. The share’s beta is 0.6, the risk-free rate 1% and the market return 6.5%. Find the WACC.
β = 0.6rf = 1%rm = 6.5%rd = 4.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 60, 21, 24, 49, 17.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 20. What proportion P of values is above 130?
μ = 170σ = 20x = 130
P =
11.ProbabilityMedium
A bag holds 7 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 7
P =
12.ProbabilityHard
A game costs €3 to play. You win €9 with probability 0.3 and €60 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €3€9: p = 0.3€60: p = 0.01
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.9031. Find z to two decimals.
A = 0.9031
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
A 99% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 2.2. What is the margin of error E₂ if the same study uses 16 observations instead?
99%n₁ = 64E₁ = 2.2n₂ = 16
E₂ =
16.Confidence intervalsHard
A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 99% confidence interval for the proportion has a margin of error of at most 0.01?
p ≈ 0.8E = 0.0199%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 51.2 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 51.2s = 5n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 80%. The standard deviation is 12. Find the minimum sample size n.
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 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 10.
n = 5x = 10
ŷ =
3Solutions
1.Time value of money
Answer: PV = 4856 €
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.
Annuity factor
(1 − (1 + i)−n)/i = (1 − 0.4173)/0.06 = 9.712
Present value
PV = 500·9.712 = €4 856.12
Answer
PV = €4 856.12
2.Time value of money
Answer: FV = 71934 €
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 €36 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = −1230 €
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 + 0.08)1 = 7 000/1.08 = 6 481.48
CF2/(1 + 0.08)2 = 6 000/1.166 = 5 144.03
CF3/(1 + 0.08)3 = 9 000/1.26 = 7 144.49
Add them up
NPV = €−1 230.00
Answer
NPV = €−1 230.00
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 27.97 %
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%) = 944.00
NPV(30%) = −603.10
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 25 + 5·944.00/(944.00 − (−603.10)) = 28.05%
exact: NPV(27.97%) = 0 → IRR = 27.97%
Answer
IRR = 27.97%
5.Bond pricing & yield
Answer: ΔP = −43.64 €
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.
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
1281 = 70·(1 − (1 + y)−3)/y + 1070/(1 + y)3
Solve numerically
y = 50%: €415.56
y = 50%: €415.56
YTC = 50%
Answer
YTC = 50%
The yield to maturity is 4.499%. 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 = 8.78 %
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 = €5 000 000, E/V = 0.6, D/V = 0.4
WACC
WACC = 0.6·12% + 0.4·5%·(1 − 0.21) = 8.78%
Answer
WACC = 8.78%
8.CAPM & WACC
Answer: WACC = 4.032 %
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·€47.00 = €94 000 000
D = 40 000·€960 = €38 400 000
Cost of equity (CAPM)
re = 1% + 0.6·5.5% = 4.3%
Weights
V = €132 400 000, E/V = 0.71, D/V = 0.29
WACC
WACC = 0.71·4.3% + 0.29·4.5%·(1 − 0.25) = 4.032%
Answer
WACC = 4.032%
9.Descriptive statistics
Answer: s = 19.1
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 1 − 0.02275 = 0.9772
Answer
P = 0.9772
11.Probability
Answer: P = 0.2308
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 = 7/14 · 6/13 = 42/182 = 0.2308
Answer
P = 0.2308
12.Probability
Answer: E(G) = 0.3 €
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: 6 (p = 0.3), 57 (p = 0.01), −3 (p = 0.69)
Expected value
E(G) = 6·0.3 + 57·0.01 − 3·0.69 = 0.3 €
Answer
E(G) = 0.3 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = 1.66
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 + 0.9031)/2 = 0.9515
Read the z-table backwards
Find 0.9515 inside the table and read off z.
z = 1.66
Answer
z = 1.66
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: E₂ = 4.4
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.
Ratio of margins
E₂/E₁ = (2.576/2.576)·√(64/16) = 2
New margin
E₂ = 2.2·2 = 4.4
Answer
E₂ = 4.4
16.Confidence intervals
Answer: n = 10618
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 = 2.576²·0.8·0.2/0.01² = 10617
Round up
n = 10618
Answer
n = 10618
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 = 5/√100 = 5/10 = 0.5
Test statistic
t = (51.2 − 50)/0.5 = 2.4
Answer
t = 2.4
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 223
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((1.645 + 0.842)·12/2)² = 222.7
Round up
n = 223
Answer
n = 223
19.Linear regression & correlation
Answer: r² = 0.9675
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 = 129/6 = 21.5
ȳ = 336/6 = 56
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 609
Syy = Σ(y − ȳ)² = 2434
Coefficient of determination
r² = 609²/(157.5·2434) = 0.9675
Answer
r² = 0.9675
About 96.75% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 94.7
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 = 35/5 = 7
ȳ = 394/5 = 78.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 212
Regression line
b = 212/40 = 5.3
a = 78.8 − 5.3·7 = 41.7
ŷ = 41.7 + 5.3·x
Prediction
ŷ = 41.7 + 5.3·10 = 94.7
Answer
ŷ = 94.7
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