An annuity pays €500 at the end of each year for 10 years. The interest rate is 5% per year. Find the present value PV of the annuity.
A = €500i = 5%n = 10
PV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 20 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.
P = €200 0002.4% / 12N = 240
A = €
3.NPV & IRRMedium
A project costs €50 000 today and returns €13 000 at the end of each year for 6 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €20 000 and returns €13 000, €12 000, €11 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 13 000CF2 = 12 000CF3 = 11 500
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 10 years to maturity (face value €1 000, annual coupon 4.5%) when its yield to maturity is 5.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 6%. Find your holding period return HPR in percent.
F = €1 000coupon = 4.5%n = 10y0 = 5.5%y1 = 6%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 7 years and sells for €1 060. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 7P = €1 060
YTM = %
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €2 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 7% and the tax rate is 20%. Find the WACC.
A company has 2 500 000 shares at €12.50 each and 40 000 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 1.8, the risk-free rate 4% and the market return 10%. Find the WACC.
β = 1.8rf = 4%rm = 10%rd = 7%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 28, 11, 33, 12, 52.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 428.
μ = 500σ = 60x = 428
z =
11.ProbabilityMedium
A bag holds 5 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
5 / 6
P =
12.ProbabilityHard
A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 3 correct answers?
n = 6p = 0.4k = 3
P =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.8990. Find z to two decimals.
A = 0.8990
z =
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 bottom 90% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm90%
x = cm
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 100 phones, mean 71 h, known standard deviation 20 h. Model 2: 50 phones, mean 62.5 h, known standard deviation 6 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
A pilot study suggests that about 0.8 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 0.05?
p ≈ 0.8E = 0.0590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 116.3 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 116.3s = 10n = 49
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 12. Find the smallest sample size n.
σ = 12E = 290%
n =
19.Linear regression & correlationMedium
The table gives 6 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 = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (26.5, 130) is made. Find its residual e = y − ŷ.
n = 6(26.5, 130)
e =
3Solutions
1.Time value of money
Answer: PV = 3861 €
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 − 0.6139)/0.05 = 7.722
Present value
PV = 500·7.722 = €3 860.87
Answer
PV = €3 860.87
2.Time value of money
Answer: A = 1050 €
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 = 0.024/12 = 0.002
N = 12·20 = 240
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 = 0.6191
A = 200 000·0.002/(1 − 0.6191) = €1 050.09
Answer
A = €1 050.09
3.NPV & IRR
Answer: DPB = 5.476 years
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 13 000/1.12 = 11 607.14 → Σ = 11 607.14
year 2: 13 000/1.254 = 10 363.52 → Σ = 21 970.66
year 3: 13 000/1.405 = 9 253.14 → Σ = 31 223.81
year 4: 13 000/1.574 = 8 261.74 → Σ = 39 485.54
year 5: 13 000/1.762 = 7 376.55 → Σ = 46 862.09
year 6: 13 000/1.974 = 6 586.20 → Σ = 53 448.30
Fraction of the last year
DPB = 5 + 3 137.91/6 586.20 = 5.476
Answer
DPB = 5.476 years
The discounted payback is always longer than the simple payback (3.846 years).
4.NPV & IRR
Answer: IRR = 38.38 %
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%) = 888.08
NPV(40%) = −400.87
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 35 + 5·888.08/(888.08 − (−400.87)) = 38.44%
exact: NPV(38.38%) = 0 → IRR = 38.38%
Answer
IRR = 38.38%
5.Bond pricing & yield
Answer: HPR = 1.985 %
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
€1 060 = €50·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷
Solve numerically
P(4%) = €1 060.02, P(5%) = €1 000.00
YTM = 4%
Answer
YTM = 4%
7.CAPM & WACC
Answer: WACC = 10.72 %
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 = 0.8, D/V = 0.2
WACC
WACC = 0.8·12% + 0.2·7%·(1 − 0.2) = 10.72%
Answer
WACC = 10.72%
8.CAPM & WACC
Answer: WACC = 9.194 %
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 = (428 − 500)/60 = −1.2
Answer
z = −1.2
11.Probability
Answer: P = 0.1818
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 = 5/11 · 4/10 = 20/110 = 0.1818
Answer
P = 0.1818
12.Probability
Answer: P = 0.2765
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 = 3) = 20·0.43·0.63 = 20·0.064·0.216 = 0.2765
Answer
P = 0.2765
13.Normal distribution
Answer: z = 1.64
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 + 0.8990)/2 = 0.9495
Read the z-table backwards
Find 0.9495 inside the table and read off z.
z = 1.64
Answer
z = 1.64
14.Normal distribution
Answer: x = 187 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) = 0.9
z from the table
z ≈ 1.28
Back to the original units
x = 178 + 1.28·7 = 187 cm
Answer
x = 187 cm
15.Confidence intervals
Answer: upper limit = 12.07 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 = 71 − 62.5 = 8.5
Standard error
SE = √(20²/100 + 6²/50) = 2.173
Margin of error
E = 1.645·2.173 = 3.574
Interval
8.5 ± 3.574 → [4.926, 12.07]
Answer
upper limit = 12.07 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 174
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 = 1.645²·0.8·0.2/0.05² = 173.2
Round up
n = 174
Answer
n = 174
17.Hypothesis tests & confidence intervals
Answer: t = −2.59
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/√49 = 10/7 = 1.429
Test statistic
t = (116.3 − 120)/1.429 = −2.59
Answer
t = −2.59
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 98
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 = (1.645·12/2)² = 97.42
Round up
n = 98
Answer
n = 98
19.Linear regression & correlation
Answer: r = 0.9672
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 = 11475 − 132·507/6 = 321
Sxx = 2974 − 132²/6 = 70
Syy = 44415 − 507²/6 = 1574
Correlation coefficient
r = 321/√(70·1574) = 0.9672
Answer
r = 0.9672
20.Linear regression & correlation
Answer: e = 1.929
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 = 675/6 = 112.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 817.5
Regression line
b = 817.5/157.5 = 5.19
a = 112.5 − 5.19·23.5 = −9.476
ŷ = −9.476 + 5.19·x
Predicted value
ŷ = −9.476 + 5.19·26.5 = 128.1
Residual
e = 130 − 128.1 = 1.929
Answer
e = 1.929
The point lies above the line.
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