You need €25 000 in 3 years. An account pays 4% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €25 0004% / 12n = 3
PV = €
2.Time value of moneyHard
A loan of €8 000 is repaid with equal monthly payments over 6 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.
P = €8 0004.8% / 12N = 72
A = €
3.NPV & IRRMedium
A machine costs €40 000 today, lasts 4 years and costs €1 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.
C0 = €40 000M = €1 000n = 4r = 12%
EAC = €
4.NPV & IRRHard
An investment of €25 000 returns €7 000 at the end of each year for 5 years. Find its internal rate of return IRR in percent.
C0 = €25 000CF = €7 000n = 5
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €957 when the yield to maturity is 5.5%. Find the coupon rate c in percent.
F = €1 000n = 12P = €957YTM = 5.5%
c = %
6.Bond pricing & yieldHard
A bond trades at €1 070 with a yield to maturity of 7% (annual compounding) and a Macaulay duration of 3.6 years. Use the modified duration to estimate the new price P if the yield falls by 150 basis points.
P0 = €1 070y = 7%D = 3.6Δy = −150 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €2 000 000 and debt worth €2 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 7% and the tax rate is 30%. Find the WACC.
A company has 1 000 000 shares at €31.00 each and 40 000 bonds trading at €1 000 each. The pre-tax cost of debt is 7% and the tax rate is 30%. The share’s beta is 1.5, the risk-free rate 1.5% and the market return 7.5%. Find the WACC.
β = 1.5rf = 1.5%rm = 7.5%rd = 7%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 50, 15, 54, 48, 31, 29.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 10. Find the z-score z of a score of 65.
μ = 70σ = 10x = 65
z =
11.ProbabilityMedium
A bag holds 7 red and 5 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 5
P =
12.ProbabilityHard
A test has 12 multiple-choice questions. A student who guesses gets each one right with probability 0.1, independently. What is the probability P of exactly 1 correct answers?
n = 12p = 0.1k = 1
P =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.8324. Find z to two decimals.
A = 0.8324
z =
14.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
A sample of 25 items has mean 100.5 and standard deviation s = 15. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 25x̄ = 100.5s = 1595%t(24) = 2.064
lower limit =
16.Confidence intervalsHard
The diameters of 20 machined shafts have a sample standard deviation of 3 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 19, 2.5% of the χ² distribution lies below 8.907 and 2.5% lies above 32.852.
n = 20s = 3 µmχ² = 8.907, 32.852
σ = µm
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 8 students, mean score 33.5, standard deviation 11.5. Group 2: 13 students, mean score 24, standard deviation 4.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 8x̄₁ = 33.5s₁ = 11.5n₂ = 13x̄₂ = 24s₂ = 4.5
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 3? The standard deviation is 4. Find the smallest sample size n.
σ = 4E = 390%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (5, 3) is made. Find its residual e = y − ŷ.
n = 5(5, 3)
e =
3Solutions
1.Time value of money
Answer: PV = 22177 €
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 = 0.04/12 = 0.003333
N = 12·3 = 36
Present value
PV = FV/(1 + i)N = 25 000/1.127 = €22 177.44
Answer
PV = €22 177.44
2.Time value of money
Answer: A = 128.1 €
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 = 0.048/12 = 0.004
N = 12·6 = 72
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 = 0.7502
A = 8 000·0.004/(1 − 0.7502) = €128.10
Answer
A = €128.10
3.NPV & IRR
Answer: EAC = 14169 €
Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.
Annuity factor
(1 − 1.12−4)/0.12 = 3.037
Annualised purchase price
40 000/3.037 = 13 169.38
EAC
EAC = 13 169.38 + 1 000 = €14 169.38
Answer
EAC = €14 169.38
4.NPV & IRR
Answer: IRR = 12.38 %
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.
Required annuity factor
(1 − (1 + i)−5)/i = 25 000/7 000 = 3.571
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 12%: 3.605
i = 13%: 3.517
IRR = 12.38%
Answer
IRR = 12.38%
5.Bond pricing & yield
Answer: c = 5.001 %
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.
Present value of the face value
1000·(1 + 0.055)−12 = €525.98
Coupon
C = (957 − 525.98)/8.619 = €50.01
Coupon rate
c = 50.01/1000 = 5.001%
Answer
c = 5.001%
6.Bond pricing & yield
Answer: P = 1124 €
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 = 3.6/1.07 = 3.364
Estimated price change
%ΔP ≈ −3.364·(−0.015) = 5.047%
P ≈ 1070·(1 + 0.05047) = €1 124.00
Answer
P = €1 124.00
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 8.95 %
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 = 0.5, D/V = 0.5
WACC
WACC = 0.5·13% + 0.5·7%·(1 − 0.3) = 8.95%
Answer
WACC = 8.95%
8.CAPM & WACC
Answer: WACC = 7.345 %
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 = (65 − 70)/10 = −0.5
Answer
z = −0.5
11.Probability
Answer: P = 0.3182
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 = 7/12 · 6/11 = 42/132 = 0.3182
Answer
P = 0.3182
12.Probability
Answer: P = 0.3766
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 = 1) = 12·0.11·0.911 = 12·0.1·0.3138 = 0.3766
Answer
P = 0.3766
13.Normal distribution
Answer: z = 1.38
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 + 0.8324)/2 = 0.9162
Read the z-table backwards
Find 0.9162 inside the table and read off z.
z = 1.38
Answer
z = 1.38
14.Normal distribution
Answer: P = 0.9868
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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm
Standardise
z = (0 − 0.04)/0.01803 = −2.22
Probability
P = 1 − Φ(−2.22) = 0.9868
Answer
P = 0.9868
15.Confidence intervals
Answer: lower limit = 94.31
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 = 24, t = 2.064
Margin of error
E = 2.064·15/√25 = 6.192
Interval
100.5 ± 6.192 → [94.31, 106.7]
Answer
lower limit = 94.31
16.Confidence intervals
Answer: σ = 2.281 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 = 19·3² = 171
Limits for σ²
171/32.852 = 5.205
171/8.907 = 19.2
Limit for σ
σ = √5.205 = 2.281 µm
Answer
σ = 2.281 µm
95% interval for σ: [2.281, 4.382] µm.
17.Hypothesis tests & confidence intervals
Answer: t = 2.696
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.
Pooled variance
sp² = (7·11.5² + 12·4.5²)/19 = 61.51
sp = 7.843
Standard error
SE = 7.843·√(1/8 + 1/13) = 3.524
Test statistic
t = (33.5 − 24)/3.524 = 2.696
Answer
t = 2.696
With df = 19, the two-sided 5% critical value is 2.093; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: n = 5
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 = (1.645·4/3)² = 4.811
Round up
n = 5
Answer
n = 5
19.Linear regression & correlation
Answer: r = 0.9862
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 = 105/5 = 21
ȳ = 550/5 = 110
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 465
Syy = Σ(y − ȳ)² = 2470
Correlation coefficient
r = 465/√(90·2470) = 0.9862
Answer
r = 0.9862
20.Linear regression & correlation
Answer: e = −4.5
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 = 46/5 = 9.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 68
Regression line
b = 68/40 = 1.7
a = 9.2 − 1.7·6 = −1
ŷ = −1 + 1.7·x
Predicted value
ŷ = −1 + 1.7·5 = 7.5
Residual
e = 3 − 7.5 = −4.5
Answer
e = −4.5
The point lies below the line.
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