You invest €20 000 at 6% per year with continuous compounding. Find the value FV after 9 years.
PV = €20 000r = 6%t = 9
FV = €
2.Time value of moneyHard
You save €250 at the end of every month for 27 years. The account pays 5.4% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €2505.4% / 12N = 324
FV = €
3.NPV & IRRMedium
A project costs €500 000 today. It generates €65 000 at the end of the first year, and the cash flow then grows by 1% per year forever. The cost of capital is 11%. Find the net present value NPV.
C0 = €500 000CF1 = €65 000g = 1%r = 11%
NPV = €
4.NPV & IRRHard
A project costs €40 000 and returns €14 000, €20 500, €22 000 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 matures in 5 years and pays annual coupons. It sells for €1 229 when the yield to maturity is 3%. Find the coupon rate c in percent.
F = €1 000n = 5P = €1 229YTM = 3%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 10 years. It sells for €961. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 4.5% (2×)n = 10P = €961
YTM = %
7.CAPM & WACCMedium
A share with a beta of 0.9 has a required return of 8.3%. The risk-free rate is 2%. What expected market return rm does CAPM imply?
β = 0.9re = 8.3%rf = 2%
rm = %
8.CAPM & WACCHard
A company has 1 000 000 shares trading at €37.00; the book value of its equity is €22 200 000. It just paid a dividend of €2.90 per share, and dividends grow by 4% per year. Its debt has a market value of €18 500 000 and a pre-tax cost of 5%. The tax rate is 25.8%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 40, 31, 57, 11, 21.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 25. Find the z-score z of a score of 510.
μ = 500σ = 25x = 510
z =
11.ProbabilityMedium
A bag holds 4 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
4 / 6
P =
12.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 2 correct answers?
n = 5p = 0.5k = 2
P =
13.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 26.3 and 33.5 min? Give P as a decimal.
μ = 32σ = 6x1 = 26.3x2 = 33.5
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 25% of them lie (the 25th percentile).
μ = 500σ = 40P = 25%
x = h
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 80 phones, mean 86 h, known standard deviation 18 h. Model 2: 80 phones, mean 97 h, known standard deviation 9 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
Two fertilisers are tested. Plot group 1: 7 plots, mean yield 19.5 kg, standard deviation 5.5 kg. Plot group 2: 14 plots, mean 13 kg, standard deviation 3 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.
A sample of 81 students has a mean score of 200.5 and a sample standard deviation of 4. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 200.5s = 4n = 81
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 75 of 250 patients on treatment A recover, and 34 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 75n₁ = 250x₂ = 34n₂ = 80
z =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 42Σy = 185Σx² = 364Σy² = 6955Σxy = 1590
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. Predict ŷ for x = 3.
n = 5x = 3
ŷ =
3Solutions
1.Time value of money
Answer: FV = 34320 €
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.
Exponent
r·t = 0.06·9 = 0.54
e0.54 = 1.716
Future value
FV = 20 000·1.716 = €34 320.14
Answer
FV = €34 320.14
2.Time value of money
Answer: FV = 182407 €
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 €81 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 150000 €
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.
Value of the cash flows
PV = 65 000/(0.11 − 0.01) = 650 000.00
NPV
NPV = 650 000.00 − 500 000 = €150 000.00
Answer
NPV = €150 000.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: MIRR = 14.77 %
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
14 000·1.082 = 16 329.60
20 500·1.081 = 22 140.00
22 000·1.080 = 22 000.00
TV = 60 469.60
MIRR
MIRR = (60 469.60/40 000)1/3 − 1 = 14.77%
Answer
MIRR = 14.77%
5.Bond pricing & yield
Answer: c = 8 %
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 + 0.03)−5 = €862.61
Coupon
C = (1229 − 862.61)/4.58 = €80.00
Coupon rate
c = 80/1000 = 8%
Answer
c = 8%
6.Bond pricing & yield
Answer: YTM = 5 %
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, N = 20
Solve numerically
i = 2.5% per half-year
YTM
YTM = 2·2.5% = 5%
Answer
YTM = 5%
7.CAPM & WACC
Answer: rm = 9 %
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 risk premium
rm − rf = (8.3% − 2%)/0.9 = 7%
Market return
rm = 2% + 7% = 9%
Answer
rm = 9%
8.CAPM & WACC
Answer: WACC = 9.338 %
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 = (510 − 500)/25 = 0.4
Answer
z = 0.4
11.Probability
Answer: P = 0.1333
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 = 4/10 · 3/9 = 12/90 = 0.1333
Answer
P = 0.1333
12.Probability
Answer: P = 0.3125
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 = 2) = 10·0.52·0.53 = 10·0.25·0.125 = 0.3125
Answer
P = 0.3125
13.Normal distribution
Answer: P = 0.4277
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (26.3 − 32)/6 = −0.95
z₂ = (33.5 − 32)/6 = 0.25
Area between
P = Φ(0.25) − Φ(−0.95) = 0.5987 − 0.1711 = 0.4277
Answer
P = 0.4277
14.Normal distribution
Answer: x = 473 h
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z for the percentile
Φ(z) = 0.25 → z = −0.6745
Back to x
x = 500 + −0.6745·40 = 473.02 h
Answer
x = 473.02 h
15.Confidence intervals
Answer: lower limit = −14.7 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 = 86 − 97 = −11
Standard error
SE = √(18²/80 + 9²/80) = 2.25
Margin of error
E = 1.645·2.25 = 3.701
Interval
−11 ± 3.701 → [−14.7, −7.299]
Answer
lower limit = −14.7 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: upper limit = 10.34 kg
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.
Pooled standard deviation
sp² = (6·5.5² + 13·3²)/19 = 15.71
sp = 3.964
Margin of error
E = 2.093·3.964·√(1/7 + 1/14) = 3.84
Interval
6.5 ± 3.84 → [2.66, 10.34]
Answer
upper limit = 10.34 kg
17.Hypothesis tests & confidence intervals
Answer: t = 1.125
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 = 4/√81 = 4/9 = 0.4444
Test statistic
t = (200.5 − 200)/0.4444 = 1.125
Answer
t = 1.125
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: z = −2.069
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.
Sample proportions
p̂₁ = 75/250 = 0.3
p̂₂ = 34/80 = 0.425
p̂ = 109/330 = 0.3303
Standard error under H₀
SE = √(0.3303·0.6697·(1/250 + 1/80)) = 0.06041
Test statistic
z = (0.3 − 0.425)/0.06041 = −2.069
Answer
z = −2.069
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9969
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 = 1590 − 42·185/6 = 295
Sxx = 364 − 42²/6 = 70
Syy = 6955 − 185²/6 = 1251
Correlation coefficient
r = 295/√(70·1251) = 0.9969
Answer
r = 0.9969
20.Linear regression & correlation
Answer: ŷ = 5.55
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 = 63/5 = 12.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 94
Regression line
b = 94/40 = 2.35
a = 12.6 − 2.35·6 = −1.5
ŷ = −1.5 + 2.35·x
Prediction
ŷ = −1.5 + 2.35·3 = 5.55
Answer
ŷ = 5.55
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