You need €15 000 in 3 years. An account pays 4.8% per year, compounded quarterly. How much must you deposit today? Find the present value PV.
FV = €15 0004.8% / 4n = 3
PV = €
2.Time value of moneyHard
A mortgage of €300 000 is repaid with equal monthly payments over 30 years at 4.2% per year, compounded monthly. How much is still owed just after the payment at the end of year 17? Find the outstanding balance B.
P = €300 0004.2% / 12N = 360k = 204
B = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 4: €19 000, €13 000, €13 000, €10 000. The cost of capital is 15%. Find the net present value NPV.
A project costs €10 000 and returns €7 000, €7 500, €5 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 7 000CF2 = 7 500CF3 = 5 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 4.5% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 6% per year. Find the price P of the bond.
F = €1 000coupon = 4.5%YTM = 6%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 6 years. The yield to maturity is 3.5%. Find the Macaulay duration D in years.
F = €1 000coupon = 4%n = 6YTM = 3.5%
D = years
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €10 000 000, preferred shares worth €500 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 7% and the pre-tax cost of debt 4%. The tax rate is 25.8%. Find the WACC.
A company has 1 000 000 shares trading at €30.50; the book value of its equity is €15 300 000. It just paid a dividend of €0.90 per share, and dividends grow by 5% per year. Its debt has a market value of €18 300 000 and a pre-tax cost of 6%. The tax rate is 20%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 31, 33, 43, 10, 39.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 15. What proportion P of values is above 32?
μ = 50σ = 15x = 32
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2, 3p(x): 0.15, 0.35, 0.3, 0.2
Var(X) =
12.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?
n = 9p = 0.4k = 1
P =
13.Normal distributionMedium
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 23 and 35.3 min? Give P as a decimal.
μ = 32σ = 6x1 = 23x2 = 35.3
P =
14.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
A sample of 8 items has mean 109.5 and standard deviation s = 12. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 8x̄ = 109.5s = 1290%t(7) = 1.895
lower limit =
16.Confidence intervalsHard
The population standard deviation is 21. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 3?
σ = 21E = 390%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 18 students, mean score 112, standard deviation 10. Group 2: 13 students, mean score 104.1, standard deviation 2. 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 = 18x̄₁ = 112s₁ = 10n₂ = 13x̄₂ = 104.1s₂ = 2
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 201.5. The population standard deviation is 20. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 201.5σ = 20n = 6495%
L =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
20.Linear regression & correlationHard
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, 126) is made. Find its residual e = y − ŷ.
n = 6(26.5, 126)
e =
3Solutions
1.Time value of money
Answer: PV = 12999 €
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 = 0.048/4 = 0.012
N = 4·3 = 12
Present value
PV = FV/(1 + i)N = 15 000/1.154 = €12 999.45
Answer
PV = €12 999.45
2.Time value of money
Answer: B = 176124 €
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.
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 360 − 204 = 156
B = 1467·(1 − (1 + 0.0035)−156)/0.0035 = €176 123.86
Answer
B = €176 123.86
3.NPV & IRR
Answer: NPV = −4383 €
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 = −45 000
CF1/(1 + 0.15)1 = 19 000/1.15 = 16 521.74
CF2/(1 + 0.15)2 = 13 000/1.322 = 9 829.87
CF3/(1 + 0.15)3 = 13 000/1.521 = 8 547.71
CF4/(1 + 0.15)4 = 10 000/1.749 = 5 717.53
Add them up
NPV = €−4 383.15
Answer
NPV = €−4 383.15
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 46.69 %
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(45%) = 198.86
NPV(50%) = −370.37
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 45 + 5·198.86/(198.86 − (−370.37)) = 46.75%
exact: NPV(46.69%) = 0 → IRR = 46.69%
Answer
IRR = 46.69%
5.Bond pricing & yield
Answer: P = 873 €
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 = 0.03, N = 24
Present value of the coupons
€22.50·(1 − 0.4919)/0.03 = €381.05
Present value of the face value
€1 000·0.4919 = €491.93
Price
P = €872.98
Answer
P = €872.98
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: D = 5.46 years
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.
Present value of each cash flow
t = 1: 40/1.035 = 38.65, t·PV = 38.65
t = 2: 40/1.071 = 37.34, t·PV = 74.68
t = 3: 40/1.109 = 36.08, t·PV = 108.23
t = 4: 40/1.148 = 34.86, t·PV = 139.43
t = 5: 40/1.188 = 33.68, t·PV = 168.39
t = 6: 1040/1.229 = 846.04, t·PV = 5 076.24
Duration
P = 1 026.64
D = 5 605.63/1 026.64 = 5.46
Answer
D = 5.46 years
7.CAPM & WACC
Answer: WACC = 9.57 %
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 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 = 0.9·1.05 = €0.95
re = 0.945/30.5 + 0.05 = 8.098%
Market values
E = 1 000 000·€30.50 = €30 500 000
V = €48 800 000, E/V = 0.625, D/V = 0.375
WACC
WACC = 0.625·8.098% + 0.375·6%·(1 − 0.2) = 6.861%
Answer
WACC = 6.861%
9.Descriptive statistics
Answer: s = 12.77
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849
Answer
P = 0.8849
11.Probability
Answer: Var(X) = 0.9475
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.
Mean
μ = 0·0.15 + 1·0.35 + 2·0.3 + 3·0.2 = 1.55
Mean of the squares
E(X²) = 0·0.15 + 1·0.35 + 4·0.3 + 9·0.2 = 3.35
Variance
Var(X) = 3.35 − 1.55² = 0.9475
Answer
Var(X) = 0.9475
12.Probability
Answer: P = 0.06047
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) = 9·0.41·0.68 = 9·0.4·0.0168 = 0.06047
Answer
P = 0.06047
13.Normal distribution
Answer: P = 0.642
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 = (23 − 32)/6 = −1.5
z₂ = (35.3 − 32)/6 = 0.55
Area between
P = Φ(0.55) − Φ(−1.5) = 0.7088 − 0.06681 = 0.642
Answer
P = 0.642
14.Normal distribution
Answer: P = 0.7881
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 = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
15.Confidence intervals
Answer: lower limit = 101.5
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 = 7, t = 1.895
Margin of error
E = 1.895·12/√8 = 8.04
Interval
109.5 ± 8.04 → [101.5, 117.5]
Answer
lower limit = 101.5
16.Confidence intervals
Answer: n = 133
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.
Solve for n
n = (1.645·21/3)² = 132.6
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 133
Answer
n = 133
17.Hypothesis tests & confidence intervals
Answer: t = 2.796
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 = (17·10² + 12·2²)/29 = 60.28
sp = 7.764
Standard error
SE = 7.764·√(1/18 + 1/13) = 2.826
Test statistic
t = (112 − 104.1)/2.826 = 2.796
Answer
t = 2.796
With df = 29, the two-sided 5% critical value is 2.045; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: L = 196.6
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 = 1.96·20/√64 = 4.9
Interval limit
L = 201.5 − 4.9 = 196.6
Answer
L = 196.6
19.Linear regression & correlation
Answer: r² = 0.9876
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 = 42/6 = 7
ȳ = 113/6 = 18.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 193
Syy = Σ(y − ȳ)² = 538.8
Coefficient of determination
r² = 193²/(70·538.8) = 0.9876
Answer
r² = 0.9876
About 98.76% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = −2.267
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 = 664/6 = 110.7
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 924
Regression line
b = 924/157.5 = 5.867
a = 110.7 − 5.867·23.5 = −27.2
ŷ = −27.2 + 5.867·x
Predicted value
ŷ = −27.2 + 5.867·26.5 = 128.3
Residual
e = 126 − 128.3 = −2.267
Answer
e = −2.267
The point lies below the line.
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