A credit card charges a nominal interest rate of 22% per year, compounded daily. Find the effective annual rate EAR in percent.
r = 22%m = 365
EAR = %
2.Time value of moneyHard
A mortgage of €250 000 is repaid with equal monthly payments over 25 years at 5.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 11? Find the outstanding balance B.
P = €250 0005.4% / 12N = 300k = 132
B = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 3: €31 000, €37 000, €17 000. The cost of capital is 9%. Find the net present value NPV.
A project costs €150 000 today. Its cash flow is €27 500 at the end of year 1 and grows by 5% per year until the last cash flow at the end of year 6. The cost of capital is 10%. Find the net present value NPV.
C0 = €150 000CF1 = €27 500g = 5%n = 6r = 10%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 3% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 2.5% per year. Find the price P of the bond.
F = €1 000coupon = 3%YTM = 2.5%n = 5
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 6 years and sells for €925. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 6P = €925
YTM = %
7.CAPM & WACCMedium
A share with a beta of 1.25 has a required return of 9.75%. The risk-free rate is 3.5%. What expected market return rm does CAPM imply?
β = 1.25re = 9.75%rf = 3.5%
rm = %
8.CAPM & WACCHard
A company has 2 500 000 shares at €25.00 each and 20 000 bonds trading at €980 each. The pre-tax cost of debt is 4% and the tax rate is 30%. The share’s beta is 0.8, the risk-free rate 3.5% and the market return 10.5%. Find the WACC.
β = 0.8rf = 3.5%rm = 10.5%rd = 4%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 34, 18, 40, 53, 42, 37, 45.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is above 166?
μ = 170σ = 8x = 166
P =
11.ProbabilityMedium
A fair die is rolled 5 times. What is the probability P of getting at least one six?
n = 5
P =
12.ProbabilityHard
X has the distribution shown, and Y = 2X − 3. Find the standard deviation σY of Y.
The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 6 of them exceeds 148.2 kg. Round z to two decimals.
μ = 25 kgσ = 0.5 kgn = 6> 148.2 kg
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 10% is to the right of it. Find the cut-off x.
μ = 250 gσ = 5 g10%
x = g
15.Confidence intervalsMedium
A sample of 12 items has mean 72 and standard deviation s = 12. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 12x̄ = 72s = 1299%t(11) = 3.106
E =
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 95% confidence interval for the proportion has a margin of error of at most 0.04?
p ≈ 0.8E = 0.0495%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 16 students has a mean score of 124.2 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 124.2s = 8n = 16
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 0.5? The standard deviation is 20. Find the smallest sample size n.
σ = 20E = 0.590%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 18?
n = 6ŷ = 18
x =
3Solutions
1.Time value of money
Answer: EAR = 24.6 %
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 per period
r/m = 0.22/365 = 0.0006027
Compound over one year
EAR = (1 + 0.0006027)365 − 1 = 1.246 − 1 = 24.6%
Answer
EAR = 24.6%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: B = 178946 €
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 = 300 − 132 = 168
B = 1520·(1 − (1 + 0.0045)−168)/0.0045 = €178 946.34
Answer
B = €178 946.34
3.NPV & IRR
Answer: NPV = −2290 €
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 = −75 000
CF1/(1 + 0.09)1 = 31 000/1.09 = 28 440.37
CF2/(1 + 0.09)2 = 37 000/1.188 = 31 142.16
CF3/(1 + 0.09)3 = 17 000/1.295 = 13 127.12
Add them up
NPV = €−2 290.35
Answer
NPV = €−2 290.35
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: NPV = −16047 €
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.
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: P = 1023 €
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 = €15.00, i = 0.0125, N = 10
Present value of the coupons
€15.00·(1 − 0.8832)/0.0125 = €140.18
Present value of the face value
€1 000·0.8832 = €883.18
Price
P = €1 023.36
Answer
P = €1 023.36
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 6.552 %
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
€925 = €50·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶
Solve numerically
P(6%) = €950.83, P(7%) = €904.67
YTM = 6.552%
Answer
YTM = 6.552%
7.CAPM & WACC
Answer: rm = 8.5 %
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 = (9.75% − 3.5%)/1.25 = 5%
Market return
rm = 3.5% + 5% = 8.5%
Answer
rm = 8.5%
8.CAPM & WACC
Answer: WACC = 7.596 %
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 = 2 500 000·€25.00 = €62 500 000
D = 20 000·€980 = €19 600 000
Cost of equity (CAPM)
re = 3.5% + 0.8·7% = 9.1%
Weights
V = €82 100 000, E/V = 0.7613, D/V = 0.2387
WACC
WACC = 0.7613·9.1% + 0.2387·4%·(1 − 0.3) = 7.596%
Answer
WACC = 7.596%
9.Descriptive statistics
Answer: s = 10.88
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
11.Probability
Answer: P = 0.5981
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.
Complement
P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981
Answer
P = 0.5981
12.Probability
Answer: σY = 3.066
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.
Spread of X
μ = 3.5, Var(X) = 2.35, σX = 1.533
Linear change
σY = |2|·1.533 = 3.066
Answer
σY = 3.066
The mean does move: E(Y) = 2·3.5 − 3 = 4.
13.Normal distribution
Answer: P = 0.9292
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 total
μT = 150 kg, σT = 0.5·√6 = 1.225 kg
Standardise
z = (148.2 − 150)/1.225 = −1.47
Probability
P = 1 − Φ(−1.47) = 0.9292
Answer
P = 0.9292
14.Normal distribution
Answer: x = 256.4 g
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 = 250 + 1.28·5 = 256.4 g
Answer
x = 256.4 g
15.Confidence intervals
Answer: E = 10.76
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 = 11, t = 3.106
Margin of error
E = 3.106·12/√12 = 10.76
Answer
E = 10.76
16.Confidence intervals
Answer: n = 385
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 = 1.96²·0.8·0.2/0.04² = 384.2
Round up
n = 385
Answer
n = 385
17.Hypothesis tests & confidence intervals
Answer: t = 2.1
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 = 8/√16 = 8/4 = 2
Test statistic
t = (124.2 − 120)/2 = 2.1
Answer
t = 2.1
t0.025; df = 15 = 2.131
|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 4330
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 = (1.645·20/0.5)² = 4330
Round up
n = 4330
Answer
n = 4330
19.Linear regression & correlation
Answer: r = 0.9669
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 = 51/5 = 10.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 62
Syy = Σ(y − ȳ)² = 102.8
Correlation coefficient
r = 62/√(40·102.8) = 0.9669
Answer
r = 0.9669
20.Linear regression & correlation
Answer: x = 9.121
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 = 84/6 = 14
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 132
Regression line
b = 132/70 = 1.886
a = 14 − 1.886·7 = 0.8
ŷ = 0.8 + 1.886·x
Solve for x
x = (18 − 0.8)/1.886 = 9.121
Answer
x = 9.121
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