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
You borrow €10 000 at 3% per year, compounded monthly, and repay €450 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.
P = €10 000A = €4503% / 12
N = payments
3.NPV & IRRMedium
A project costs €40 000 today and returns €16 000, €14 000, €15 000, €12 000, €13 000, €12 000 at the end of years 1 to 6. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €50 000 and returns €36 500, €22 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 36 500CF2 = 22 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 10 years and pays annual coupons. It sells for €1 154 when the yield to maturity is 5%. Find the coupon rate c in percent.
F = €1 000n = 10P = €1 154YTM = 5%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 7% matures in 10 years and sells for €1 155. Find its yield to maturity YTM in percent.
F = €1 000coupon = 7%n = 10P = €1 155
YTM = %
7.CAPM & WACCMedium
A company has equity worth €12 000 000 and debt worth €6 000 000. Its cost of equity is 11%, its pre-tax cost of debt is 4% and the tax rate is 20%. Find the WACC.
A company has 2 000 000 shares at €15.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 4% and the tax rate is 20%. The share’s beta is 0.9, the risk-free rate 3% and the market return 10%. Find the WACC.
β = 0.9rf = 3%rm = 10%rd = 4%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 60, 15, 54, 58, 31, 15.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 10. What proportion P of values is above 72.5?
μ = 60σ = 10x = 72.5
P =
11.ProbabilityMedium
A bag holds 8 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 9
P =
12.ProbabilityHard
5 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?
n = 5p = 0.7
P =
13.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 62 and 85.5 points? Give P as a decimal.
μ = 65σ = 10x1 = 62x2 = 85.5
P =
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 top 20% is to the right of it. Find the cut-off x.
μ = 178 cmσ = 7 cm20%
x = cm
15.Confidence intervalsMedium
A sample of 8 items has mean 44.5 and standard deviation s = 10. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 8x̄ = 44.5s = 1099%t(7) = 3.499
upper limit =
16.Confidence intervalsHard
Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.04?
p = 0.5E = 0.0490%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 116.1 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 116.1s = 12n = 64
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.10 with a sample of 64; the standard deviation is known to be 20. In reality the mean is 56.5. Find the power 1 − β of the test in percent.
μ₀ = 50μ = 56.5σ = 20n = 64α = 0.10
1 − β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (25, 117) is made. Find its residual e = y − ŷ.
n = 5(25, 117)
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: N = 23 payments
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.
Rearrange the annuity formula
i = 0.0025
(1 + i)−N = 1 − 10 000·0.0025/450 = 0.9444
Solve with logarithms
N = −ln(0.9444)/ln(1.002) = 22.89
Round up to whole payments
22 full payments leave a small balance, which the last, smaller payment clears.
N = 23
Answer
N = 23 payments
3.NPV & IRR
Answer: DPB = 3.145 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: 16 000/1.08 = 14 814.81 → Σ = 14 814.81
year 2: 14 000/1.166 = 12 002.74 → Σ = 26 817.56
year 3: 15 000/1.26 = 11 907.48 → Σ = 38 725.04
year 4: 12 000/1.36 = 8 820.36 → Σ = 47 545.40
Fraction of the last year
DPB = 3 + 1 274.96/8 820.36 = 3.145
Answer
DPB = 3.145 years
The discounted payback is always longer than the simple payback (2.667 years).
4.NPV & IRR
Answer: IRR = 12.21 %
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(10%) = 1 363.64
NPV(15%) = −1 625.71
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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 + 0.05)−10 = €613.91
Coupon
C = (1154 − 613.91)/7.722 = €69.94
Coupon rate
c = 69.94/1000 = 6.994%
Answer
c = 6.994%
6.Bond pricing & yield
Answer: YTM = 4.993 %
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.
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 = €18 000 000, E/V = 0.6667, D/V = 0.3333
WACC
WACC = 0.6667·11% + 0.3333·4%·(1 − 0.2) = 8.4%
Answer
WACC = 8.4%
8.CAPM & WACC
Answer: WACC = 5.875 %
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 = 2 000 000·€15.00 = €30 000 000
D = 40 000·€960 = €38 400 000
Cost of equity (CAPM)
re = 3% + 0.9·7% = 9.3%
Weights
V = €68 400 000, E/V = 0.4386, D/V = 0.5614
WACC
WACC = 0.4386·9.3% + 0.5614·4%·(1 − 0.2) = 5.875%
Answer
WACC = 5.875%
9.Descriptive statistics
Answer: s = 21.18
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
11.Probability
Answer: P = 0.2059
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 = 8/17 · 7/16 = 56/272 = 0.2059
Answer
P = 0.2059
12.Probability
Answer: P = 0.03078
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.
Zero and one failure
P(X = 0) = 0.35 = 0.00243
P(X = 1) = 5·0.7·0.34 = 0.02835
Add them
P = 0.00243 + 0.02835 = 0.03078
Answer
P = 0.03078
13.Normal distribution
Answer: P = 0.5977
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 = (62 − 65)/10 = −0.3
z₂ = (85.5 − 65)/10 = 2.05
Area between
P = Φ(2.05) − Φ(−0.3) = 0.9798 − 0.3821 = 0.5977
Answer
P = 0.5977
14.Normal distribution
Answer: x = 183.9 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) = 0.8
z from the table
z ≈ 0.84
Back to the original units
x = 178 + 0.84·7 = 183.9 cm
Answer
x = 183.9 cm
15.Confidence intervals
Answer: upper limit = 56.87
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 = 3.499
Margin of error
E = 3.499·10/√8 = 12.37
Interval
44.5 ± 12.37 → [32.13, 56.87]
Answer
upper limit = 56.87
16.Confidence intervals
Answer: n = 423
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 = 1.645²·0.5·0.5/0.04² = 422.8
Round up
n = 423
Answer
n = 423
17.Hypothesis tests & confidence intervals
Answer: t = −2.6
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).
Standard error
SE = 12/√64 = 12/8 = 1.5
Test statistic
t = (116.1 − 120)/1.5 = −2.6
Answer
t = −2.6
t0.025; df = 63 = 1.998
|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 90.62 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 20/√64 = 2.5
x̄c = 50 + 1.282·2.5 = 53.2
z under the true mean
z = (53.2 − 56.5)/2.5 = −1.318
Power
power = P(Z > −1.318) = 90.62%
1 − β = 90.62%
Answer
1 − β = 90.62%
19.Linear regression & correlation
Answer: r = 0.9847
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 = 135/6 = 22.5
ȳ = 439/6 = 73.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 691.5
Syy = Σ(y − ȳ)² = 3131
Correlation coefficient
r = 691.5/√(157.5·3131) = 0.9847
Answer
r = 0.9847
20.Linear regression & correlation
Answer: e = −1.6
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 = 110/5 = 22
ȳ = 521/5 = 104.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 192
Regression line
b = 192/40 = 4.8
a = 104.2 − 4.8·22 = −1.4
ŷ = −1.4 + 4.8·x
Predicted value
ŷ = −1.4 + 4.8·25 = 118.6
Residual
e = 117 − 118.6 = −1.6
Answer
e = −1.6
The point lies below the line.
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