You invest €5 000 at 7% per year with continuous compounding. Find the value FV after 11 years.
PV = €5 000r = 7%t = 11
FV = €
2.Time value of moneyHard
You save €100 at the end of every month for 8 years. The account pays 3% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €1003% / 12N = 96
FV = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 4: €21 000, €16 000, €18 000, €15 000. The cost of capital is 10%. Find the net present value NPV.
A project costs €50 000 and returns €19 000, €24 500, €23 000, €22 500 at the end of years 1 to 4. 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 pays an 8% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 7% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 7%n = 10
P = €
6.Bond pricing & yieldHard
A bond trades at €1 055 with a yield to maturity of 7% (annual compounding) and a Macaulay duration of 3.1 years. Use the modified duration to estimate the new price P if the yield rises by 50 basis points.
P0 = €1 055y = 7%D = 3.1Δy = +50 bp
P = €
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €12 000 000, preferred shares worth €500 000 and debt worth €6 000 000. The cost of equity is 11%, the cost of preferred stock 7% and the pre-tax cost of debt 4.5%. The tax rate is 25%. Find the WACC.
A company has 4 000 000 shares at €23.50 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 7% and the tax rate is 21%. The share’s beta is 1.8, the risk-free rate 1% and the market return 7%. Find the WACC.
β = 1.8rf = 1%rm = 7%rd = 7%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 30, 28, 45, 27, 29, 20.
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 48?
μ = 60σ = 10x = 48
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.2, 0.3, 0.15, 0.35
Var(X) =
12.ProbabilityHard
X has the distribution shown, and Y = 10X + 5. Find the standard deviation σY of Y.
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 12 of them exceeds 957.6 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 12> 957.6 kg
P =
14.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
A sample of 10 items has mean 27.5 and standard deviation s = 20. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 10x̄ = 27.5s = 2099%t(9) = 3.25
upper limit =
16.Confidence intervalsHard
The population standard deviation is 25. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 1.5?
σ = 25E = 1.599%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 81 students has a mean score of 201.6 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 201.6s = 12n = 81
t =
18.Hypothesis tests & confidence intervalsHard
In a trial, 100 of 200 patients on treatment A recover, and 47 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 100n₁ = 200x₂ = 47n₂ = 80
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 4.5.
n = 5x = 4.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 10799 €
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 = 0.07·11 = 0.77
e0.77 = 2.16
Future value
FV = 5 000·2.16 = €10 798.83
Answer
FV = €10 798.83
2.Time value of money
Answer: FV = 10835 €
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 €9 600 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = −3917 €
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 = −60 000
CF1/(1 + 0.1)1 = 21 000/1.1 = 19 090.91
CF2/(1 + 0.1)2 = 16 000/1.21 = 13 223.14
CF3/(1 + 0.1)3 = 18 000/1.331 = 13 523.67
CF4/(1 + 0.1)4 = 15 000/1.464 = 10 245.20
Add them up
NPV = €−3 917.08
Answer
NPV = €−3 917.08
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: MIRR = 18.88 %
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
19 000·1.083 = 23 934.53
24 500·1.082 = 28 576.80
23 000·1.081 = 24 840.00
22 500·1.080 = 22 500.00
TV = 99 851.33
MIRR
MIRR = (99 851.33/50 000)1/4 − 1 = 18.88%
Answer
MIRR = 18.88%
5.Bond pricing & yield
Answer: P = 1071 €
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 = €40.00, i = 0.035, N = 20
Present value of the coupons
€40.00·(1 − 0.5026)/0.035 = €568.50
Present value of the face value
€1 000·0.5026 = €502.57
Price
P = €1 071.06
Answer
P = €1 071.06
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: P = 1040 €
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 = 3.1/1.07 = 2.897
Estimated price change
%ΔP ≈ −2.897·(0.005) = −1.449%
P ≈ 1055·(1 − 0.01449) = €1 039.72
Answer
P = €1 039.72
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 8.419 %
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).
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) = 1.327
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 = 1·0.2 + 2·0.3 + 3·0.15 + 4·0.35 = 2.65
Mean of the squares
E(X²) = 1·0.2 + 4·0.3 + 9·0.15 + 16·0.35 = 8.35
Variance
Var(X) = 8.35 − 2.65² = 1.327
Answer
Var(X) = 1.327
12.Probability
Answer: σY = 11.61
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 = 2.55, Var(X) = 1.348, σX = 1.161
Linear change
σY = |10|·1.161 = 11.61
Answer
σY = 11.61
The mean does move: E(Y) = 10·2.55 + 5 = 30.5.
13.Normal distribution
Answer: P = 0.9332
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 = 1020 kg, σT = 12·√12 = 41.57 kg
Standardise
z = (957.6 − 1020)/41.57 = −1.5
Probability
P = 1 − Φ(−1.5) = 0.9332
Answer
P = 0.9332
14.Normal distribution
Answer: P = 0.8413
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 = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
15.Confidence intervals
Answer: upper limit = 48.05
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 = 9, t = 3.25
Margin of error
E = 3.25·20/√10 = 20.55
Interval
27.5 ± 20.55 → [6.945, 48.05]
Answer
upper limit = 48.05
16.Confidence intervals
Answer: n = 1844
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 = (2.576·25/1.5)² = 1843
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1844
Answer
n = 1844
17.Hypothesis tests & confidence intervals
Answer: t = 1.2
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/√81 = 12/9 = 1.333
Test statistic
t = (201.6 − 200)/1.333 = 1.2
Answer
t = 1.2
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 = −1.325
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 = 100/200 = 0.5
p̂₂ = 47/80 = 0.5875
p̂ = 147/280 = 0.525
Standard error under H₀
SE = √(0.525·0.475·(1/200 + 1/80)) = 0.06606
Test statistic
z = (0.5 − 0.5875)/0.06606 = −1.325
Answer
z = −1.325
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = 0.9178
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 = 15/5 = 3
ȳ = 268/5 = 53.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 46
Syy = Σ(y − ȳ)² = 251.2
Correlation coefficient
r = 46/√(10·251.2) = 0.9178
Answer
r = 0.9178
20.Linear regression & correlation
Answer: ŷ = 18.2
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 = 15/5 = 3
ȳ = 106/5 = 21.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −20
Regression line
b = −20/10 = −2
a = 21.2 − (−2)·3 = 27.2
ŷ = 27.2 − 2·x
Prediction
ŷ = 27.2 + (−2)·4.5 = 18.2
Answer
ŷ = 18.2
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