You need €15 000 in 3 years. An account pays 5% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €15 0005% / 12n = 3
PV = €
2.Time value of moneyHard
A loan of €20 000 is repaid with equal monthly payments over 5 years at 3% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €20 0003% / 12N = 60
I = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 4: €10 000, €6 000, €10 000, €8 000. The cost of capital is 7.000000000000001%. Find the net present value NPV.
A project costs €20 000 and returns €12 000, €9 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 12 000CF2 = 9 500
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 8 years to maturity (face value €1 000, annual coupon 4%) when its yield to maturity is 5.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 6%. Find your holding period return HPR in percent.
F = €1 000coupon = 4%n = 8y0 = 5.5%y1 = 6%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 6 years. It sells for €1 076. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.
F = €1 000coupon = 7% (2×)n = 6P = €1 076
YTM = %
7.CAPM & WACCMedium
A company has equity worth €6 000 000 and debt worth €5 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the WACC.
A company has equity with a market value of €12 000 000 and a cost of equity of 13%. Its debt consists of 5 000 bonds (face value €1 000, annual coupon 4%, 10 years to maturity) that trade at €887 each. The tax rate is 28%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.
Find the sample standard deviation s of the data: 56, 16, 40, 45, 11.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 8. Find the z-score z of a score of 110.
μ = 100σ = 8x = 110
z =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 1, 2p(x): 0.35, 0.35, 0.3
σ =
12.ProbabilityHard
10 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?
n = 10p = 0.3
P =
13.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 482 and 532 h? Give P as a decimal.
μ = 500σ = 40x1 = 482x2 = 532
P =
14.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
A sample of 7 items has mean 111.5 and standard deviation s = 16.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.
n = 7x̄ = 111.5s = 16.595%t(6) = 2.447
upper limit =
16.Confidence intervalsHard
The population standard deviation is 40. 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?
σ = 40E = 199%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 100 parts has mean length 128 mm, and the line's known standard deviation is 11 mm. Line 2: a sample of 60 parts has mean 129 mm, known standard deviation 10 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
In a trial, 40 of 100 patients on treatment A recover, and 88 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 40n₁ = 100x₂ = 88n₂ = 250
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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: PV = 12915 €
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 = 0.05/12 = 0.004167
N = 12·3 = 36
Present value
PV = FV/(1 + i)N = 15 000/1.161 = €12 914.64
Answer
PV = €12 914.64
2.Time value of money
Answer: I = 1562 €
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.
Monthly payment
i = 0.0025, N = 60
A = 20 000·0.0025/(1 − (1 + 0.0025)−60) = €359.37
Total interest
I = N·A − P = 60·359.4 − 20 000 = €1 562.43
Answer
I = €1 562.43
3.NPV & IRR
Answer: NPV = 8853 €
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 = −20 000
CF1/(1 + 0.07)1 = 10 000/1.07 = 9 345.79
CF2/(1 + 0.07)2 = 6 000/1.145 = 5 240.63
CF3/(1 + 0.07)3 = 10 000/1.225 = 8 162.98
CF4/(1 + 0.07)4 = 8 000/1.311 = 6 103.16
Add them up
NPV = €8 852.57
Answer
NPV = €8 852.57
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 5.166 %
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(5%) = 45.35
NPV(10%) = −1 239.67
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 5 + 5·45.35/(45.35 − (−1 239.67)) = 5.176%
exact: NPV(5.166%) = 0 → IRR = 5.166%
Answer
IRR = 5.166%
5.Bond pricing & yield
Answer: HPR = 2.582 %
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.
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 = €35.00, N = 12
Solve numerically
i = 2.748% per half-year
YTM
YTM = 2·2.748% = 5.496%
Answer
YTM = 5.496%
7.CAPM & WACC
Answer: WACC = 8.455 %
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 = €11 000 000, E/V = 0.5455, D/V = 0.4545
WACC
WACC = 0.5455·13% + 0.4545·4%·(1 − 0.25) = 8.455%
Answer
WACC = 8.455%
8.CAPM & WACC
Answer: WACC = 10.56 %
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 = (110 − 100)/8 = 1.25
Answer
z = 1.25
11.Probability
Answer: σ = 0.8047
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 = 0·0.35 + 1·0.35 + 2·0.3 = 0.95
Mean of the squares
E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55
Variance
Var(X) = 1.55 − 0.95² = 0.6475
σ = √0.6475 = 0.8047
Answer
σ = 0.8047
12.Probability
Answer: P = 0.1493
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) = 0.710 = 0.02825
P(X = 1) = 10·0.3·0.79 = 0.1211
Add them
P = 0.02825 + 0.1211 = 0.1493
Answer
P = 0.1493
13.Normal distribution
Answer: P = 0.4618
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 = (482 − 500)/40 = −0.45
z₂ = (532 − 500)/40 = 0.8
Area between
P = Φ(0.8) − Φ(−0.45) = 0.7881 − 0.3264 = 0.4618
Answer
P = 0.4618
14.Normal distribution
Answer: P = 0.7389
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 = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
15.Confidence intervals
Answer: upper limit = 126.8
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 = 6, t = 2.447
Margin of error
E = 2.447·16.5/√7 = 15.26
Interval
111.5 ± 15.26 → [96.24, 126.8]
Answer
upper limit = 126.8
16.Confidence intervals
Answer: n = 10618
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 = (2.576·40/1)² = 10617
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 10618
Answer
n = 10618
17.Hypothesis tests & confidence intervals
Answer: z = −0.5896
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.
Standard error of the difference
SE = √(11²/100 + 10²/60) = √2.877 = 1.696
Test statistic
z = (128 − 129)/1.696 = −0.5896
Answer
z = −0.5896
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = 0.8423
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 = 40/100 = 0.4
p̂₂ = 88/250 = 0.352
p̂ = 128/350 = 0.3657
Standard error under H₀
SE = √(0.3657·0.6343·(1/100 + 1/250)) = 0.05699
Test statistic
z = (0.4 − 0.352)/0.05699 = 0.8423
Answer
z = 0.8423
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r² = 0.9844
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 = 121/5 = 24.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 168
Syy = Σ(y − ȳ)² = 716.8
Coefficient of determination
r² = 168²/(40·716.8) = 0.9844
Answer
r² = 0.9844
About 98.44% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 22.7
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 = 124/5 = 24.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −14
Regression line
b = −14/10 = −1.4
a = 24.8 − (−1.4)·3 = 29
ŷ = 29 − 1.4·x
Prediction
ŷ = 29 + (−1.4)·4.5 = 22.7
Answer
ŷ = 22.7
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