You invest €10 000 at 8% per year with continuous compounding. Find the value FV after 9 years.
PV = €10 000r = 8%t = 9
FV = €
2.Time value of moneyHard
You deposit €6 000 at the start of every year for 24 years, the first deposit today. The account pays 8% per year. How much is in the account at the end of year 24? Find FV.
A = €6 000i = 8%n = 24
FV = €
3.NPV & IRRMedium
A machine costs €40 000 today. It generates €8 000 at the end of each year for 5 years and can then be sold for €6 000. The cost of capital is 6%. Find the net present value NPV.
C0 = €40 000CF = €8 000S = €6 000n = 5r = 6%
NPV = €
4.NPV & IRRHard
A project costs €10 000 and returns €6 500, €6 500, €5 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 6 500CF2 = 6 500CF3 = 5 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 11 years and pays annual coupons. It sells for €1 278 when the yield to maturity is 3%. Find the coupon rate c in percent.
F = €1 000n = 11P = €1 278YTM = 3%
c = %
6.Bond pricing & yieldHard
A bond with 14 years to maturity (face value €1 000, annual coupon 5%) sells for €1 106. The issuer can call it in 3 years at a call price of €1 050. Find the yield to call YTC in percent.
F = €1 000coupon = 5%P = €1 106call price = €1 050call in 3 years
YTC = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €6 000 000, preferred shares worth €2 000 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 8% and the pre-tax cost of debt 5%. The tax rate is 30%. Find the WACC.
A company has 2 000 000 shares trading at €49.50; the book value of its equity is €39 600 000. It just paid a dividend of €2.30 per share, and dividends grow by 2% per year. Its debt has a market value of €59 400 000 and a pre-tax cost of 4.5%. The tax rate is 21%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 35, 33, 20, 45, 27, 49.
n = 6
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 60.
μ = 100σ = 20x = 60
z =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.3, 0.1, 0.05, 0.25, 0.3
σ =
12.ProbabilityHard
X has the distribution shown, and Y = 0.5X + 10. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 0.5X + 10
σY =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.8064. Find z to two decimals.
A = 0.8064
z =
14.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
A sample of 20 items has mean 54 and standard deviation s = 2.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 20x̄ = 54s = 2.590%t(19) = 1.729
lower limit =
16.Confidence intervalsHard
The population standard deviation is 23. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 4?
σ = 23E = 495%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 80; the population standard deviation is known to be 2.5. A random sample of 25 observations has a mean of 78.81. Test H₀: μ = 80 against H₁: μ ≠ 80. Find the p-value p.
μ₀ = 80σ = 2.5n = 25x̄ = 78.81H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 498.2. The population standard deviation is 15. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 498.2σ = 15n = 2590%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (5.5, 15) is made. Find its residual e = y − ŷ.
n = 6(5.5, 15)
e =
3Solutions
1.Time value of money
Answer: FV = 20544 €
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 = 0.08·9 = 0.72
e0.72 = 2.054
Future value
FV = 10 000·2.054 = €20 544.33
Answer
FV = €20 544.33
2.Time value of money
Answer: FV = 432636 €
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.
Ordinary annuity factor
((1 + 0.08)24 − 1)/0.08 = 66.76
Annuity due
Each deposit is invested one year longer.
FV = 6 000·66.76·1.08 = €432 635.64
Answer
FV = €432 635.64
3.NPV & IRR
Answer: NPV = −1818 €
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 the yearly cash flows
8 000·4.212 = 33 698.91
Present value of the salvage value
6 000/1.338 = 4 483.55
NPV
NPV = −40 000 + 33 698.91 + 4 483.55 = −€1 817.54
Answer
NPV = −€1 817.54
4.NPV & IRR
Answer: IRR = 38.21 %
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(35%) = 413.55
NPV(40%) = −218.66
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 35 + 5·413.55/(413.55 − (−218.66)) = 38.27%
exact: NPV(38.21%) = 0 → IRR = 38.21%
Answer
IRR = 38.21%
5.Bond pricing & yield
Answer: c = 6.005 %
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 + 0.03)−11 = €722.42
Coupon
C = (1278 − 722.42)/9.253 = €60.05
Coupon rate
c = 60.05/1000 = 6.005%
Answer
c = 6.005%
6.Bond pricing & yield
Answer: YTC = 2.881 %
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
1106 = 50·(1 − (1 + y)−3)/y + 1050/(1 + y)3
Solve numerically
y = 2%: €1 133.63
y = 3%: €1 102.33
YTC = 2.881%
Answer
YTC = 2.881%
The yield to maturity is 3.997%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.
7.CAPM & WACC
Answer: WACC = 8.577 %
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 = (60 − 100)/20 = −2
Answer
z = −2
11.Probability
Answer: σ = 1.652
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.
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 = 1, Var(X) = 0.7, σX = 0.8367
Linear change
σY = |0.5|·0.8367 = 0.4183
Answer
σY = 0.4183
The mean does move: E(Y) = 0.5·1 + 10 = 10.5.
13.Normal distribution
Answer: z = 1.3
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 of z
Φ(z) = (1 + 0.8064)/2 = 0.9032
Read the z-table backwards
Find 0.9032 inside the table and read off z.
z = 1.30
Answer
z = 1.30
14.Normal distribution
Answer: P = 0.9868
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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm
Standardise
z = (0 − 0.04)/0.01803 = −2.22
Probability
P = 1 − Φ(−2.22) = 0.9868
Answer
P = 0.9868
15.Confidence intervals
Answer: lower limit = 53.03
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 = 19, t = 1.729
Margin of error
E = 1.729·2.5/√20 = 0.9665
Interval
54 ± 0.9665 → [53.03, 54.97]
Answer
lower limit = 53.03
16.Confidence intervals
Answer: n = 128
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 = (1.96·23/4)² = 127
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 128
Answer
n = 128
17.Hypothesis tests & confidence intervals
Answer: p = 0.01731
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.
Test statistic
SE = σ/√n = 2.5/√25 = 0.5
z = (78.81 − 80)/0.5 = −2.38
Tail area
Φ(2.38) = 0.9913
1 − Φ(2.38) = 0.0087
p-value
p = 2·0.0087 = 0.01731
Answer
p = 0.01731
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: L = 493.3
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 = 1.645·15/√25 = 4.935
Interval limit
L = 498.2 − 4.935 = 493.3
Answer
L = 493.3
19.Linear regression & correlation
Answer: r² = 0.595
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 = 279/5 = 55.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 24
Syy = Σ(y − ȳ)² = 96.8
Coefficient of determination
r² = 24²/(10·96.8) = 0.595
Answer
r² = 0.595
About 59.5% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = −2.681
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 = 21/6 = 3.5
ȳ = 127/6 = 21.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −30.5
Regression line
b = −30.5/17.5 = −1.743
a = 21.17 − (−1.743)·3.5 = 27.27
ŷ = 27.27 − 1.743·x
Predicted value
ŷ = 27.27 + (−1.743)·5.5 = 17.68
Residual
e = 15 − 17.68 = −2.681
Answer
e = −2.681
The point lies below the line.
This is independent practice material. ExamProblems is not affiliated with or endorsed by the organisation that runs this exam, and the exam name is used only to describe the content. CFA® and Chartered Financial Analyst® are trademarks of CFA Institute, which is not affiliated with ExamProblems and does not endorse it or vouch for its accuracy.