You need €15 000 in 2 years. An account pays 3% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €15 0003% / 12n = 2
PV = €
2.Time value of moneyHard
You deposit €2 000 at the start of every year for 7 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 7? Find FV.
A = €2 000i = 5%n = 7
FV = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 3: €8 000, €6 000, €6 000. The cost of capital is 10%. Find the net present value NPV.
A project costs €10 000 and returns €7 000, €7 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 7 000CF2 = 7 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 8 years and pays annual coupons. It sells for €1 179 when the yield to maturity is 2.5%. Find the coupon rate c in percent.
F = €1 000n = 8P = €1 179YTM = 2.5%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 3% matures in 5 years and sells for €895. Find its yield to maturity YTM in percent.
F = €1 000coupon = 3%n = 5P = €895
YTM = %
7.CAPM & WACCMedium
A share with a beta of 0.8 has a required return of 7.2%. The risk-free rate is 2%. What expected market return rm does CAPM imply?
β = 0.8re = 7.2%rf = 2%
rm = %
8.CAPM & WACCHard
A company has 2 000 000 shares trading at €37.50; the book value of its equity is €22 500 000. It just paid a dividend of €1.60 per share, and dividends grow by 4% per year. Its debt has a market value of €22 500 000 and a pre-tax cost of 4%. The tax rate is 21%. Find the WACC, using the dividend growth model for the cost of equity.
The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 9 of them exceeds 356.4 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 9> 356.4 g
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 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 11.6. What is the margin of error E₂ of a 90% interval from the same data?
99%n₁ = 50E₁ = 11.690%
E₂ =
16.Confidence intervalsHard
The diameters of 30 machined shafts have a sample standard deviation of 1.1 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.
n = 30s = 1.1 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 60 parts has mean length 75.5 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 80 parts has mean 80.3 mm, known standard deviation 5 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A sample of 36 measurements has mean 114.8. The population standard deviation is 12. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 114.8σ = 12n = 3690%
L =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (7, 20) is made. Find its residual e = y − ŷ.
n = 5(7, 20)
e =
3Solutions
1.Time value of money
Answer: PV = 14128 €
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 = 0.03/12 = 0.0025
N = 12·2 = 24
Present value
PV = FV/(1 + i)N = 15 000/1.062 = €14 127.53
Answer
PV = €14 127.53
2.Time value of money
Answer: FV = 17098 €
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 + 0.05)7 − 1)/0.05 = 8.142
Annuity due
Each deposit is invested one year longer.
FV = 2 000·8.142·1.05 = €17 098.22
Answer
FV = €17 098.22
3.NPV & IRR
Answer: NPV = −3261 €
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 + 0.1)1 = 8 000/1.1 = 7 272.73
CF2/(1 + 0.1)2 = 6 000/1.21 = 4 958.68
CF3/(1 + 0.1)3 = 6 000/1.331 = 4 507.89
Add them up
NPV = €−3 260.71
Answer
NPV = €−3 260.71
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 28.41 %
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(25%) = 400.00
NPV(30%) = −177.51
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 25 + 5·400.00/(400.00 − (−177.51)) = 28.46%
exact: NPV(28.41%) = 0 → IRR = 28.41%
Answer
IRR = 28.41%
5.Bond pricing & yield
Answer: c = 4.996 %
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 + 0.025)−8 = €820.75
Coupon
C = (1179 − 820.75)/7.17 = €49.96
Coupon rate
c = 49.96/1000 = 4.996%
Answer
c = 4.996%
6.Bond pricing & yield
Answer: YTM = 5.456 %
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
€895 = €30·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(5%) = €913.41, P(6%) = €873.63
YTM = 5.456%
Answer
YTM = 5.456%
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 = (7.2% − 2%)/0.8 = 6.5%
Market return
rm = 2% + 6.5% = 8.5%
Answer
rm = 8.5%
8.CAPM & WACC
Answer: WACC = 7.219 %
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 = (165 − 170)/5 = −1
Answer
z = −1
11.Probability
Answer: P = 0.4242
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 = 8/12 · 7/11 = 56/132 = 0.4242
Answer
P = 0.4242
12.Probability
Answer: σY = 1.195
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 = 2.7, Var(X) = 5.71, σX = 2.39
Linear change
σY = |0.5|·2.39 = 1.195
Answer
σY = 1.195
The mean does move: E(Y) = 0.5·2.7 + 5 = 6.35.
13.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 total
μT = 360 g, σT = 1.2·√9 = 3.6 g
Standardise
z = (356.4 − 360)/3.6 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
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: E₂ = 7.408
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.
Ratio of margins
E₂/E₁ = (1.645/2.576)·√(50/50) = 0.6386
New margin
E₂ = 11.6·0.6386 = 7.408
Answer
E₂ = 7.408
16.Confidence intervals
Answer: σ = 0.8761 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 29·1.1² = 35.09
Limits for σ²
35.09/45.722 = 0.7675
35.09/16.047 = 2.187
Limit for σ
σ = √0.7675 = 0.8761 µm
Answer
σ = 0.8761 µm
95% interval for σ: [0.8761, 1.479] µm.
17.Hypothesis tests & confidence intervals
Answer: z = −2.914
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 = √(12²/60 + 5²/80) = √2.712 = 1.647
Test statistic
z = (75.5 − 80.3)/1.647 = −2.914
Answer
z = −2.914
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: L = 111.5
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 = 1.645·12/√36 = 3.29
Interval limit
L = 114.8 − 3.29 = 111.5
Answer
L = 111.5
19.Linear regression & correlation
Answer: r² = 0.9618
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 = 88/6 = 14.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 148
Syy = Σ(y − ȳ)² = 325.3
Coefficient of determination
r² = 148²/(70·325.3) = 0.9618
Answer
r² = 0.9618
About 96.18% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: e = 4.25
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 = 68/5 = 13.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 86
Regression line
b = 86/40 = 2.15
a = 13.6 − 2.15·6 = 0.7
ŷ = 0.7 + 2.15·x
Predicted value
ŷ = 0.7 + 2.15·7 = 15.75
Residual
e = 20 − 15.75 = 4.25
Answer
e = 4.25
The point lies above 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.