A credit card charges a nominal interest rate of 24% per year, compounded semi-annually. Find the effective annual rate EAR in percent.
r = 24%m = 2
EAR = %
2.Time value of moneyHard
A loan of €15 000 is repaid with equal monthly payments over 7 years at 7.2% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €15 0007.2% / 12N = 84
I = €
3.NPV & IRRMedium
An investment of €30 000 gives these cash flows at the end of years 1 to 4: €12 000, €9 000, €10 000, €12 000. The cost of capital is 15%. Find the net present value NPV.
An investment of €10 000 returns €4 000 at the end of each year for 3 years. Find its internal rate of return IRR in percent.
C0 = €10 000CF = €4 000n = 3
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 7 years and pays annual coupons. It sells for €648 when the yield to maturity is 9%. Find the coupon rate c in percent.
F = €1 000n = 7P = €648YTM = 9%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 8% matures in 3 years and sells for €975. Find its yield to maturity YTM in percent.
F = €1 000coupon = 8%n = 3P = €975
YTM = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the WACC.
A company has 2 500 000 shares at €45.00 each and 10 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3.5% and the tax rate is 20%. The share’s beta is 0.8, the risk-free rate 3% and the market return 7%. Find the WACC.
β = 0.8rf = 3%rm = 7%rd = 3.5%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 23, 22, 18, 46, 53, 28, 57.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is below 82.5?
μ = 60σ = 15x = 82.5
P =
11.ProbabilityMedium
In how many ways N can a committee of 2 people be chosen from a group of 6? The order does not matter.
n = 6k = 2
N =
12.ProbabilityHard
X has the distribution shown, and Y = 2X + 4. Find the standard deviation σY of Y.
Drag z so that the shaded area between −z and z under the standard normal curve is 0.4381. Find z to two decimals.
A = 0.4381
z =
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 bottom 25% is to the left of it. Find the cut-off x.
μ = 178 cmσ = 7 cm25%
x = cm
15.Confidence intervalsMedium
A sample of 13 items has mean 27 and standard deviation s = 16. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 13x̄ = 27s = 1695%t(12) = 2.179
lower limit =
16.Confidence intervalsHard
The population standard deviation is 39. 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 0.5?
σ = 39E = 0.599%
n =
17.Hypothesis tests & confidence intervalsMedium
The mean of a population is claimed to be 100; the population standard deviation is known to be 10. A random sample of 100 observations has a mean of 97.82. Test H₀: μ = 100 against H₁: μ < 100. Find the p-value p.
μ₀ = 100σ = 10n = 100x̄ = 97.82H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 57. The population standard deviation is 20. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 57σ = 20n = 6499%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.
n = 5x = 3.5
ŷ =
3Solutions
1.Time value of money
Answer: EAR = 25.44 %
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 per period
r/m = 0.24/2 = 0.12
Compound over one year
EAR = (1 + 0.12)2 − 1 = 1.254 − 1 = 25.44%
Answer
EAR = 25.44%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: I = 4140 €
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 = 0.006, N = 84
A = 15 000·0.006/(1 − (1 + 0.006)−84) = €227.86
Total interest
I = N·A − P = 84·227.9 − 15 000 = €4 140.20
Answer
I = €4 140.20
3.NPV & IRR
Answer: NPV = 676.3 €
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 = −30 000
CF1/(1 + 0.15)1 = 12 000/1.15 = 10 434.78
CF2/(1 + 0.15)2 = 9 000/1.322 = 6 805.29
CF3/(1 + 0.15)3 = 10 000/1.521 = 6 575.16
CF4/(1 + 0.15)4 = 12 000/1.749 = 6 861.04
Add them up
NPV = €676.28
Answer
NPV = €676.28
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 9.701 %
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.
Required annuity factor
(1 − (1 + i)−3)/i = 10 000/4 000 = 2.5
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 9%: 2.531
i = 10%: 2.487
IRR = 9.701%
Answer
IRR = 9.701%
5.Bond pricing & yield
Answer: c = 2.006 %
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 + 0.09)−7 = €547.03
Coupon
C = (648 − 547.03)/5.033 = €20.06
Coupon rate
c = 20.06/1000 = 2.006%
Answer
c = 2.006%
6.Bond pricing & yield
Answer: YTM = 8.987 %
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
€975 = €80·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³
Solve numerically
P(8%) = €1 000.00, P(9%) = €974.69
YTM = 8.987%
Answer
YTM = 8.987%
7.CAPM & WACC
Answer: WACC = 5.8 %
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 = €10 000 000, E/V = 0.4, D/V = 0.6
WACC
WACC = 0.4·10% + 0.6·4%·(1 − 0.25) = 5.8%
Answer
WACC = 5.8%
8.CAPM & WACC
Answer: WACC = 5.91 %
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 = 1.5
Standard normal table
Φ(1.5) = 0.9332
P = 0.9332
Answer
P = 0.9332
11.Probability
Answer: N = 15
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.
Combinations
C(6, 2) = (6·5)/(2·1) = 15
Answer
N = 15
12.Probability
Answer: σY = 2.304
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 = 3.35, Var(X) = 1.328, σX = 1.152
Linear change
σY = |2|·1.152 = 2.304
Answer
σY = 2.304
The mean does move: E(Y) = 2·3.35 + 4 = 10.7.
13.Normal distribution
Answer: z = 0.58
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 + 0.4381)/2 = 0.7190
Read the z-table backwards
Find 0.7190 inside the table and read off z.
z = 0.58
Answer
z = 0.58
14.Normal distribution
Answer: x = 173.3 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) = 0.25
z from the table
z ≈ −0.67
Back to the original units
x = 178 − 0.67·7 = 173.3 cm
Answer
x = 173.3 cm
15.Confidence intervals
Answer: lower limit = 17.33
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 = 12, t = 2.179
Margin of error
E = 2.179·16/√13 = 9.67
Interval
27 ± 9.67 → [17.33, 36.67]
Answer
lower limit = 17.33
16.Confidence intervals
Answer: n = 40373
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 = (2.576·39/0.5)² = 40372
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 40373
Answer
n = 40373
17.Hypothesis tests & confidence intervals
Answer: p = 0.01463
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 = 10/√100 = 1
z = (97.82 − 100)/1 = −2.18
Tail area
Φ(2.18) = 0.9854
1 − Φ(2.18) = 0.0146
p-value
p = 0.01463
Answer
p = 0.01463
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: L = 50.56
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 = 2.576·20/√64 = 6.44
Interval limit
L = 57 − 6.44 = 50.56
Answer
L = 50.56
19.Linear regression & correlation
Answer: r = 0.9829
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 = 56/5 = 11.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 70
Syy = Σ(y − ȳ)² = 126.8
Correlation coefficient
r = 70/√(40·126.8) = 0.9829
Answer
r = 0.9829
20.Linear regression & correlation
Answer: ŷ = 46.25
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 = 25/5 = 5
ȳ = 274/5 = 54.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 57
Regression line
b = 57/10 = 5.7
a = 54.8 − 5.7·5 = 26.3
ŷ = 26.3 + 5.7·x
Prediction
ŷ = 26.3 + 5.7·3.5 = 46.25
Answer
ŷ = 46.25
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.