You invest €1 000 at 5% per year with continuous compounding. Find the value FV after 4 years.
PV = €1 000r = 5%t = 4
FV = €
2.Time value of moneyHard
A loan of €20 000 is repaid with equal monthly payments over 4 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.
P = €20 0004.8% / 12N = 48
A = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 3: €21 000, €9 000, €11 000. The cost of capital is 6%. Find the net present value NPV.
Project A costs €40 000 and returns €22 500, €16 500, €9 000 at the end of years 1 to 3. Project B costs €45 000 and returns €14 000, €18 000, €23 500. Find the crossover rate r* at which both projects have the same NPV, in percent.
A bond with a face value of €1 000 matures in 15 years and pays annual coupons. It sells for €436 when the yield to maturity is 9%. Find the coupon rate c in percent.
F = €1 000n = 15P = €436YTM = 9%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 8% and matures in 5 years. The yield to maturity is 5%. Find the Macaulay duration D in years.
F = €1 000coupon = 8%n = 5YTM = 5%
D = years
7.CAPM & WACCMedium
A company has equity worth €10 000 000 and debt worth €1 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 3.5% and the tax rate is 25.8%. Find the WACC.
A company has 4 000 000 shares at €15.00 each and 40 000 bonds trading at €1 050 each. The pre-tax cost of debt is 6% and the tax rate is 25.8%. The share’s beta is 1.8, the risk-free rate 2.5% and the market return 8.5%. Find the WACC.
β = 1.8rf = 2.5%rm = 8.5%rd = 6%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 35, 34, 40, 41, 46, 34, 26.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 15. What proportion P of values is above 109?
μ = 100σ = 15x = 109
P =
11.ProbabilityMedium
A bag holds 4 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
4 / 7
P =
12.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 3 correct answers?
n = 9p = 0.7k = 3
P =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.3594. Find z to two decimals.
A = 0.3594
z =
14.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
A sample of 15 items has mean 113.5 and standard deviation s = 16.5. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.
n = 15x̄ = 113.5s = 16.595%t(14) = 2.145
E =
16.Confidence intervalsHard
The population standard deviation is 30. 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 2.5?
σ = 30E = 2.595%
n =
17.Hypothesis tests & confidence intervalsMedium
11 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 1.93 and the standard deviation of the differences is 3.2. Find the paired test statistic t for H₀: μd = 0.
n = 11d̄ = 1.93sd = 3.2
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 36 measurements has mean 199.4. The population standard deviation is 12. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 199.4σ = 12n = 3695%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. 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 = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.
n = 5x = 2.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 1221 €
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 = 0.05·4 = 0.2
e0.2 = 1.221
Future value
FV = 1 000·1.221 = €1 221.40
Answer
FV = €1 221.40
2.Time value of money
Answer: A = 458.8 €
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 rate and number of payments
i = 0.048/12 = 0.004
N = 12·4 = 48
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.8256
A = 20 000·0.004/(1 − 0.8256) = €458.78
Answer
A = €458.78
3.NPV & IRR
Answer: NPV = −7943 €
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 = −45 000
CF1/(1 + 0.06)1 = 21 000/1.06 = 19 811.32
CF2/(1 + 0.06)2 = 9 000/1.124 = 8 009.97
CF3/(1 + 0.06)3 = 11 000/1.191 = 9 235.81
Add them up
NPV = €−7 942.90
Answer
NPV = €−7 942.90
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: r* = 7.729 %
Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.
Difference in cash flows
ΔCF0 = −45 000 − (−40 000) = −5 000
ΔCF1 = 14 000 − 22 500 = −8 500
ΔCF2 = 18 000 − 16 500 = 1 500
ΔCF3 = 23 500 − 9 000 = 14 500
IRR of the difference
NPVΔ(7%) = 202.55
NPVΔ(8%) = −73.79
r* = 7.729%
Answer
r* = 7.729%
Check: at 7.729% both NPVs are €2 301.71.
5.Bond pricing & yield
Answer: c = 2.003 %
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)−15 = €274.54
Coupon
C = (436 − 274.54)/8.061 = €20.03
Coupon rate
c = 20.03/1000 = 2.003%
Answer
c = 2.003%
6.Bond pricing & yield
Answer: D = 4.357 years
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.
Present value of each cash flow
t = 1: 80/1.05 = 76.19, t·PV = 76.19
t = 2: 80/1.103 = 72.56, t·PV = 145.12
t = 3: 80/1.158 = 69.11, t·PV = 207.32
t = 4: 80/1.216 = 65.82, t·PV = 263.26
t = 5: 1080/1.276 = 846.21, t·PV = 4 231.04
Duration
P = 1 129.88
D = 4 922.94/1 129.88 = 4.357
Answer
D = 4.357 years
7.CAPM & WACC
Answer: WACC = 12.05 %
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 = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 1 − 0.7257 = 0.2743
Answer
P = 0.2743
11.Probability
Answer: P = 0.1091
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 = 4/11 · 3/10 = 12/110 = 0.1091
Answer
P = 0.1091
12.Probability
Answer: P = 0.021
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.
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 − 0.3594 = 0.6406
Read the z-table backwards
Find 0.6406 inside the table and read off z.
z = 0.36
Answer
z = 0.36
14.Normal distribution
Answer: P = 0.7881
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 = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
15.Confidence intervals
Answer: E = 9.138
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 = 14, t = 2.145
Margin of error
E = 2.145·16.5/√15 = 9.138
Answer
E = 9.138
16.Confidence intervals
Answer: n = 554
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 = (1.96·30/2.5)² = 553.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 554
Answer
n = 554
17.Hypothesis tests & confidence intervals
Answer: t = 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 of the mean difference
SE = 3.2/√11 = 0.9648
Test statistic
t = 1.93/0.9648 = 2
Answer
t = 2
With df = 10, the two-sided 5% critical value is 2.228; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: L = 195.5
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 = 1.96·12/√36 = 3.92
Interval limit
L = 199.4 − 3.92 = 195.5
Answer
L = 195.5
19.Linear regression & correlation
Answer: r² = 0.9595
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 = 100/5 = 20
ȳ = 393/5 = 78.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 312
Syy = Σ(y − ȳ)² = 1127
Coefficient of determination
r² = 312²/(90·1127) = 0.9595
Answer
r² = 0.9595
About 95.95% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 43.05
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 = 223/5 = 44.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 31
Regression line
b = 31/10 = 3.1
a = 44.6 − 3.1·3 = 35.3
ŷ = 35.3 + 3.1·x
Prediction
ŷ = 35.3 + 3.1·2.5 = 43.05
Answer
ŷ = 43.05
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.