You invest €2 000 at a nominal rate of 4.8% per year, compounded monthly. Find the value FV of the investment after 9 years.
PV = €2 0004.8% / 12n = 9
FV = €
2.Time value of moneyHard
You want €75 000 in 9 years. You save the same amount at the end of every month in an account paying 7.2% per year, compounded monthly. Find the monthly deposit A.
FV = €75 0007.2% / 12N = 108
A = €
3.NPV & IRRMedium
A project costs €60 000 today and returns €16 500 at the end of each year for 6 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €40 000 and returns €16 000, €21 000, €16 500, €14 000 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 8% per year. Find the price P of the bond.
F = €1 000coupon = 5%YTM = 8%n = 7
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 6 years. The yield to maturity is 3%. Find the Macaulay duration D in years.
F = €1 000coupon = 4%n = 6YTM = 3%
D = years
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €5 000 000. Its cost of equity is 14%, its pre-tax cost of debt is 3% and the tax rate is 25%. Find the WACC.
A company has 2 000 000 shares at €32.00 each and 10 000 bonds trading at €1 000 each. The pre-tax cost of debt is 4% and the tax rate is 21%. The share’s beta is 1.3, the risk-free rate 1.5% and the market return 5.5%. Find the WACC.
β = 1.3rf = 1.5%rm = 5.5%rd = 4%T = 21%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 56, 22, 54, 36, 56.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 80.
μ = 70σ = 5x = 80
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.15, 0.3, 0.3, 0.15, 0.1
Var(X) =
12.ProbabilityHard
7 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?
n = 7p = 0.6
P =
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.6579. Find z to two decimals.
A = 0.6579
z =
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 30 items has mean 66.5 and standard deviation s = 11.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.
n = 30x̄ = 66.5s = 11.599%t(29) = 2.756
lower limit =
16.Confidence intervalsHard
The population standard deviation is 36. 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 2?
σ = 36E = 299%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 122.9 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 120.
μ₀ = 120x̄ = 122.9s = 5n = 25
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 10. Find the smallest sample size n.
σ = 10E = 1.595%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 5Σx = 20Σy = 148Σx² = 90Σy² = 4488Σxy = 619
r =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.
n = 6x = 3.5
ŷ =
3Solutions
1.Time value of money
Answer: FV = 3078 €
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.
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.
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 16 500/1.12 = 14 732.14 → Σ = 14 732.14
year 2: 16 500/1.254 = 13 153.70 → Σ = 27 885.84
year 3: 16 500/1.405 = 11 744.37 → Σ = 39 630.22
year 4: 16 500/1.574 = 10 486.05 → Σ = 50 116.26
year 5: 16 500/1.762 = 9 362.54 → Σ = 59 478.81
year 6: 16 500/1.974 = 8 359.41 → Σ = 67 838.22
Fraction of the last year
DPB = 5 + 521.19/8 359.41 = 5.062
Answer
DPB = 5.062 years
The discounted payback is always longer than the simple payback (3.636 years).
4.NPV & IRR
Answer: MIRR = 17.59 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
16 000·1.083 = 20 155.39
21 000·1.082 = 24 494.40
16 500·1.081 = 17 820.00
14 000·1.080 = 14 000.00
TV = 76 469.79
MIRR
MIRR = (76 469.79/40 000)1/4 − 1 = 17.59%
Answer
MIRR = 17.59%
5.Bond pricing & yield
Answer: P = 841.6 €
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 = €25.00, i = 0.04, N = 14
Present value of the coupons
€25.00·(1 − 0.5775)/0.04 = €264.08
Present value of the face value
€1 000·0.5775 = €577.48
Price
P = €841.55
Answer
P = €841.55
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: D = 5.468 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: 40/1.03 = 38.83, t·PV = 38.83
t = 2: 40/1.061 = 37.70, t·PV = 75.41
t = 3: 40/1.093 = 36.61, t·PV = 109.82
t = 4: 40/1.126 = 35.54, t·PV = 142.16
t = 5: 40/1.159 = 34.50, t·PV = 172.52
t = 6: 1040/1.194 = 870.98, t·PV = 5 225.90
Duration
P = 1 054.17
D = 5 764.64/1 054.17 = 5.468
Answer
D = 5.468 years
7.CAPM & WACC
Answer: WACC = 6.656 %
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 = €8 000 000, E/V = 0.375, D/V = 0.625
WACC
WACC = 0.375·14% + 0.625·3%·(1 − 0.25) = 6.656%
Answer
WACC = 6.656%
8.CAPM & WACC
Answer: WACC = 6.222 %
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 = (80 − 70)/5 = 2
Answer
z = 2
11.Probability
Answer: Var(X) = 1.387
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.
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) = 0.47 = 0.001638
P(X = 1) = 7·0.6·0.46 = 0.0172
Add them
P = 0.001638 + 0.0172 = 0.01884
Answer
P = 0.01884
13.Normal distribution
Answer: z = 0.95
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 + 0.6579)/2 = 0.8289
Read the z-table backwards
Find 0.8289 inside the table and read off z.
z = 0.95
Answer
z = 0.95
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: lower limit = 60.71
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 = 29, t = 2.756
Margin of error
E = 2.756·11.5/√30 = 5.787
Interval
66.5 ± 5.787 → [60.71, 72.29]
Answer
lower limit = 60.71
16.Confidence intervals
Answer: n = 2150
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 = (2.576·36/2)² = 2150
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 2150
Answer
n = 2150
17.Hypothesis tests & confidence intervals
Answer: t = 2.9
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
SE = 5/√25 = 5/5 = 1
Test statistic
t = (122.9 − 120)/1 = 2.9
Answer
t = 2.9
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 171
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).
Solve for n
n = (1.96·10/1.5)² = 170.7
Round up
n = 171
Answer
n = 171
19.Linear regression & correlation
Answer: r = 0.8246
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.
Centred sums
Sxy = 619 − 20·148/5 = 27
Sxx = 90 − 20²/5 = 10
Syy = 4488 − 148²/5 = 107.2
Correlation coefficient
r = 27/√(10·107.2) = 0.8246
Answer
r = 0.8246
20.Linear regression & correlation
Answer: ŷ = 58.79
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 = 33/6 = 5.5
ȳ = 398/6 = 66.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 66
Regression line
b = 66/17.5 = 3.771
a = 66.33 − 3.771·5.5 = 45.59
ŷ = 45.59 + 3.771·x
Prediction
ŷ = 45.59 + 3.771·3.5 = 58.79
Answer
ŷ = 58.79
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.