You invest €5 000 at a nominal rate of 3% per year, compounded monthly. Find the value FV of the investment after 8 years.
PV = €5 0003% / 12n = 8
FV = €
2.Time value of moneyHard
You save €500 at the end of every month for 27 years. The account pays 6% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €5006% / 12N = 324
FV = €
3.NPV & IRRMedium
A machine costs €50 000 today. It generates €11 000 at the end of each year for 5 years and can then be sold for €10 000. The cost of capital is 8%. Find the net present value NPV.
C0 = €50 000CF = €11 000S = €10 000n = 5r = 8%
NPV = €
4.NPV & IRRHard
A project costs €40 000 and returns €17 500, €20 000, €19 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −40 000CF1 = 17 500CF2 = 20 000CF3 = 19 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 13 years and pays annual coupons. It sells for €1 220 when the yield to maturity is 2.5%. Find the coupon rate c in percent.
F = €1 000n = 13P = €1 220YTM = 2.5%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 7 years and sells for €1 060. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 7P = €1 060
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €6 000 000, preferred shares worth €500 000 and debt worth €6 000 000. The cost of equity is 11%, the cost of preferred stock 6% and the pre-tax cost of debt 5%. The tax rate is 25%. Find the WACC.
A company has 4 000 000 shares trading at €24.00; the book value of its equity is €28 800 000. It just paid a dividend of €1.00 per share, and dividends grow by 5% per year. Its debt has a market value of €57 600 000 and a pre-tax cost of 3%. The tax rate is 28%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 19, 43, 49, 57, 43, 30, 24.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 100. Find the z-score z of a score of 450.
μ = 500σ = 100x = 450
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
12.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 2 correct answers?
n = 9p = 0.3k = 2
P =
13.Normal distributionMedium
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 16 of them is below 649.6 g. Round z to two decimals.
μ = 40 gσ = 1.2 gn = 16< 649.6 g
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.02 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 = 25.02 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
A sample of 6 items has mean 92.5 and standard deviation s = 1. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 6x̄ = 92.5s = 199%t(5) = 4.032
upper limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 7 plots, mean yield 94 kg, standard deviation 6.5 kg. Plot group 2: 7 plots, mean 93 kg, standard deviation 3.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 95% confidence interval for μ₁ − μ₂.
A sample of 81 students has a mean score of 497.2 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 497.2s = 8n = 81
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 8. Find the smallest sample size n.
σ = 8E = 290%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 5
r =
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, 20) is made. Find its residual e = y − ŷ.
n = 6(5.5, 20)
e =
3Solutions
1.Time value of money
Answer: FV = 6354 €
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.
You paid in €162 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 725.6 €
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
11 000·3.993 = 43 919.81
Present value of the salvage value
10 000/1.469 = 6 805.83
NPV
NPV = −50 000 + 43 919.81 + 6 805.83 = €725.64
Answer
NPV = €725.64
4.NPV & IRR
Answer: IRR = 19.16 %
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(15%) = 2 833.07
NPV(20%) = −532.41
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
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)−13 = €725.42
Coupon
C = (1220 − 725.42)/10.98 = €45.03
Coupon rate
c = 45.03/1000 = 4.503%
Answer
c = 4.503%
6.Bond pricing & yield
Answer: YTM = 3.036 %
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
€1 060 = €40·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷
Solve numerically
P(3%) = €1 062.30, P(4%) = €1 000.00
YTM = 3.036%
Answer
YTM = 3.036%
7.CAPM & WACC
Answer: WACC = 7.32 %
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 = (450 − 500)/100 = −0.5
Answer
z = −0.5
11.Probability
Answer: Var(X) = 0.7
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.
Mean
μ = 0·0.35 + 1·0.3 + 2·0.35 = 1
Mean of the squares
E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7
Variance
Var(X) = 1.7 − 1² = 0.7
Answer
Var(X) = 0.7
12.Probability
Answer: P = 0.2668
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.
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 = 640 g, σT = 1.2·√16 = 4.8 g
Standardise
z = (649.6 − 640)/4.8 = 2
Probability
P = Φ(2) = 0.9772
Answer
P = 0.9772
14.Normal distribution
Answer: P = 0.8665
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 = 25.02 − 25 = 0.02 mm, σ = 0.01803 mm
Standardise
z = (0 − 0.02)/0.01803 = −1.11
Probability
P = 1 − Φ(−1.11) = 0.8665
Answer
P = 0.8665
15.Confidence intervals
Answer: upper limit = 94.15
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 = 5, t = 4.032
Margin of error
E = 4.032·1/√6 = 1.646
Interval
92.5 ± 1.646 → [90.85, 94.15]
Answer
upper limit = 94.15
16.Confidence intervals
Answer: lower limit = −5.08 kg
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Pooled standard deviation
sp² = (6·6.5² + 6·3.5²)/12 = 27.25
sp = 5.22
Margin of error
E = 2.179·5.22·√(1/7 + 1/7) = 6.08
Interval
1 ± 6.08 → [−5.08, 7.08]
Answer
lower limit = −5.08 kg
17.Hypothesis tests & confidence intervals
Answer: t = −3.15
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 = 8/√81 = 8/9 = 0.8889
Test statistic
t = (497.2 − 500)/0.8889 = −3.15
Answer
t = −3.15
t0.025; df = 80 = 1.99
|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 44
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 = (1.645·8/2)² = 43.3
Round up
n = 44
Answer
n = 44
19.Linear regression & correlation
Answer: r = 0.9924
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 = 378/5 = 75.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 206
Syy = Σ(y − ȳ)² = 1077
Correlation coefficient
r = 206/√(40·1077) = 0.9924
Answer
r = 0.9924
20.Linear regression & correlation
Answer: e = 4.81
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 = 122/6 = 20.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −45
Regression line
b = −45/17.5 = −2.571
a = 20.33 − (−2.571)·3.5 = 29.33
ŷ = 29.33 − 2.571·x
Predicted value
ŷ = 29.33 + (−2.571)·5.5 = 15.19
Residual
e = 20 − 15.19 = 4.81
Answer
e = 4.81
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.