You invest €1 000 at a nominal rate of 2% per year, compounded monthly. Find the value FV of the investment after 6 years.
PV = €1 0002% / 12n = 6
FV = €
2.Time value of moneyHard
You save €100 at the end of every month for 24 years. The account pays 5.4% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €1005.4% / 12N = 288
FV = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 3: €6 000, €6 000, €6 000. The cost of capital is 6%. Find the net present value NPV.
An investment of €80 000 returns €12 000 at the end of each year for 9 years. Find its internal rate of return IRR in percent.
C0 = €80 000CF = €12 000n = 9
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 7 years to maturity (face value €1 000, annual coupon 4%) when its yield to maturity is 5.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 4.5%. Find your holding period return HPR in percent.
F = €1 000coupon = 4%n = 7y0 = 5.5%y1 = 4.5%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 3% matures in 6 years and sells for €925. Find its yield to maturity YTM in percent.
F = €1 000coupon = 3%n = 6P = €925
YTM = %
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 6% and the tax rate is 25.8%. Find the WACC.
A company has equity with a market value of €15 000 000 and a cost of equity of 10%. Its debt consists of 2 000 bonds (face value €1 000, annual coupon 5%, 12 years to maturity) that trade at €878 each. The tax rate is 20%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.
Find the sample standard deviation s of the data: 44, 44, 34, 14, 41, 20.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is below 36?
μ = 60σ = 20x = 36
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.05, 0.3
Var(X) =
12.ProbabilityHard
A game costs €2 to play. You win €4 with probability 0.2 and €20 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €2€4: p = 0.2€20: p = 0.05
E(G) = €
13.Normal distributionMedium
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 88 and 118.75 points? Give P as a decimal.
μ = 100σ = 15x1 = 88x2 = 118.75
P =
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 8 items has mean 99 and standard deviation s = 4.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 8x̄ = 99s = 4.590%t(7) = 1.895
lower limit =
16.Confidence intervalsHard
The diameters of 30 machined shafts have a sample standard deviation of 9.7 µm. Assuming the diameters are normally distributed, find the upper 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 = 9.7 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 143 mm, and the line's known standard deviation is 8 mm. Line 2: a sample of 30 parts has mean 139.7 mm, known standard deviation 3 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 12. Find the smallest sample size n.
σ = 12E = 190%
n =
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 6 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 = 6.
n = 6x = 6
ŷ =
3Solutions
1.Time value of money
Answer: FV = 1127 €
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 €28 800 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = −3962 €
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 + 0.06)1 = 6 000/1.06 = 5 660.38
CF2/(1 + 0.06)2 = 6 000/1.124 = 5 339.98
CF3/(1 + 0.06)3 = 6 000/1.191 = 5 037.72
Add them up
NPV = €−3 961.93
Answer
NPV = €−3 961.93
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 6.463 %
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)−9)/i = 80 000/12 000 = 6.667
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 6%: 6.802
i = 7%: 6.515
IRR = 6.463%
Answer
IRR = 6.463%
5.Bond pricing & yield
Answer: HPR = 10.87 %
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.
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
€925 = €30·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶
Solve numerically
P(4%) = €947.58, P(5%) = €898.49
YTM = 4.452%
Answer
YTM = 4.452%
7.CAPM & WACC
Answer: WACC = 7.622 %
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 = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 0.1151
Answer
P = 0.1151
11.Probability
Answer: Var(X) = 1.51
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 = 1·0.35 + 2·0.3 + 3·0.05 + 4·0.3 = 2.3
Mean of the squares
E(X²) = 1·0.35 + 4·0.3 + 9·0.05 + 16·0.3 = 6.8
Variance
Var(X) = 6.8 − 2.3² = 1.51
Answer
Var(X) = 1.51
12.Probability
Answer: E(G) = −0.2 €
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.
Net outcomes
G: 2 (p = 0.2), 18 (p = 0.05), −2 (p = 0.75)
Expected value
E(G) = 2·0.2 + 18·0.05 − 2·0.75 = −0.2 €
Answer
E(G) = −0.2 €
A negative value means you lose money on average.
13.Normal distribution
Answer: P = 0.6825
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (88 − 100)/15 = −0.8
z₂ = (118.75 − 100)/15 = 1.25
Area between
P = Φ(1.25) − Φ(−0.8) = 0.8944 − 0.2119 = 0.6825
Answer
P = 0.6825
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: lower limit = 95.99
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 = 7, t = 1.895
Margin of error
E = 1.895·4.5/√8 = 3.015
Interval
99 ± 3.015 → [95.99, 102]
Answer
lower limit = 95.99
16.Confidence intervals
Answer: σ = 13.04 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 = 29·9.7² = 2729
Limits for σ²
2729/45.722 = 59.68
2729/16.047 = 170
Limit for σ
σ = √170 = 13.04 µm
Answer
σ = 13.04 µm
95% interval for σ: [7.725, 13.04] µm.
17.Hypothesis tests & confidence intervals
Answer: z = 2.823
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 = √(8²/60 + 3²/30) = √1.367 = 1.169
Test statistic
z = (143 − 139.7)/1.169 = 2.823
Answer
z = 2.823
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 390
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 = (1.645·12/1)² = 389.7
Round up
n = 390
Answer
n = 390
19.Linear regression & correlation
Answer: r² = 0.9766
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 = 79/6 = 13.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 141
Syy = Σ(y − ȳ)² = 290.8
Coefficient of determination
r² = 141²/(70·290.8) = 0.9766
Answer
r² = 0.9766
About 97.66% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 52.83
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 = 36/6 = 6
ȳ = 317/6 = 52.83
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 301
Regression line
b = 301/70 = 4.3
a = 52.83 − 4.3·6 = 27.03
ŷ = 27.03 + 4.3·x
Prediction
ŷ = 27.03 + 4.3·6 = 52.83
Answer
ŷ = 52.83
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.