An annuity pays €2 000 at the end of each year for 6 years. The interest rate is 3% per year. Find the present value PV of the annuity.
A = €2 000i = 3%n = 6
PV = €
2.Time value of moneyHard
A loan of €8 000 is repaid with equal monthly payments over 5 years. The interest rate is 4.8% per year, compounded monthly. Find the monthly payment A.
P = €8 0004.8% / 12N = 60
A = €
3.NPV & IRRMedium
A project costs €250 000 today. It generates €14 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 10%. Find the net present value NPV.
C0 = €250 000CF1 = €14 000g = 3%r = 10%
NPV = €
4.NPV & IRRHard
A project costs €25 000 and returns €10 500, €18 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 10 500CF2 = 18 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 9 years. The yield to maturity is 8% per year. Find the price P of the bond.
F = €1 000coupon = 2%YTM = 8%n = 9
P = €
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 €1 055. Find its yield to maturity YTM in percent.
F = €1 000coupon = 8%n = 3P = €1 055
YTM = %
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 0.75. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 25.8%. Find the WACC.
D/E = 0.75re = 10%rd = 4.5%T = 25.8%
WACC = %
8.CAPM & WACCHard
A company has 2 500 000 shares at €20.00 each and 20 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 30%. The share’s beta is 0.9, the risk-free rate 4% and the market return 9%. Find the WACC.
β = 0.9rf = 4%rm = 9%rd = 3%T = 30%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 52, 37, 56, 51, 52, 49.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 8. What proportion P of values is below 60.4?
μ = 70σ = 8x = 60.4
P =
11.ProbabilityMedium
A bag holds 8 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
8 / 4
P =
12.ProbabilityHard
X has the distribution shown, and Y = 5X + 10. Find the standard deviation σY of Y.
X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 29.5 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 29.5 min
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g40%
x = g
15.Confidence intervalsMedium
A sample of 9 items has mean 96 and standard deviation s = 11. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 9x̄ = 96s = 1190%t(8) = 1.86
lower limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 5 plots, mean yield 89 kg, standard deviation 8 kg. Plot group 2: 16 plots, mean 88 kg, standard deviation 4.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.
The mean of a population is claimed to be 500; the population standard deviation is known to be 9. A random sample of 36 observations has a mean of 503.6. Test H₀: μ = 500 against H₁: μ > 500. Find the p-value p.
μ₀ = 500σ = 9n = 36x̄ = 503.6H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
A sample of 100 measurements has mean 47.5. The population standard deviation is 15. Find the lower limit L of the 90% confidence interval for the mean.
x̄ = 47.5σ = 15n = 10090%
L =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 6
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. For which value x does the line predict ŷ = 43?
n = 5ŷ = 43
x =
3Solutions
1.Time value of money
Answer: PV = 10834 €
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.
Annuity factor
(1 − (1 + i)−n)/i = (1 − 0.8375)/0.03 = 5.417
Present value
PV = 2 000·5.417 = €10 834.38
Answer
PV = €10 834.38
2.Time value of money
Answer: A = 150.2 €
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 = 12·5 = 60
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 = 0.787
A = 8 000·0.004/(1 − 0.787) = €150.24
Answer
A = €150.24
3.NPV & IRR
Answer: NPV = −50000 €
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.
Value of the cash flows
PV = 14 000/(0.1 − 0.03) = 200 000.00
NPV
NPV = 200 000.00 − 250 000 = −€50 000.00
Answer
NPV = −€50 000.00
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 9.549 %
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(5%) = 1 780.05
NPV(10%) = −165.29
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.
Per period
C = €10.00, i = 0.04, N = 18
Present value of the coupons
€10.00·(1 − 0.4936)/0.04 = €126.59
Present value of the face value
€1 000·0.4936 = €493.63
Price
P = €620.22
Answer
P = €620.22
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 5.945 %
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
€1 055 = €80·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³
Solve numerically
P(5%) = €1 081.70, P(6%) = €1 053.46
YTM = 5.945%
Answer
YTM = 5.945%
7.CAPM & WACC
Answer: WACC = 7.145 %
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).
Market values
E = 2 500 000·€20.00 = €50 000 000
D = 20 000·€1 050 = €21 000 000
Cost of equity (CAPM)
re = 4% + 0.9·5% = 8.5%
Weights
V = €71 000 000, E/V = 0.7042, D/V = 0.2958
WACC
WACC = 0.7042·8.5% + 0.2958·3%·(1 − 0.3) = 6.607%
Answer
WACC = 6.607%
9.Descriptive statistics
Answer: s = 6.535
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: P = 0.4242
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 = 8/12 · 7/11 = 56/132 = 0.4242
Answer
P = 0.4242
12.Probability
Answer: σY = 6.519
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, Var(X) = 1.7, σX = 1.304
Linear change
σY = |5|·1.304 = 6.519
Answer
σY = 6.519
The mean does move: E(Y) = 5·3 + 10 = 25.
13.Normal distribution
Answer: P = 0.8413
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 sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (29.5 − 32)/2.5 = −1
Upper tail
P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
14.Normal distribution
Answer: x = 248.8 g
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) = 0.4
z from the table
z ≈ −0.25
Back to the original units
x = 250 − 0.25·5 = 248.8 g
Answer
x = 248.8 g
15.Confidence intervals
Answer: lower limit = 89.18
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 = 8, t = 1.86
Margin of error
E = 1.86·11/√9 = 6.82
Interval
96 ± 6.82 → [89.18, 102.8]
Answer
lower limit = 89.18
16.Confidence intervals
Answer: E = 5.82 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 = (4·8² + 15·4.5²)/19 = 29.46
sp = 5.428
Margin of error
E = 2.093·5.428·√(1/5 + 1/16) = 5.82
Answer
E = 5.82 kg
17.Hypothesis tests & confidence intervals
Answer: p = 0.008198
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 = 9/√36 = 1.5
z = (503.6 − 500)/1.5 = 2.4
Tail area
Φ(2.4) = 0.9918
1 − Φ(2.4) = 0.0082
p-value
p = 0.008198
Answer
p = 0.008198
p ≤ 0.05, so H₀ is rejected at the 5% level.
18.Hypothesis tests & confidence intervals
Answer: L = 45.03
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 = 1.645·15/√100 = 2.468
Interval limit
L = 47.5 − 2.468 = 45.03
Answer
L = 45.03
19.Linear regression & correlation
Answer: r = −0.9872
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 = 111/6 = 18.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −51.5
Syy = Σ(y − ȳ)² = 155.5
Correlation coefficient
r = −51.5/√(17.5·155.5) = −0.9872
Answer
r = −0.9872
20.Linear regression & correlation
Answer: x = 3.509
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 = 20/5 = 4
ȳ = 228/5 = 45.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 53
Regression line
b = 53/10 = 5.3
a = 45.6 − 5.3·4 = 24.4
ŷ = 24.4 + 5.3·x
Solve for x
x = (43 − 24.4)/5.3 = 3.509
Answer
x = 3.509
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.