You invest €10 000 at 7% per year with continuous compounding. Find the value FV after 15 years.
PV = €10 000r = 7%t = 15
FV = €
2.Time value of moneyHard
You want €30 000 in 9 years. You save the same amount at the end of every month in an account paying 2.4% per year, compounded monthly. Find the monthly deposit A.
FV = €30 0002.4% / 12N = 108
A = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 4: €22 000, €27 000, €21 000, €15 000. The cost of capital is 15%. Find the net present value NPV.
A project costs €10 000 and returns €4 500, €4 500, €5 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 4 500CF2 = 4 500CF3 = 5 000
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 10 years to maturity (face value €1 000, annual coupon 2%) when its yield to maturity is 8%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 7.5%. Find your holding period return HPR in percent.
F = €1 000coupon = 2%n = 10y0 = 8%y1 = 7.5%
HPR = %
6.Bond pricing & yieldHard
A bond trades at €1 025 with a yield to maturity of 5% (annual compounding) and a Macaulay duration of 9.3 years. Use the modified duration to estimate the new price P if the yield rises by 150 basis points.
P0 = €1 025y = 5%D = 9.3Δy = +150 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €1 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 28%. Find the WACC.
A company's shares have a beta of 1 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 0.75. The tax rate is 21%, the risk-free rate 3.5% and the market risk premium 4.5%. Find the cost of equity re after the change.
Find the sample standard deviation s of the data: 59, 58, 21, 16, 35.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is above 580?
μ = 500σ = 40x = 580
P =
11.ProbabilityMedium
A bag holds 7 red and 8 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
7 / 8
P =
12.ProbabilityHard
A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.5, independently. What is the probability P of exactly 3 correct answers?
n = 9p = 0.5k = 3
P =
13.Normal distributionMedium
The masses of passengers with luggage are independent and normal with mean 85 kg and standard deviation 12 kg. Find the probability P that the total mass of 4 of them exceeds 388 kg. Round z to two decimals.
μ = 85 kgσ = 12 kgn = 4> 388 kg
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 75% is to the left of it. Find the cut-off x.
μ = 250 gσ = 5 g75%
x = g
15.Confidence intervalsMedium
A sample of 21 items has mean 106 and standard deviation s = 17. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 21x̄ = 106s = 1790%t(20) = 1.725
upper limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 10 plots, mean yield 64.5 kg, standard deviation 8 kg. Plot group 2: 21 plots, mean 76 kg, standard deviation 10 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.
A sample of 16 students has a mean score of 103 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 103s = 5n = 16
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 2 units with a power of 80%. The standard deviation is 7. Find the minimum sample size n.
α = 0.01power = 80%δ = 2σ = 7zα = 2.326zβ = 0.842
n =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 5 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 (2.5, 16) is made. Find its residual e = y − ŷ.
n = 5(2.5, 16)
e =
3Solutions
1.Time value of money
Answer: FV = 28577 €
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 = 0.07·15 = 1.05
e1.05 = 2.858
Future value
FV = 10 000·2.858 = €28 576.51
Answer
FV = €28 576.51
2.Time value of money
Answer: A = 249.1 €
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.
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 = −60 000
CF1/(1 + 0.15)1 = 22 000/1.15 = 19 130.43
CF2/(1 + 0.15)2 = 27 000/1.322 = 20 415.88
CF3/(1 + 0.15)3 = 21 000/1.521 = 13 807.84
CF4/(1 + 0.15)4 = 15 000/1.749 = 8 576.30
Add them up
NPV = €1 930.45
Answer
NPV = €1 930.45
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 18.54 %
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%) = 603.27
NPV(20%) = −231.48
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 15 + 5·603.27/(603.27 − (−231.48)) = 18.61%
exact: NPV(18.54%) = 0 → IRR = 18.54%
Answer
IRR = 18.54%
5.Bond pricing & yield
Answer: HPR = 12.01 %
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.
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.
Modified duration
Dmod = 9.3/1.05 = 8.857
Estimated price change
%ΔP ≈ −8.857·(0.015) = −13.29%
P ≈ 1025·(1 − 0.1329) = €888.82
Answer
P = €888.82
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 9.249 %
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).
Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.
Unlever the beta
βU = 1/(1 + 0.79·0.2) = 0.8636
Relever at the new D/E
βL = 0.8636·(1 + 0.79·0.75) = 1.375
Cost of equity (CAPM)
re = 3.5% + 1.375·4.5% = 9.688%
Answer
re = 9.688%
9.Descriptive statistics
Answer: s = 20.14
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 = 2
Standard normal table
Φ(2) = 0.9772
P = 1 − 0.9772 = 0.02275
Answer
P = 0.02275
11.Probability
Answer: P = 0.2
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 = 7/15 · 6/14 = 42/210 = 0.2
Answer
P = 0.2
12.Probability
Answer: P = 0.1641
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 = 340 kg, σT = 12·√4 = 24 kg
Standardise
z = (388 − 340)/24 = 2
Probability
P = 1 − Φ(2) = 0.02275
Answer
P = 0.02275
14.Normal distribution
Answer: x = 253.4 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) = 0.75
z from the table
z ≈ 0.67
Back to the original units
x = 250 + 0.67·5 = 253.4 g
Answer
x = 253.4 g
15.Confidence intervals
Answer: upper limit = 112.4
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 = 20, t = 1.725
Margin of error
E = 1.725·17/√21 = 6.399
Interval
106 ± 6.399 → [99.6, 112.4]
Answer
upper limit = 112.4
16.Confidence intervals
Answer: upper limit = −4.095 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 = (9·8² + 20·10²)/29 = 88.83
sp = 9.425
Margin of error
E = 2.045·9.425·√(1/10 + 1/21) = 7.405
Interval
−11.5 ± 7.405 → [−18.91, −4.095]
Answer
upper limit = −4.095 kg
17.Hypothesis tests & confidence intervals
Answer: t = 2.4
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/√16 = 5/4 = 1.25
Test statistic
t = (103 − 100)/1.25 = 2.4
Answer
t = 2.4
t0.025; df = 15 = 2.131
|t| > 2.131, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 123
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((2.326 + 0.842)·7/2)² = 122.9
Round up
n = 123
Answer
n = 123
19.Linear regression & correlation
Answer: r = 0.9875
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 = 27/6 = 4.5
ȳ = 390/6 = 65
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 105
Syy = Σ(y − ȳ)² = 646
Correlation coefficient
r = 105/√(17.5·646) = 0.9875
Answer
r = 0.9875
20.Linear regression & correlation
Answer: e = −4.6
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 = 96/5 = 19.2
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −28
Regression line
b = −28/10 = −2.8
a = 19.2 − (−2.8)·3 = 27.6
ŷ = 27.6 − 2.8·x
Predicted value
ŷ = 27.6 + (−2.8)·2.5 = 20.6
Residual
e = 16 − 20.6 = −4.6
Answer
e = −4.6
The point lies below 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.