You invest €10 000 at 4% per year with continuous compounding. Find the value FV after 2 years.
PV = €10 000r = 4%t = 2
FV = €
2.Time value of moneyHard
A loan of €10 000 is repaid with equal monthly payments over 5 years at 3% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €10 0003% / 12N = 60
I = €
3.NPV & IRRMedium
A project costs €500 000 today. It generates €60 000 at the end of the first year, and the cash flow then grows by 2% per year forever. The cost of capital is 12%. Find the net present value NPV.
C0 = €500 000CF1 = €60 000g = 2%r = 12%
NPV = €
4.NPV & IRRHard
A project costs €100 000 today. Its cash flow is €13 000 at the end of year 1 and grows by 5% per year until the last cash flow at the end of year 7. The cost of capital is 8%. Find the net present value NPV.
C0 = €100 000CF1 = €13 000g = 5%n = 7r = 8%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 7% per year. Find the price P of the bond.
F = €1 000coupon = 8%YTM = 7%n = 12
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 4% matures in 8 years and sells for €1 110. Find its yield to maturity YTM in percent.
F = €1 000coupon = 4%n = 8P = €1 110
YTM = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €1 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 7% and the tax rate is 30%. Find the WACC.
A company has 5 000 000 shares at €50.50 each and 20 000 bonds trading at €980 each. The pre-tax cost of debt is 3.5% and the tax rate is 25%. The share’s beta is 1.2, the risk-free rate 1.5% and the market return 6.5%. Find the WACC.
β = 1.2rf = 1.5%rm = 6.5%rd = 3.5%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 29, 17, 60, 40, 20, 52, 10.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 15. What proportion P of values is below 75?
μ = 60σ = 15x = 75
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 0, 2, 4, 6p(x): 0.35, 0.35, 0.2, 0.1
σ =
12.ProbabilityHard
A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 3 correct answers?
n = 10p = 0.3k = 3
P =
13.Normal distributionMedium
The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 10 of them is below 116.2 kg. Round z to two decimals.
μ = 12 kgσ = 0.8 kgn = 10< 116.2 kg
P =
14.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 70% is to the left of it. Find the cut-off x.
μ = 500 hσ = 40 h70%
x = h
15.Confidence intervalsMedium
A sample of 11 items has mean 65 and standard deviation s = 14.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 11x̄ = 65s = 14.595%t(10) = 2.228
lower limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 5 plots, mean yield 58.5 kg, standard deviation 1 kg. Plot group 2: 7 plots, mean 65.5 kg, standard deviation 2.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.
A sample of 36 students has a mean score of 503.9 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.
μ₀ = 500x̄ = 503.9s = 8n = 36
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 73.8. The population standard deviation is 5. Find the lower limit L of the 99% confidence interval for the mean.
x̄ = 73.8σ = 5n = 6499%
L =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 26.
n = 6x = 26
ŷ =
3Solutions
1.Time value of money
Answer: FV = 10833 €
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 = 0.04·2 = 0.08
e0.08 = 1.083
Future value
FV = 10 000·1.083 = €10 832.87
Answer
FV = €10 832.87
2.Time value of money
Answer: I = 781.2 €
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 payment
i = 0.0025, N = 60
A = 10 000·0.0025/(1 − (1 + 0.0025)−60) = €179.69
Total interest
I = N·A − P = 60·179.7 − 10 000 = €781.21
Answer
I = €781.21
3.NPV & IRR
Answer: NPV = 100000 €
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 = 60 000/(0.12 − 0.02) = 600 000.00
NPV
NPV = 600 000.00 − 500 000 = €100 000.00
Answer
NPV = €100 000.00
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: NPV = −22446 €
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 growing annuity
((1.05)/(1.08))7 = 0.821
PV = 13 000/(0.08 − 0.05)·(1 − 0.821) = 77 553.85
NPV
NPV = 77 553.85 − 100 000 = −€22 446.15
Answer
NPV = −€22 446.15
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: P = 1080 €
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 = €40.00, i = 0.035, N = 24
Present value of the coupons
€40.00·(1 − 0.438)/0.035 = €642.33
Present value of the face value
€1 000·0.438 = €437.96
Price
P = €1 080.29
Answer
P = €1 080.29
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 2.468 %
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
€1 110 = €40·(1 − (1 + y)⁻⁸)/y + €1 000/(1 + y)⁸
Solve numerically
P(2%) = €1 146.51, P(3%) = €1 070.20
YTM = 2.468%
Answer
YTM = 2.468%
7.CAPM & WACC
Answer: WACC = 10.58 %
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 = €5 000 000, E/V = 0.8, D/V = 0.2
WACC
WACC = 0.8·12% + 0.2·7%·(1 − 0.3) = 10.58%
Answer
WACC = 10.58%
8.CAPM & WACC
Answer: WACC = 7.149 %
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
Standard normal table
Φ(1) = 0.8413
P = 0.8413
Answer
P = 0.8413
11.Probability
Answer: σ = 1.947
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 + 2·0.35 + 4·0.2 + 6·0.1 = 2.1
Mean of the squares
E(X²) = 0·0.35 + 4·0.35 + 16·0.2 + 36·0.1 = 8.2
Variance
Var(X) = 8.2 − 2.1² = 3.79
σ = √3.79 = 1.947
Answer
σ = 1.947
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 = 120 kg, σT = 0.8·√10 = 2.53 kg
Standardise
z = (116.2 − 120)/2.53 = −1.5
Probability
P = Φ(−1.5) = 0.06681
Answer
P = 0.06681
14.Normal distribution
Answer: x = 521 h
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) = 0.7
z from the table
z ≈ 0.52
Back to the original units
x = 500 + 0.52·40 = 521 h
Answer
x = 521 h
15.Confidence intervals
Answer: lower limit = 55.26
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 = 10, t = 2.228
Margin of error
E = 2.228·14.5/√11 = 9.741
Interval
65 ± 9.741 → [55.26, 74.74]
Answer
lower limit = 55.26
16.Confidence intervals
Answer: E = 2.161 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 = (4·1² + 6·2.5²)/10 = 4.15
sp = 2.037
Margin of error
E = 1.812·2.037·√(1/5 + 1/7) = 2.161
Answer
E = 2.161 kg
17.Hypothesis tests & confidence intervals
Answer: t = 2.925
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/√36 = 8/6 = 1.333
Test statistic
t = (503.9 − 500)/1.333 = 2.925
Answer
t = 2.925
t0.025; df = 35 = 2.03
|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: L = 72.19
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 = 2.576·5/√64 = 1.61
Interval limit
L = 73.8 − 1.61 = 72.19
Answer
L = 72.19
19.Linear regression & correlation
Answer: r² = 0.9957
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 = 30/5 = 6
ȳ = 297/5 = 59.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 228
Syy = Σ(y − ȳ)² = 1305
Coefficient of determination
r² = 228²/(40·1305) = 0.9957
Answer
r² = 0.9957
About 99.57% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 150.8
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 = 132/6 = 22
ȳ = 771/6 = 128.5
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 391
Regression line
b = 391/70 = 5.586
a = 128.5 − 5.586·22 = 5.614
ŷ = 5.614 + 5.586·x
Prediction
ŷ = 5.614 + 5.586·26 = 150.8
Answer
ŷ = 150.8
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.