You need €20 000 in 5 years. An account pays 4% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €20 0004% / 12n = 5
PV = €
2.Time value of moneyHard
A loan of €20 000 is repaid with equal monthly payments over 5 years. The interest rate is 2.4% per year, compounded monthly. Find the monthly payment A.
P = €20 0002.4% / 12N = 60
A = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 4: €17 000, €17 000, €14 000, €28 000. The cost of capital is 8%. Find the net present value NPV.
A project costs €25 000 and returns €10 000, €17 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −25 000CF1 = 10 000CF2 = 17 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 9 years and pays annual coupons. It sells for €1 104 when the yield to maturity is 5.5%. Find the coupon rate c in percent.
F = €1 000n = 9P = €1 104YTM = 5.5%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 7 years and sells for €920. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 7P = €920
YTM = %
7.CAPM & WACCMedium
A company has equity worth €5 000 000 and debt worth €6 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 4% and the tax rate is 25.8%. Find the WACC.
A company has 2 000 000 shares at €35.50 each and 40 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 20%. The share’s beta is 1.2, the risk-free rate 2.5% and the market return 8%. Find the WACC.
β = 1.2rf = 2.5%rm = 8%rd = 3%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 18, 27, 34, 14, 28, 19, 34.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 20. What proportion P of values is below 76?
μ = 100σ = 20x = 76
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4p(x): 0.35, 0.2, 0.3, 0.15
Var(X) =
12.ProbabilityHard
5 components are tested, and each one fails with probability 0.4, independently. What is the probability P that at most one fails?
n = 5p = 0.4
P =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9222. Find z to two decimals.
A = 0.9222
z =
14.Normal distributionHard
The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 99% of them lie (the 99th percentile).
μ = 100σ = 15P = 99%
x = points
15.Confidence intervalsMedium
A sample of 7 items has mean 83 and standard deviation s = 11.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.
n = 7x̄ = 83s = 11.599%t(6) = 3.707
E =
16.Confidence intervalsHard
The population standard deviation is 21. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 2.5?
σ = 21E = 2.590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 25 students has a mean score of 86.6 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 86.6s = 20n = 25
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 44. The population standard deviation is 12. Find the upper limit U of the 95% confidence interval for the mean.
x̄ = 44σ = 12n = 2595%
U =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.
n = 6x = 5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 16380 €
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.
Rate and number of periods
i = 0.04/12 = 0.003333
N = 12·5 = 60
Present value
PV = FV/(1 + i)N = 20 000/1.221 = €16 380.06
Answer
PV = €16 380.06
2.Time value of money
Answer: A = 354.1 €
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.
Monthly rate and number of payments
i = 0.024/12 = 0.002
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 = 0.887
A = 20 000·0.002/(1 − 0.887) = €354.07
Answer
A = €354.07
3.NPV & IRR
Answer: NPV = 2010 €
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 + 0.08)1 = 17 000/1.08 = 15 740.74
CF2/(1 + 0.08)2 = 17 000/1.166 = 14 574.76
CF3/(1 + 0.08)3 = 14 000/1.26 = 11 113.65
CF4/(1 + 0.08)4 = 28 000/1.36 = 20 580.84
Add them up
NPV = €2 009.99
Answer
NPV = €2 009.99
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 4.853 %
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(0%) = 2 000.00
NPV(5%) = −56.69
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 + 0.055)−9 = €617.63
Coupon
C = (1104 − 617.63)/6.952 = €69.96
Coupon rate
c = 69.96/1000 = 6.996%
Answer
c = 6.996%
6.Bond pricing & yield
Answer: YTM = 6.456 %
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
€920 = €50·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷
Solve numerically
P(6%) = €944.18, P(7%) = €892.21
YTM = 6.456%
Answer
YTM = 6.456%
7.CAPM & WACC
Answer: WACC = 7.073 %
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 = 2 000 000·€35.50 = €71 000 000
D = 40 000·€1 050 = €42 000 000
Cost of equity (CAPM)
re = 2.5% + 1.2·5.5% = 9.1%
Weights
V = €113 000 000, E/V = 0.6283, D/V = 0.3717
WACC
WACC = 0.6283·9.1% + 0.3717·3%·(1 − 0.2) = 6.61%
Answer
WACC = 6.61%
9.Descriptive statistics
Answer: s = 7.967
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) = 1.188
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 + 2·0.2 + 3·0.3 + 4·0.15 = 2.25
Mean of the squares
E(X²) = 1·0.35 + 4·0.2 + 9·0.3 + 16·0.15 = 6.25
Variance
Var(X) = 6.25 − 2.25² = 1.188
Answer
Var(X) = 1.188
12.Probability
Answer: P = 0.337
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.
Zero and one failure
P(X = 0) = 0.65 = 0.07776
P(X = 1) = 5·0.4·0.64 = 0.2592
Add them
P = 0.07776 + 0.2592 = 0.337
Answer
P = 0.337
13.Normal distribution
Answer: z = −1.42
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 of z
Φ(z) = 1 − 0.9222 = 0.0778
Read the z-table backwards
Find 0.0778 inside the table and read off z.
z = −1.42
Answer
z = −1.42
14.Normal distribution
Answer: x = 134.9 points
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 for the percentile
Φ(z) = 0.99 → z = 2.326
Back to x
x = 100 + 2.326·15 = 134.89 points
Answer
x = 134.89 points
15.Confidence intervals
Answer: E = 16.11
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 = 6, t = 3.707
Margin of error
E = 3.707·11.5/√7 = 16.11
Answer
E = 16.11
16.Confidence intervals
Answer: n = 191
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.
Solve for n
n = (1.645·21/2.5)² = 190.9
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 191
Answer
n = 191
17.Hypothesis tests & confidence intervals
Answer: t = 2.9
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 = 20/√25 = 20/5 = 4
Test statistic
t = (86.6 − 75)/4 = 2.9
Answer
t = 2.9
t0.025; df = 24 = 2.064
|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 48.7
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 = 1.96·12/√25 = 4.704
Interval limit
U = 44 + 4.704 = 48.7
Answer
U = 48.7
19.Linear regression & correlation
Answer: r = 0.9542
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.
Centred sums
Sxy = 8250 − 105·377/5 = 333
Sxx = 2295 − 105²/5 = 90
Syy = 29779 − 377²/5 = 1353
Correlation coefficient
r = 333/√(90·1353) = 0.9542
Answer
r = 0.9542
20.Linear regression & correlation
Answer: ŷ = 10.18
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 = 82/6 = 13.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 122
Regression line
b = 122/70 = 1.743
a = 13.67 − 1.743·7 = 1.467
ŷ = 1.467 + 1.743·x
Prediction
ŷ = 1.467 + 1.743·5 = 10.18
Answer
ŷ = 10.18
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.