You need €40 000 in 6 years. An account pays 3% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €40 0003% / 12n = 6
PV = €
2.Time value of moneyHard
You save €200 at the end of every month for 16 years. The account pays 4.8% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €2004.8% / 12N = 192
FV = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 3: €35 000, €17 000, €28 000. The cost of capital is 8%. Find the net present value NPV.
An investment of €10 000 returns €2 500 at the end of each year for 6 years. Find its internal rate of return IRR in percent.
C0 = €10 000CF = €2 500n = 6
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 6 years. The yield to maturity is 7% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 7%n = 6
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 10 years and sells for €1 080. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 10P = €1 080
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €12 000 000, preferred shares worth €1 500 000 and debt worth €3 000 000. The cost of equity is 10%, the cost of preferred stock 8% and the pre-tax cost of debt 4.5%. The tax rate is 28%. Find the WACC.
A company has equity with a market value of €6 000 000 and a cost of equity of 10%. Its debt consists of 4 000 bonds (face value €1 000, annual coupon 5%, 6 years to maturity) that trade at €1 080 each. The tax rate is 28%. 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: 42, 33, 42, 11, 55, 29, 12.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 52.
μ = 60σ = 10x = 52
z =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.25, 0.3, 0.3, 0.15
Var(X) =
12.ProbabilityHard
A game costs €1 to play. You win €5 with probability 0.15 and €10 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€5: p = 0.15€10: p = 0.05
E(G) = €
13.Normal distributionMedium
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 30 min. Round z to two decimals.
μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30 min
P =
14.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
A sample of 15 items has mean 26.5 and standard deviation s = 7. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.
n = 15x̄ = 26.5s = 790%t(14) = 1.761
upper limit =
16.Confidence intervalsHard
A pilot study suggests that about 0.15 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.03?
p ≈ 0.15E = 0.0395%
n =
17.Hypothesis tests & confidence intervalsMedium
8 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −3.25 and the standard deviation of the differences is 10.2. Find the paired test statistic t for H₀: μd = 0.
n = 8d̄ = −3.25sd = 10.2
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 5. Find the smallest sample size n.
σ = 5E = 2.599%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.
n = 5
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. Predict ŷ for x = 2.5.
n = 5x = 2.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 33418 €
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 = 0.03/12 = 0.0025
N = 12·6 = 72
Present value
PV = FV/(1 + i)N = 40 000/1.197 = €33 418.31
Answer
PV = €33 418.31
2.Time value of money
Answer: FV = 57608 €
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 €38 400 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = −5791 €
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 = −75 000
CF1/(1 + 0.08)1 = 35 000/1.08 = 32 407.41
CF2/(1 + 0.08)2 = 17 000/1.166 = 14 574.76
CF3/(1 + 0.08)3 = 28 000/1.26 = 22 227.30
Add them up
NPV = €−5 790.53
Answer
NPV = €−5 790.53
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: IRR = 12.98 %
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)−6)/i = 10 000/2 500 = 4
Solve for the rate
The annuity factor falls as the rate rises; bracket the rate and narrow down.
i = 12%: 4.111
i = 13%: 3.998
IRR = 12.98%
Answer
IRR = 12.98%
5.Bond pricing & yield
Answer: P = 951.7 €
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 = €30.00, i = 0.035, N = 12
Present value of the coupons
€30.00·(1 − 0.6618)/0.035 = €289.90
Present value of the face value
€1 000·0.6618 = €661.78
Price
P = €951.68
Answer
P = €951.68
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 4.013 %
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.
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 = (52 − 60)/10 = −0.8
Answer
z = −0.8
11.Probability
Answer: Var(X) = 4.11
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 = 0·0.25 + 2·0.3 + 4·0.3 + 6·0.15 = 2.7
Mean of the squares
E(X²) = 0·0.25 + 4·0.3 + 16·0.3 + 36·0.15 = 11.4
Variance
Var(X) = 11.4 − 2.7² = 4.11
Answer
Var(X) = 4.11
12.Probability
Answer: E(G) = 0.25 €
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: 4 (p = 0.15), 9 (p = 0.05), −1 (p = 0.8)
Expected value
E(G) = 4·0.15 + 9·0.05 − 1·0.8 = 0.25 €
Answer
E(G) = 0.25 €
A positive value means you win money on average.
13.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 sum
μ = 32 min, σ = √(1.5² + 2²) = 2.5 min
Standardise
z = (30 − 32)/2.5 = −0.8
Upper tail
P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881
Answer
P = 0.7881
14.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 difference
μ = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
15.Confidence intervals
Answer: upper limit = 29.68
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 = 14, t = 1.761
Margin of error
E = 1.761·7/√15 = 3.183
Interval
26.5 ± 3.183 → [23.32, 29.68]
Answer
upper limit = 29.68
16.Confidence intervals
Answer: n = 545
Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.
Solve for n
n = 1.96²·0.15·0.85/0.03² = 544.2
Round up
n = 545
Answer
n = 545
17.Hypothesis tests & confidence intervals
Answer: t = −0.9012
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 of the mean difference
SE = 10.2/√8 = 3.606
Test statistic
t = −3.25/3.606 = −0.9012
Answer
t = −0.9012
With df = 7, the two-sided 5% critical value is 2.365; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: n = 27
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 = (2.576·5/2.5)² = 26.54
Round up
n = 27
Answer
n = 27
19.Linear regression & correlation
Answer: r = 0.9909
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 = 115/5 = 23
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 144
Syy = Σ(y − ȳ)² = 528
Correlation coefficient
r = 144/√(40·528) = 0.9909
Answer
r = 0.9909
20.Linear regression & correlation
Answer: ŷ = 20.5
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 = 97/5 = 19.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −22
Regression line
b = −22/10 = −2.2
a = 19.4 − (−2.2)·3 = 26
ŷ = 26 − 2.2·x
Prediction
ŷ = 26 + (−2.2)·2.5 = 20.5
Answer
ŷ = 20.5
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.