A credit card charges a nominal interest rate of 15% per year, compounded semi-annually. Find the effective annual rate EAR in percent.
r = 15%m = 2
EAR = %
2.Time value of moneyHard
You want €30 000 in 7 years. You save the same amount at the end of every month in an account paying 4.8% per year, compounded monthly. Find the monthly deposit A.
FV = €30 0004.8% / 12N = 84
A = €
3.NPV & IRRMedium
A project costs €20 000 today and returns €6 500 at the end of each year for 5 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €40 000 and returns €18 500, €13 000, €22 500, €19 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays a 4% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 5.5% per year. Find the price P of the bond.
F = €1 000coupon = 4%YTM = 5.5%n = 7
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 8% matures in 5 years and sells for €1 060. Find its yield to maturity YTM in percent.
F = €1 000coupon = 8%n = 5P = €1 060
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 3% and the tax rate is 25%. Find the WACC.
A company has 2 000 000 shares at €54.00 each and 10 000 bonds trading at €980 each. The pre-tax cost of debt is 4% and the tax rate is 20%. The share’s beta is 0.8, the risk-free rate 1% and the market return 7.5%. Find the WACC.
β = 0.8rf = 1%rm = 7.5%rd = 4%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 12, 38, 17, 53, 47, 16, 56.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 5. What proportion P of values is below 62?
μ = 60σ = 5x = 62
P =
11.ProbabilityMedium
In how many ways N can a committee of 4 people be chosen from a group of 14? The order does not matter.
n = 14k = 4
N =
12.ProbabilityHard
A test has 7 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 5 correct answers?
n = 7p = 0.4k = 5
P =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0239. Find z to two decimals.
A = 0.0239
z =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded bottom 40% is to the left of it. Find the cut-off x.
μ = 65 pointsσ = 10 points40%
x = points
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 60 phones, mean 122.5 h, known standard deviation 14 h. Model 2: 60 phones, mean 114 h, known standard deviation 14 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 28. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1.5?
σ = 28E = 1.595%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 80 parts has mean length 148.5 mm, and the line's known standard deviation is 4 mm. Line 2: a sample of 50 parts has mean 146.1 mm, known standard deviation 9 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
In a trial, 68 of 150 patients on treatment A recover, and 61 of 120 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 68n₁ = 150x₂ = 61n₂ = 120
z =
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 = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (9, 31) is made. Find its residual e = y − ŷ.
n = 5(9, 31)
e =
3Solutions
1.Time value of money
Answer: EAR = 15.56 %
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 per period
r/m = 0.15/2 = 0.075
Compound over one year
EAR = (1 + 0.075)2 − 1 = 1.156 − 1 = 15.56%
Answer
EAR = 15.56%
The EAR is always above the nominal rate when interest compounds more than once a year.
2.Time value of money
Answer: A = 301.2 €
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.
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 6 500/1.12 = 5 803.57 → Σ = 5 803.57
year 2: 6 500/1.254 = 5 181.76 → Σ = 10 985.33
year 3: 6 500/1.405 = 4 626.57 → Σ = 15 611.90
year 4: 6 500/1.574 = 4 130.87 → Σ = 19 742.77
year 5: 6 500/1.762 = 3 688.27 → Σ = 23 431.05
Fraction of the last year
DPB = 4 + 257.23/3 688.27 = 4.07
Answer
DPB = 4.07 years
The discounted payback is always longer than the simple payback (3.077 years).
4.NPV & IRR
Answer: MIRR = 18.92 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
18 500·1.063 = 22 033.80
13 000·1.062 = 14 606.80
22 500·1.061 = 23 850.00
19 500·1.060 = 19 500.00
TV = 79 990.60
MIRR
MIRR = (79 990.60/40 000)1/4 − 1 = 18.92%
Answer
MIRR = 18.92%
5.Bond pricing & yield
Answer: P = 913.8 €
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 = €20.00, i = 0.0275, N = 14
Present value of the coupons
€20.00·(1 − 0.684)/0.0275 = €229.82
Present value of the face value
€1 000·0.684 = €684.00
Price
P = €913.82
Answer
P = €913.82
The yield is above the coupon rate, so the bond sells at a discount.
6.Bond pricing & yield
Answer: YTM = 6.554 %
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
€1 060 = €80·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵
Solve numerically
P(6%) = €1 084.25, P(7%) = €1 041.00
YTM = 6.554%
Answer
YTM = 6.554%
7.CAPM & WACC
Answer: WACC = 10.05 %
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% + 0.2·3%·(1 − 0.25) = 10.05%
Answer
WACC = 10.05%
8.CAPM & WACC
Answer: WACC = 5.95 %
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·€54.00 = €108 000 000
D = 10 000·€980 = €9 800 000
Cost of equity (CAPM)
re = 1% + 0.8·6.5% = 6.2%
Weights
V = €117 800 000, E/V = 0.9168, D/V = 0.08319
WACC
WACC = 0.9168·6.2% + 0.08319·4%·(1 − 0.2) = 5.95%
Answer
WACC = 5.95%
9.Descriptive statistics
Answer: s = 18.83
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 = 0.4
Standard normal table
Φ(0.4) = 0.6554
P = 0.6554
Answer
P = 0.6554
11.Probability
Answer: N = 1001
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.
Combinations
C(14, 4) = (14·13·12·11)/(4·3·2·1) = 1001
Answer
N = 1001
12.Probability
Answer: P = 0.07741
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.
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 − 0.0239 = 0.9761
Read the z-table backwards
Find 0.9761 inside the table and read off z.
z = 1.98
Answer
z = 1.98
14.Normal distribution
Answer: x = 62.5 points
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 = 65 − 0.25·10 = 62.5 points
Answer
x = 62.5 points
15.Confidence intervals
Answer: upper limit = 12.7 h
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.
Point estimate
x̄₁ − x̄₂ = 122.5 − 114 = 8.5
Standard error
SE = √(14²/60 + 14²/60) = 2.556
Margin of error
E = 1.645·2.556 = 4.205
Interval
8.5 ± 4.205 → [4.295, 12.7]
Answer
upper limit = 12.7 h
The interval does not contain 0, so the means differ significantly.
16.Confidence intervals
Answer: n = 1339
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 = (1.96·28/1.5)² = 1339
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 1339
Answer
n = 1339
17.Hypothesis tests & confidence intervals
Answer: z = 1.779
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.
Standard error of the difference
SE = √(4²/80 + 9²/50) = √1.82 = 1.349
Test statistic
z = (148.5 − 146.1)/1.349 = 1.779
Answer
z = 1.779
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −0.899
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.
Sample proportions
p̂₁ = 68/150 = 0.4533
p̂₂ = 61/120 = 0.5083
p̂ = 129/270 = 0.4778
Standard error under H₀
SE = √(0.4778·0.5222·(1/150 + 1/120)) = 0.06118
Test statistic
z = (0.4533 − 0.5083)/0.06118 = −0.899
Answer
z = −0.899
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r = −0.9728
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 = 79/6 = 13.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −53.5
Syy = Σ(y − ȳ)² = 172.8
Correlation coefficient
r = −53.5/√(17.5·172.8) = −0.9728
Answer
r = −0.9728
20.Linear regression & correlation
Answer: e = 2.85
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 = 95/5 = 19
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 122
Regression line
b = 122/40 = 3.05
a = 19 − 3.05·6 = 0.7
ŷ = 0.7 + 3.05·x
Predicted value
ŷ = 0.7 + 3.05·9 = 28.15
Residual
e = 31 − 28.15 = 2.85
Answer
e = 2.85
The point lies above 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.