CFA Level I quantitative methods and fixed income practice: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Time value of money

Future value
FV = PV·(1 + i)n
Present value
PV = FV/(1 + i)n
Several periods per year
i = r/m, N = m·n
Annuity, present value
PV = PMT·[1 − (1 + i)−n]/i
Annuity, future value
FV = PMT·[(1 + i)n − 1]/i

Method

  1. Draw a timeline with every cash flow and its timing.
  2. Match the rate to the period: a monthly rate for monthly payments.
  3. Use FV = PV·(1 + i)ⁿ for single amounts and the annuity formulas for equal payments.
  4. Check the sign and the order of magnitude of the answer.

NPV & IRR

Net present value
NPV = −C0 + Σ CFt/(1 + r)t
Equal cash flows
NPV = −C0 + CF·[1 − (1 + r)−n]/r
Internal rate of return
the r at which NPV = 0
Decision
accept when NPV > 0 (or IRR > required return)

Method

  1. List all cash flows with their year; the investment at year 0 is negative.
  2. Discount each cash flow: CFt/(1 + r)t.
  3. Add them up to get the NPV. Positive means the project earns more than r.
  4. For the IRR, find the rate that makes NPV zero by trial or interpolation.

Bond pricing & yield

Bond price
P = C·(1 − (1 + y)−n)/y + F/(1 + y)n
Semi-annual
C/2, y/2 and 2n periods
Current yield
C/P
Price and yield
yield up → price down

Method

  1. Find the coupon per period and the yield per period.
  2. Price = present value of the coupons (an annuity) + present value of the face value.
  3. For semi-annual coupons use half the coupon, half the yield and twice the periods.
  4. Check: if the yield is above the coupon rate, the price is below face value.

CAPM & WACC

CAPM
re = rf + β·(rm − rf)
WACC
E/V·re + D/V·rd·(1 − T)
Firm value
V = E + D

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

2Practice problems

1.Time value of moneyMedium
periods0123411 + EAR = ?19.5% / 4

A credit card charges a nominal interest rate of 19.5% per year, compounded quarterly. Find the effective annual rate EAR in percent.

r = 19.5%m = 4
EAR = %
2.Time value of moneyHard
months0123…168€150€150€150FV = ?

You save €150 at the end of every month for 14 years. The account pays 3.6% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €1503.6% / 12N = 168
FV = €
3.NPV & IRRMedium
years01234€80 000€5 000€5 000€5 000€5 000r = 12%, EAC = ?

A machine costs €80 000 today, lasts 4 years and costs €5 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.

C0 = €80 000M = €5 000n = 4r = 12%
EAC = €
4.NPV & IRRHard
years0123456−150 00027 500+5%+5%+5%+5%+5%r = 10%, NPV = ?

A project costs €150 000 today. Its cash flow is €27 500 at the end of year 1 and grows by 5% per year until the last cash flow at the end of year 6. The cost of capital is 10%. Find the net present value NPV.

C0 = €150 000CF1 = €27 500g = 5%n = 6r = 10%
NPV = €
5.Bond pricing & yieldMedium
years0123…12€1 188C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 12 years and pays annual coupons. It sells for €1 188 when the yield to maturity is 4%. Find the coupon rate c in percent.

F = €1 000n = 12P = €1 188YTM = 4%
c = %
6.Bond pricing & yieldHard
half-years0123…18€757€25€25€25 + €1 000

A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 9 years. It sells for €757. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 5% (2×)n = 9P = €757
YTM = %
7.CAPM & WACCMedium
WACC = ?equityEre​ = 12%debtD = 0.6·Erd​ = 3%T = 25.8%

A company has a debt-to-equity ratio (at market values) of 0.6. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.

D/E = 0.6re = 12%rd = 3%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €49.00β = 1.2rf​ = 1%, rm​ = 5%debt5 000 × €960rd​ = 4.5%T = 28%

A company has 5 000 000 shares at €49.00 each and 5 000 bonds trading at €960 each. The pre-tax cost of debt is 4.5% and the tax rate is 28%. The share’s beta is 1.2, the risk-free rate 1% and the market return 5%. Find the WACC.

β = 1.2rf = 1%rm = 5%rd = 4.5%T = 28%
WACC = %
9.Descriptive statisticsMedium
202530354045505560

Find the sample standard deviation s of the data: 60, 32, 17, 56, 33.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 5. What proportion P of values is below 175?

μ = 170σ = 5x = 175
P =
11.ProbabilityMedium
xp(x)00.0510.320.3530.3μ = 1.9

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2, 3p(x): 0.05, 0.3, 0.35, 0.3
Var(X) =
12.ProbabilityHard
€p(x)−30.6830.3270.02

A game costs €3 to play. You win €6 with probability 0.3 and €30 with probability 0.02; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€6: p = 0.3€30: p = 0.02
E(G) = €
13.Normal distributionMedium
750 g754.9 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 754.9 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 754.9 g
P =
14.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
μ₁ − μ₂0LUx̄₁ − x̄₂E = ?90%

The battery life of two phone models is compared. Model 1: 50 phones, mean 114 h, known standard deviation 13 h. Model 2: 30 phones, mean 105 h, known standard deviation 4 h. Find the margin of error E of the 90% confidence interval for μ₁ − μ₂.

n₁ = 50x̄₁ = 114 hσ₁ = 13 hn₂ = 30x̄₂ = 105 hσ₂ = 4 h90%
E = h
16.Confidence intervalsHard
σ = 17, n = ?x̄ − Ex̄ + Ex̄E = 299%

The population standard deviation is 17. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 2?

σ = 17E = 299%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

Two teaching methods are compared. Group 1: 5 students, mean score 24.5, standard deviation 3. Group 2: 11 students, mean score 20.4, standard deviation 5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 5x̄₁ = 24.5s₁ = 3n₂ = 11x̄₂ = 20.4s₂ = 5
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 20 of 100 patients on treatment A recover, and 60 of 250 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 20n₁ = 100x₂ = 60n₂ = 250
z =
19.Linear regression & correlationMedium
xy1533615717739801191xy5106080

The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 36Σy = 429Σx² = 286Σy² = 31581Σxy = 2823
r =
20.Linear regression & correlationHard
xy35557077898911101xy5106080100x = 6

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (6, 76) is made. Find its residual e = y − ŷ.

n = 5(6, 76)
e =

3Solutions

1.Time value of money

Answer: EAR = 20.97 %

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 = 0.195/4 = 0.04875

Compound over one year

EAR = (1 + 0.04875)4 − 1 = 1.21 − 1 = 20.97%

Answer

EAR = 20.97%

The EAR is always above the nominal rate when interest compounds more than once a year.

2.Time value of money

Answer: FV = 32704 €

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 deposits

i = 0.036000000000000004/12 = 0.003
N = 12·14 = 168

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 150·(1.654 − 1)/0.003 = €32 704.05

Answer

FV = €32 704.05

You paid in €25 200 yourself; the rest is interest.

3.NPV & IRR

Answer: EAC = 31339 €

Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.

Annuity factor

(1 − 1.12−4)/0.12 = 3.037

Annualised purchase price

80 000/3.037 = 26 338.75

EAC

EAC = 26 338.75 + 5 000 = €31 338.75

Answer

EAC = €31 338.75
4.NPV & IRR

Answer: NPV = −16047 €

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)/(1.1))6 = 0.7564
PV = 27 500/(0.1 − 0.05)·(1 − 0.7564) = 133 953.02

NPV

NPV = 133 953.02 − 150 000 = −€16 046.98

Answer

NPV = −€16 046.98

NPV is negative, so the project earns less than the required return.

5.Bond pricing & yield

Answer: c = 6.003 %

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 + 0.04)−12 = €624.60

Coupon

C = (1188 − 624.60)/9.385 = €60.03

Coupon rate

c = 60.03/1000 = 6.003%

Answer

c = 6.003%
6.Bond pricing & yield

Answer: YTM = 8.996 %

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 = €25.00, N = 18

Solve numerically

i = 4.498% per half-year

YTM

YTM = 2·4.498% = 8.996%

Answer

YTM = 8.996%
7.CAPM & WACC

Answer: WACC = 8.335 %

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 from D/E

E/V = 1/(1 + 0.6) = 0.625
D/V = 0.6/(1 + 0.6) = 0.375

WACC

WACC = 0.625·12% + 0.375·3%·(1 − 0.258) = 8.335%

Answer

WACC = 8.335%
8.CAPM & WACC

Answer: WACC = 5.751 %

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 = 5 000 000·€49.00 = €245 000 000
D = 5 000·€960 = €4 800 000

Cost of equity (CAPM)

re = 1% + 1.2·4% = 5.8%

Weights

V = €249 800 000, E/V = 0.9808, D/V = 0.01922

WACC

WACC = 0.9808·5.8% + 0.01922·4.5%·(1 − 0.28) = 5.751%

Answer

WACC = 5.751%
9.Descriptive statistics

Answer: s = 18.01

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.

Mean

x̄ = 198/5 = 39.6

Squared deviations

Σ(x − x̄)² = 20.4² + (−7.6)² + (−22.6)² + 16.4² + (−6.6)² = 1297

Standard deviation

s = √(1297/4) = 18.01

Answer

s = 18.01
10.Descriptive statistics

Answer: P = 0.8413

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: Var(X) = 0.79

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 = 0·0.05 + 1·0.3 + 2·0.35 + 3·0.3 = 1.9

Mean of the squares

E(X²) = 0·0.05 + 1·0.3 + 4·0.35 + 9·0.3 = 4.4

Variance

Var(X) = 4.4 − 1.9² = 0.79

Answer

Var(X) = 0.79
12.Probability

Answer: E(G) = −0.6 €

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: 3 (p = 0.3), 27 (p = 0.02), −3 (p = 0.68)

Expected value

E(G) = 3·0.3 + 27·0.02 − 3·0.68 = −0.6 €

Answer

E(G) = −0.6 €

A negative value means you lose money on average.

13.Normal distribution

Answer: P = 0.3085

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 = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (754.9 − 750)/9.849 = 0.5

Upper tail

P = 1 − Φ(0.5) = 1 − 0.6915 = 0.3085

Answer

P = 0.3085
14.Normal distribution

Answer: P = 0.9207

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 = 25.02 − 25 = 0.02 mm, σ = 0.01414 mm

Standardise

z = (0 − 0.02)/0.01414 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207
15.Confidence intervals

Answer: E = 3.254 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 = 114 − 105 = 9

Standard error

SE = √(13²/50 + 4²/30) = 1.978

Margin of error

E = 1.645·1.978 = 3.254

Answer

E = 3.254 h

The interval does not contain 0, so the means differ significantly.

16.Confidence intervals

Answer: n = 480

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 = (2.576·17/2)² = 479.4

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 480

Answer

n = 480
17.Hypothesis tests & confidence intervals

Answer: t = 1.682

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.

Pooled variance

sp² = (4·3² + 10·5²)/14 = 20.43
sp = 4.52

Standard error

SE = 4.52·√(1/5 + 1/11) = 2.438

Test statistic

t = (24.5 − 20.4)/2.438 = 1.682

Answer

t = 1.682

With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: z = −0.8051

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 = 20/100 = 0.2
p̂₂ = 60/250 = 0.24
p̂ = 80/350 = 0.2286

Standard error under H₀

SE = √(0.2286·0.7714·(1/100 + 1/250)) = 0.04968

Test statistic

z = (0.2 − 0.24)/0.04968 = −0.8051

Answer

z = −0.8051

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

19.Linear regression & correlation

Answer: r = 0.9879

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 = 2823 − 36·429/6 = 249
Sxx = 286 − 36²/6 = 70
Syy = 31581 − 429²/6 = 907.5

Correlation coefficient

r = 249/√(70·907.5) = 0.9879

Answer

r = 0.9879
20.Linear regression & correlation

Answer: e = 2.95

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 = 35/5 = 7
ȳ = 393/5 = 78.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 222

Regression line

b = 222/40 = 5.55
a = 78.6 − 5.55·7 = 39.75
ŷ = 39.75 + 5.55·x

Predicted value

ŷ = 39.75 + 5.55·6 = 73.05

Residual

e = 76 − 73.05 = 2.95

Answer

e = 2.95

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.

ExamProblems · examproblems.com/courses/cfa-level-1-quant-and-fixed-income