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.

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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
years0123456€8 000FV = ?

You invest €8 000 at a nominal rate of 4.8% per year, compounded monthly. Find the value FV of the investment after 6 years.

PV = €8 0004.8% / 12n = 6
FV = €
2.Time value of moneyHard
months01…120…360€150 000AB = ?A2.4% / 12

A mortgage of €150 000 is repaid with equal monthly payments over 30 years at 2.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 10? Find the outstanding balance B.

P = €150 0002.4% / 12N = 360k = 120
B = €
3.NPV & IRRMedium
years0123…9−250 00027 000+3%+3%…r = 12%, NPV = ?

A project costs €250 000 today. It generates €27 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 12%. Find the net present value NPV.

C0 = €250 000CF1 = €27 000g = 3%r = 12%
NPV = €
4.NPV & IRRHard
years01234−25 0009 00011 0008 5009 000

A project costs €25 000 and returns €9 000, €11 000, €8 500, €9 000 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.

r = 8%CF0 = −25 000CF1 = 9 000CF2 = 11 000CF3 = 8 500CF4 = 9 000
MIRR = %
5.Bond pricing & yieldMedium
years0123…10P€45€45€45 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 4.5% and matures in 10 years. Its yield to maturity changes from 7% to 8%. By how much ΔP does the bond price change?

F = €1 000coupon = 4.5%n = 10y1 = 7%y2 = 8%
ΔP = €
6.Bond pricing & yieldHard
years012345P€50€50€50 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 5 years. The yield to maturity is 2.5%. Find the Macaulay duration D in years.

F = €1 000coupon = 5%n = 5YTM = 2.5%
D = years
7.CAPM & WACCMedium
WACC = ?ordinary shares€5 000 000re​ = 13%preferred shares€2 000 000rp​ = 8%debt€5 000 000rd​ = 5%T = 20%

A company is financed with ordinary shares worth €5 000 000, preferred shares worth €2 000 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 8% and the pre-tax cost of debt 5%. The tax rate is 20%. Find the WACC.

E = €5 000 000Ps = €2 000 000D = €5 000 000re = 13%rp = 8%rd = 5%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €49.00β = 0.6rf​ = 1%, rm​ = 6.5%debt40 000 × €960rd​ = 4.5%T = 30%

A company has 2 500 000 shares at €49.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 4.5% and the tax rate is 30%. The share’s beta is 0.6, the risk-free rate 1% and the market return 6.5%. Find the WACC.

β = 0.6rf = 1%rm = 6.5%rd = 4.5%T = 30%
WACC = %
9.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 24, 27, 39, 55, 56, 57.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 160.4?

μ = 170σ = 12x = 160.4
P =
11.ProbabilityMedium
xp(x)00.310.3520.35μ = 1.05

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
12.ProbabilityHard
€p(x)−50.7200.25950.05

A game costs €5 to play. You win €25 with probability 0.25 and €100 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€25: p = 0.25€100: p = 0.05
E(G) = €
13.Normal distributionMedium
460540P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 460 and 540 h? Give P as a decimal.

μ = 500σ = 40x1 = 460x2 = 540
P =
14.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
15.Confidence intervalsMedium
n = 9, s = 14L = ?Ux̄ = 64E90%

A sample of 9 items has mean 64 and standard deviation s = 14. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 9x̄ = 64s = 1490%t(8) = 1.86
lower limit =
16.Confidence intervalsHard
μ₁ − μ₂, t(20) = 2.0860LU = ?x̄₁ − x̄₂E95%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 79.5 kg, standard deviation 8 kg. Plot group 2: 16 plots, mean 86.5 kg, standard deviation 4.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the upper limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 79.5s₁ = 8n₂ = 16x̄₂ = 86.5s₂ = 4.595%t(20) = 2.086
upper limit = kg
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 30 parts has mean length 60 mm, and the line's known standard deviation is 16 mm. Line 2: a sample of 40 parts has mean 54.1 mm, known standard deviation 3 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 30x̄₁ = 60 mmσ₁ = 16 mmn₂ = 40x̄₂ = 54.1 mmσ₂ = 3 mm
z =
18.Hypothesis tests & confidence intervalsHard
σ = 10, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 10. Find the smallest sample size n.

σ = 10E = 0.599%
n =
19.Linear regression & correlationMedium
xy126220320414514612xy246152025

The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy18852188241012712630130xy202530100120ŷ = 87

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 87?

n = 5ŷ = 87
x =

3Solutions

1.Time value of money

Answer: FV = 10664 €

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 = r/m = 0.048/12 = 0.004
N = m·n = 12·6 = 72

Future value

FV = 8 000·(1 + 0.004)72 = 8 000·1.333 = €10 663.93

Answer

FV = €10 663.93
2.Time value of money

Answer: B = 111402 €

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 = 0.002, N = 360
A = 150 000·0.002/(1 − (1 + 0.002)−360) = €584.91

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 360 − 120 = 240
B = 584.9·(1 − (1 + 0.002)−240)/0.002 = €111 402.32

Answer

B = €111 402.32
3.NPV & IRR

Answer: NPV = 50000 €

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 = 27 000/(0.12 − 0.03) = 300 000.00

NPV

NPV = 300 000.00 − 250 000 = €50 000.00

Answer

NPV = €50 000.00

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: MIRR = 14.08 %

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

9 000·1.083 = 11 337.41
11 000·1.082 = 12 830.40
8 500·1.081 = 9 180.00
9 000·1.080 = 9 000.00
TV = 42 347.81

MIRR

MIRR = (42 347.81/25 000)1/4 − 1 = 14.08%

Answer

MIRR = 14.08%
5.Bond pricing & yield

Answer: ΔP = −59.26 €

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.

Price at the old yield

C = €45.00, i = 0.07, N = 10
€45.00·(1 − 0.5083)/0.07 + €1 000·0.5083 = €316.06 + €508.35 = €824.41

Price at the new yield

C = €45.00, i = 0.08, N = 10
€45.00·(1 − 0.4632)/0.08 + €1 000·0.4632 = €301.95 + €463.19 = €765.15

Change

ΔP = 765.15 − 824.41 = −€59.26

Answer

ΔP = −€59.26
6.Bond pricing & yield

Answer: D = 4.573 years

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Present value of each cash flow

t = 1: 50/1.025 = 48.78, t·PV = 48.78
t = 2: 50/1.051 = 47.59, t·PV = 95.18
t = 3: 50/1.077 = 46.43, t·PV = 139.29
t = 4: 50/1.104 = 45.30, t·PV = 181.19
t = 5: 1050/1.131 = 928.05, t·PV = 4 640.24

Duration

P = 1 116.15
D = 5 104.68/1 116.15 = 4.573

Answer

D = 4.573 years
7.CAPM & WACC

Answer: WACC = 8.417 %

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 + 2 000 000 + 5 000 000 = €12 000 000
E/V = 0.4167, Ps/V = 0.1667, D/V = 0.4167

WACC

WACC = 0.4167·13% + 0.1667·8% + 0.4167·5%·(1 − 0.2) = 8.417%

Answer

WACC = 8.417%
8.CAPM & WACC

Answer: WACC = 4.026 %

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 = 2 500 000·€49.00 = €122 500 000
D = 40 000·€960 = €38 400 000

Cost of equity (CAPM)

re = 1% + 0.6·5.5% = 4.3%

Weights

V = €160 900 000, E/V = 0.7613, D/V = 0.2387

WACC

WACC = 0.7613·4.3% + 0.2387·4.5%·(1 − 0.3) = 4.026%

Answer

WACC = 4.026%
9.Descriptive statistics

Answer: s = 15.11

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 = 258/6 = 43

Squared deviations

Σ(x − x̄)² = (−19)² + (−16)² + (−4)² + 12² + 13² + 14² = 1142

Standard deviation

s = √(1142/5) = 15.11

Answer

s = 15.11
10.Descriptive statistics

Answer: P = 0.2119

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 0.2119

Answer

P = 0.2119
11.Probability

Answer: σ = 0.8047

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 = 0·0.3 + 1·0.35 + 2·0.35 = 1.05

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75

Variance

Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047

Answer

σ = 0.8047
12.Probability

Answer: E(G) = 6.25 €

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: 20 (p = 0.25), 95 (p = 0.05), −5 (p = 0.7)

Expected value

E(G) = 20·0.25 + 95·0.05 − 5·0.7 = 6.25 €

Answer

E(G) = 6.25 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.6827

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-scores

z₁ = (460 − 500)/40 = −1
z₂ = (540 − 500)/40 = 1

Area between

P = Φ(1) − Φ(−1) = 0.8413 − 0.1587 = 0.6827

Answer

P = 0.6827
14.Normal distribution

Answer: P = 0.9452

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 = 25.04 − 25 = 0.04 mm, σ = 0.025 mm

Standardise

z = (0 − 0.04)/0.025 = −1.6

Probability

P = 1 − Φ(−1.6) = 0.9452

Answer

P = 0.9452
15.Confidence intervals

Answer: lower limit = 55.32

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 = 8, t = 1.86

Margin of error

E = 1.86·14/√9 = 8.68

Interval

64 ± 8.68 → [55.32, 72.68]

Answer

lower limit = 55.32
16.Confidence intervals

Answer: upper limit = −1.423 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 = (5·8² + 15·4.5²)/20 = 31.19
sp = 5.585

Margin of error

E = 2.086·5.585·√(1/6 + 1/16) = 5.577

Interval

−7 ± 5.577 → [−12.58, −1.423]

Answer

upper limit = −1.423 kg
17.Hypothesis tests & confidence intervals

Answer: z = 1.994

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 = √(16²/30 + 3²/40) = √8.758 = 2.959

Test statistic

z = (60 − 54.1)/2.959 = 1.994

Answer

z = 1.994

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 2655

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 = (2.576·10/0.5)² = 2654

Round up

n = 2655

Answer

n = 2655
19.Linear regression & correlation

Answer: r² = 0.9059

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 = 106/6 = 17.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −47
Syy = Σ(y − ȳ)² = 139.3

Coefficient of determination

r² = −47²/(17.5·139.3) = 0.9059

Answer

r² = 0.9059

About 90.59% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: x = 19.55

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 = 120/5 = 24
ȳ = 530/5 = 106

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 384

Regression line

b = 384/90 = 4.267
a = 106 − 4.267·24 = 3.6
ŷ = 3.6 + 4.267·x

Solve for x

x = (87 − 3.6)/4.267 = 19.55

Answer

x = 19.55

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