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
years01234567PV = ?€20 000

You need €20 000 in 7 years. An account pays 4.8% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €20 0004.8% / 12n = 7
PV = €
2.Time value of moneyHard
years0123…30€1 000€1 000€1 000FV = ?

You deposit €1 000 at the start of every year for 30 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 30? Find FV.

A = €1 000i = 5%n = 30
FV = €
3.NPV & IRRMedium
years01234−45 00011 00022 00016 0009 000

An investment of €45 000 gives these cash flows at the end of years 1 to 4: €11 000, €22 000, €16 000, €9 000. The cost of capital is 9%. Find the net present value NPV.

r = 9%CF0 = −45 000CF1 = 11 000CF2 = 22 000CF3 = 16 000CF4 = 9 000
NPV = €
4.NPV & IRRHard
years0123−10 0003 0005 0003 500

A project costs €10 000 and returns €3 000, €5 000, €3 500 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.

r = 6%CF0 = −10 000CF1 = 3 000CF2 = 5 000CF3 = 3 500
MIRR = %
5.Bond pricing & yieldMedium

You buy a bond with 12 years to maturity (face value €1 000, annual coupon 4.5%) when its yield to maturity is 5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 4%. Find your holding period return HPR in percent.

F = €1 000coupon = 4.5%n = 12y0 = 5%y1 = 4%
HPR = %
6.Bond pricing & yieldHard
years01234567€1 060€50€50 + €1 000

A bond with a face value of €1 000 and an annual coupon of 5% matures in 7 years and sells for €1 060. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 7P = €1 060
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€4 000 000re​ = 13%debt€1 000 000rd​ = 3%T = 25.8%

A company has equity worth €4 000 000 and debt worth €1 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.

E = €4 000 000D = €1 000 000re = 13%rd = 3%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €25.50β = 1.1rf​ = 2.5%, rm​ = 9%debt5 000 × €1 050rd​ = 6%T = 20%

A company has 2 000 000 shares at €25.50 each and 5 000 bonds trading at €1 050 each. The pre-tax cost of debt is 6% and the tax rate is 20%. The share’s beta is 1.1, the risk-free rate 2.5% and the market return 9%. Find the WACC.

β = 1.1rf = 2.5%rm = 9%rd = 6%T = 20%
WACC = %
9.Descriptive statisticsMedium
101520253035404550

Find the sample standard deviation s of the data: 50, 10, 48, 27, 24, 28, 43.

n = 7
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is above 593.8?

μ = 500σ = 75x = 593.8
P =
11.ProbabilityMedium
xp(x)10.2520.130.340.2550.1μ = 2.85

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

x: 1, 2, 3, 4, 5p(x): 0.25, 0.1, 0.3, 0.25, 0.1
σ =
12.ProbabilityHard
xp(x)00.220.240.2560.35μ

X has the distribution shown, and Y = 3X + 10. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.2, 0.2, 0.25, 0.35Y = 3X + 10
σY =
13.Normal distributionMedium
502532P = ?

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 502 and 532 h? Give P as a decimal.

μ = 500σ = 40x1 = 502x2 = 532
P =
14.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 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 = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
n = 13, sd​ = 5.40L = ?Ud̄ = 1.8E99%

The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 1.8 and standard deviation 5.4. Find the lower limit of the 99% confidence interval for the mean difference.

n = 13d̄ = 1.8sd = 5.499%t(12) = 3.055
lower limit = mmHg
16.Confidence intervalsHard
σ = 9, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 9. 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 4?

σ = 9E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 100 parts has mean length 87.5 mm, and the line's known standard deviation is 20 mm. Line 2: a sample of 50 parts has mean 80.2 mm, known standard deviation 14 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 100x̄₁ = 87.5 mmσ₁ = 20 mmn₂ = 50x̄₂ = 80.2 mmσ₂ = 14 mm
z =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 38 of 150 patients on treatment A recover, and 33 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 38n₁ = 150x₂ = 33n₂ = 100
z =
19.Linear regression & correlationMedium
xy127223320421518xy242025

The table gives 5 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 = 5
r² =
20.Linear regression & correlationHard
xy171041910621130231422515327157xy2025100120140160x = 26

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 26.

n = 6x = 26
ŷ =

3Solutions

1.Time value of money

Answer: PV = 14302 €

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 = 0.048/12 = 0.004
N = 12·7 = 84

Present value

PV = FV/(1 + i)N = 20 000/1.398 = €14 302.04

Answer

PV = €14 302.04
2.Time value of money

Answer: FV = 69761 €

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.

Ordinary annuity factor

((1 + 0.05)30 − 1)/0.05 = 66.44

Annuity due

Each deposit is invested one year longer.

FV = 1 000·66.44·1.05 = €69 760.79

Answer

FV = €69 760.79
3.NPV & IRR

Answer: NPV = 2339 €

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 = −45 000
CF1/(1 + 0.09)1 = 11 000/1.09 = 10 091.74
CF2/(1 + 0.09)2 = 22 000/1.188 = 18 516.96
CF3/(1 + 0.09)3 = 16 000/1.295 = 12 354.94
CF4/(1 + 0.09)4 = 9 000/1.412 = 6 375.83

Add them up

NPV = €2 339.47

Answer

NPV = €2 339.47

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

4.NPV & IRR

Answer: MIRR = 6.768 %

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

3 000·1.062 = 3 370.80
5 000·1.061 = 5 300.00
3 500·1.060 = 3 500.00
TV = 12 170.80

MIRR

MIRR = (12 170.80/10 000)1/3 − 1 = 6.768%

Answer

MIRR = 6.768%
5.Bond pricing & yield

Answer: HPR = 13.93 %

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.

Purchase price

C = €45.00, i = 0.05, N = 12
€45.00·(1 − 0.5568)/0.05 + €1 000·0.5568 = €398.85 + €556.84 = €955.68

Selling price

C = €45.00, i = 0.04, N = 11
€45.00·(1 − 0.6496)/0.04 + €1 000·0.6496 = €394.22 + €649.58 = €1 043.80

Holding period return

HPR = (45 + 1 043.80 − 955.68)/955.68 = 13.93%

Answer

HPR = 13.93%
6.Bond pricing & yield

Answer: YTM = 4 %

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 = €50·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷

Solve numerically

P(4%) = €1 060.02, P(5%) = €1 000.00
YTM = 4%

Answer

YTM = 4%
7.CAPM & WACC

Answer: WACC = 10.85 %

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 = 0.8·13% + 0.2·3%·(1 − 0.258) = 10.85%

Answer

WACC = 10.85%
8.CAPM & WACC

Answer: WACC = 9.197 %

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·€25.50 = €51 000 000
D = 5 000·€1 050 = €5 250 000

Cost of equity (CAPM)

re = 2.5% + 1.1·6.5% = 9.65%

Weights

V = €56 250 000, E/V = 0.9067, D/V = 0.09333

WACC

WACC = 0.9067·9.65% + 0.09333·6%·(1 − 0.2) = 9.197%

Answer

WACC = 9.197%
9.Descriptive statistics

Answer: s = 14.63

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 = 230/7 = 32.86

Squared deviations

Σ(x − x̄)² = 17.14² + (−22.86)² + 15.14² + (−5.857)² + (−8.857)² + (−4.857)² + 10.14² = 1285

Standard deviation

s = √(1285/6) = 14.63

Answer

s = 14.63
10.Descriptive statistics

Answer: P = 0.1056

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 = 1.25

Standard normal table

Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056

Answer

P = 0.1056
11.Probability

Answer: σ = 1.314

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 = 1·0.25 + 2·0.1 + 3·0.3 + 4·0.25 + 5·0.1 = 2.85

Mean of the squares

E(X²) = 1·0.25 + 4·0.1 + 9·0.3 + 16·0.25 + 25·0.1 = 9.85

Variance

Var(X) = 9.85 − 2.85² = 1.728
σ = √1.728 = 1.314

Answer

σ = 1.314
12.Probability

Answer: σY = 6.808

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.

Spread of X

μ = 3.5, Var(X) = 5.15, σX = 2.269

Linear change

σY = |3|·2.269 = 6.808

Answer

σY = 6.808

The mean does move: E(Y) = 3·3.5 + 10 = 20.5.

13.Normal distribution

Answer: P = 0.2682

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 = (502 − 500)/40 = 0.05
z₂ = (532 − 500)/40 = 0.8

Area between

P = Φ(0.8) − Φ(0.05) = 0.7881 − 0.5199 = 0.2682

Answer

P = 0.2682
14.Normal distribution

Answer: P = 0.9868

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 = 0.01803 mm

Standardise

z = (0 − 0.04)/0.01803 = −2.22

Probability

P = 1 − Φ(−2.22) = 0.9868

Answer

P = 0.9868
15.Confidence intervals

Answer: lower limit = −2.775 mmHg

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.

Critical value

df = 12, t = 3.055

Margin of error

E = 3.055·5.4/√13 = 4.575

Interval

1.8 ± 4.575 → [−2.775, 6.375]

Answer

lower limit = −2.775 mmHg
16.Confidence intervals

Answer: n = 34

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 = (2.576·9/4)² = 33.59

Round up

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

n = 34

Answer

n = 34
17.Hypothesis tests & confidence intervals

Answer: z = 2.594

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 = √(20²/100 + 14²/50) = √7.92 = 2.814

Test statistic

z = (87.5 − 80.2)/2.814 = 2.594

Answer

z = 2.594

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

18.Hypothesis tests & confidence intervals

Answer: z = −1.317

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 = 38/150 = 0.2533
p̂₂ = 33/100 = 0.33
p̂ = 71/250 = 0.284

Standard error under H₀

SE = √(0.284·0.716·(1/150 + 1/100)) = 0.05822

Test statistic

z = (0.2533 − 0.33)/0.05822 = −1.317

Answer

z = −1.317

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

19.Linear regression & correlation

Answer: r² = 0.8547

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 = 109/5 = 21.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −20
Syy = Σ(y − ȳ)² = 46.8

Coefficient of determination

r² = −20²/(10·46.8) = 0.8547

Answer

r² = 0.8547

About 85.47% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 155.9

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 = 132/6 = 22
ȳ = 792/6 = 132

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 418

Regression line

b = 418/70 = 5.971
a = 132 − 5.971·22 = 0.6286
ŷ = 0.6286 + 5.971·x

Prediction

ŷ = 0.6286 + 5.971·26 = 155.9

Answer

ŷ = 155.9

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