Find the sample standard deviation s of the data: 37, 14, 27, 24, 24, 37.
n = 6
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 15. What proportion P of values is below 140?
μ = 170σ = 15x = 140
P =
19.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.6772. Find z to two decimals.
A = 0.6772
z =
20.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points40%
x = points
3Solutions
1.Limits
Answer: L = 2
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
Multiply by the conjugate
f(x) = 4x/(√(x2 + 4x) + x)
Divide by x
f(x) = 4/(√(1 + 4/x) + 1)
Let x → ∞
L = 4/(1 + 1) = 2
Answer
L = 2
2.Limits
Answer: L = 4
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
L'Hôpital's rule
L = limx→0 [4/(1 + 4x)]/[cos x]
Substitute x = 0
L = 4/1 = 4
Answer
L = 4
3.Derivatives
Answer: f′(−1) = 240
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Chain rule
f′(x) = 5·(3x + 5)4·3
Substitute
f′(−1) = 5·24·3 = 240
Answer
f′(−1) = 240
4.Derivatives
Answer: x = 4
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Derivative
f′(x) = 3x2 − 21x + 36 = 3·(x2 − 7x + 12)
Solve f′(x) = 0
x = 3 or x = 4
Which one is which
f″(x) = 6x − 21: negative at a maximum, positive at a minimum.
f″(3) = −3, f″(4) = 3
x = 4
Answer
x = 4
5.Integrals
Answer: V = 28.27
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Disc method
V = π∫₀3 1x² dx
Integrate
V = π·1·3³/3 = 28.27
Answer
V = 28.27
Check: a cone of radius 3 and height 3 has (1/3)πr²h = 28.27.
6.Integrals
Answer: A = 18
Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.
Intersection points
0.5x2 = 3x → x = 0, x = 6
Integrate upper minus lower
A = [1.5x2 − 0.1667x3]06 = 18
Answer
A = 18
7.Complex numbers
Answer: θ = 75.96 °
Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.
Quadrant
a = 2, b = 8 → I
Reference angle
α = atan(8/2) = 75.96°
Argument
θ = α = 75.96°
Answer
θ = 75.96°
In polar form z = 8.246∠75.96°.
8.Complex numbers
Answer: D (3∠−45°)
De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.
Modulus and arguments of the roots
|z| = 3
θ = 45°, 135°, −135°, −45°
The point
D = 3∠−45°
Answer
D (3∠−45°)
9.Differential equations
Answer: y(0.5) = 2.951
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Characteristic roots
(r + 1)(r + 5) = 0 → r = −1, −5
Constants from the initial values
A + B = 4, −1A − 5B = 0
A = 5, B = −1
Evaluate
y(0.5) = 5·e−0.5 + (−1)·e−2.5 = 2.951
Answer
y(0.5) = 2.951
10.Differential equations
Answer: ymax = 2.416
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Steady value and roots
yss = 2, r = −1 ± 2j
Constants from the initial values
y(0) = 2 + A = 0 → A = −2
y′(0) = −1A + 2B = 0 → B = −1
Peak
t = π/2 = 1.571
ymax = 2·(1 + e−1·π/2) = 2.416
Answer
ymax = 2.416
11.Matrices & determinants
Answer: det A = −143
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
2×2 minors of the first row
M₁₁ = −19, M₁₂ = −24, M₁₃ = 21
Cofactor expansion
det A = 5·(−19) − (−2)·(−24) + 0·21 = −143
Answer
det A = −143
12.Matrices & determinants
Answer: x = −0.5909
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
Determinant of the coefficient matrix
det A = (−4)·4 − (−3)·(−2) = −22
Replace the x column
det Aₓ = 10·4 − (−3)·(−9) = 13
x = 13/(−22) = −0.5909
Answer
x = −0.5909
y = −2.545
13.Vectors: dot & cross product
Answer: θ = 29.18 °
Dot product:a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.
Dot product
a·b = (−3)·(−4) + (−4)·(−6) + 0·4 = 36
Magnitudes
|a| = √((−3)² + (−4)² + 0²) = 5
|b| = √((−4)² + (−6)² + 4²) = 8.246
Angle
cos θ = 36/(5·8.246) = 0.8731
θ = arccos(0.8731) = 29.18°
Answer
θ = 29.18°
14.Vectors: dot & cross product
Answer: A = 36.51
Cross product:a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.
Cross product
(a × b)x = aybz − azby = (−1)·(−6) − 6·3 = −12
(a × b)y = azbx − axbz = 6·1 − (−6)·(−6) = −30
(a × b)z = axby − aybx = (−6)·3 − (−1)·1 = −17
Area
A = |a × b| = √((−12)² + (−30)² + (−17)²) = 36.51
Answer
A = 36.51
15.Probability
Answer: σ = 1.152
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 = 1·0.3 + 2·0.3 + 3·0.15 + 4·0.25 = 2.35
Mean of the squares
E(X²) = 1·0.3 + 4·0.3 + 9·0.15 + 16·0.25 = 6.85
Variance
Var(X) = 6.85 − 2.35² = 1.328
σ = √1.328 = 1.152
Answer
σ = 1.152
16.Probability
Answer: σY = 2.39
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 = 2.65, Var(X) = 1.428, σX = 1.195
Linear change
σY = |−2|·1.195 = 2.39
Answer
σY = 2.39
The mean does move: E(Y) = −2·2.65 + 1 = −4.3.
17.Descriptive statistics
Answer: s = 8.796
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 0.02275
Answer
P = 0.02275
19.Normal distribution
Answer: z = −0.46
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 − 0.6772 = 0.3228
Read the z-table backwards
Find 0.3228 inside the table and read off z.
z = −0.46
Answer
z = −0.46
20.Normal distribution
Answer: x = 67.5 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.