Find the sample standard deviation s of the data: 35, 32, 20, 53, 30, 35, 59.
n = 7
s =
18.Descriptive statisticsHard
Exam scores are normally distributed with mean 500 and standard deviation 40. Find the z-score z of a score of 516.
μ = 500σ = 40x = 516
z =
19.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.6680. Find z to two decimals.
A = 0.6680
z =
20.Normal distributionHard
The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 25% of them lie (the 25th percentile).
μ = 32σ = 6P = 25%
x = min
3Solutions
1.Limits
Answer: L = −1
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).
Divide by the highest power
(−4 + (−3)/x2 + (−6)/x3)/(4 + 6/x2 + 4/x3)
Let x → ∞
L = −4/4 = −1
Answer
L = −1
For equal degrees the limit is the ratio of the leading coefficients.
2.Limits
Answer: L = 6.25
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).
First application
5·sin 5x/(4x) → 0/0
Second application
25·cos 5x/4
Substitute x = 0
L = 25/4 = 6.25
Answer
L = 6.25
3.Derivatives
Answer: f′(1) = −2.829
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.
Product and chain rule
f′(x) = sin(3x) + 3x·cos(3x)
Substitute
f′(1) = 0.1411 + 3·(−0.99) = −2.829
Answer
f′(1) = −2.829
4.Derivatives
Answer: b = 1
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.
Point and slope
f(−1) = 4
f′(x) = 6x + 3 → f′(−1) = −3
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 4 − (−3)·(−1) = 1
Answer
b = 1
5.Integrals
Answer: I = 1.5
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.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u1 du
Integrate and evaluate
I = [u2/2]12 = (4 − 1)/2 = 1.5
Answer
I = 1.5
6.Integrals
Answer: A = 10.67
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.
Zeros
4 − x² = 0 → x = ±2
Integrate
A = [4x − x³/3]−22 = 5.333 − (−5.333) = 10.67
Answer
A = 10.67
7.Complex numbers
Answer: b = −1.915
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.
Components
a = 2.5·cos(−130°) = −1.607
b = 2.5·sin(−130°) = −1.915
Rectangular form
z = −1.607 − j1.915
b = −1.915
Answer
b = −1.915
8.Complex numbers
Answer: Re(z⁵) = −4
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.
Powers step by step
z² = (1 + j)(1 + j) = j2
z³ = (j2)(1 + j) = −2 + j2
z⁴ = (−2 + j2)(1 + j) = −4
z⁵ = (−4)(1 + j) = −4 − j4
Check with De Moivre
z = 1.414∠45° → z⁵ = 5.657∠−135°
Re(z⁵) = −4
Answer
Re(z⁵) = −4
9.Differential equations
Answer: y(2) = 0.009728
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.
Double root
(r + 4)² = 0 → r = −4
Constants
A = y(0) = 3
y′(0) = B − 4A → B = 1 + 12 = 13
Evaluate
y(2) = (3 + 13·2)·e−8 = 0.009728
Answer
y(2) = 0.009728
10.Differential equations
Answer: y(0.25) = 2.551
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.
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.
Row times column
c11 = 5·2 + (−6)·3 = −8
Answer
c11 = −8
12.Matrices & determinants
Answer: x = −2.056
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 = (−1)·3 − 5·3 = −18
Replace the x column
det Aₓ = (−1)·3 − 5·(−8) = 37
x = 37/(−18) = −2.056
Answer
x = −2.056
y = −0.6111
13.Vectors: dot & cross product
Answer: compb a = 8.485
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 = (−7)·0 + 7·4 + (−5)·(−4) = 48
Length of b
|b| = √(0² + 4² + (−4)²) = 5.657
Scalar projection
compb a = a·b/|b| = 48/5.657 = 8.485
Answer
compb a = 8.485
14.Vectors: dot & cross product
Answer: A = 9.11 m²
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.
Edge vectors
AB = B − A = (0, 2, 2)
AC = C − A = (1, 3, −6)
Cross product
(AB × AC)x = AByACz − ABzACy = 2·(−6) − 2·3 = −18
(AB × AC)y = ABzACx − ABxACz = 2·1 − 0·(−6) = 2
(AB × AC)z = ABxACy − AByACx = 0·3 − 2·1 = −2
|AB × AC| = √((−18)² + 2² + (−2)²) = 18.22
Area of the triangle
A = ½·18.22 = 9.11 m²
Answer
A = 9.11 m²
15.Probability
Answer: N = 165
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(11, 3) = (11·10·9)/(3·2·1) = 165
Answer
N = 165
16.Probability
Answer: σY = 46.1
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 = 25, Var(X) = 85, σX = 9.22
Linear change
σY = |5|·9.22 = 46.1
Answer
σY = 46.1
The mean does move: E(Y) = 5·25 + 1 = 126.
17.Descriptive statistics
Answer: s = 13.59
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 = (516 − 500)/40 = 0.4
Answer
z = 0.4
19.Normal distribution
Answer: z = 0.97
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 + 0.6680)/2 = 0.8340
Read the z-table backwards
Find 0.8340 inside the table and read off z.
z = 0.97
Answer
z = 0.97
20.Normal distribution
Answer: x = 27.95 min
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.