Comprehensive Study Guide: Statistics, Probability, and Complex Numbers

Measures of Central Tendency and Data Spread

  • Median Calculation: The median is defined as the average of the two middle values after the data set has been sorted in ascending or descending order.

  • Specific Problem (Q47):     * Dataset provided: $6, 68, 93, 82, 76, 110$. (Based on scores of 6 students).     * Sorting the data: $6, 68, 76, 82, 93, 110$.     * The middle two values are $76$ and $82$.     * Calculation: Median=76+822=79\text{Median} = \frac{76 + 82}{2} = 79.

  • Measures of Dispersion:     * These include the Range, Variance, and Standard Deviation.     * The Arithmetic Mean is a measure of central tendency, not dispersion.

  • Comparing Standard Deviation (Q49):     * To find the data set with the largest standard deviation, look for the set with the greatest spread (Range).     * A useful technique is to delete identical values among choices to identify which range is largest.     * Data Sets:         * A: $14, 10, 12, 11, 13, 13$         * B: $14, 10, 15, 11, 13, 13$         * C: $11, 10, 20, 11, 13, 13$         * D: $14, 10, 30, 11, 13, 13$     * Dataset D has the largest outlier (3030), resulting in the largest range and thus the largest standard deviation.

Normal Distribution Analysis

  • Empirical Rule Application (Q50):     * Total population: 10,00010,000 batteries.     * Mean (μ\mu): 300300 days.     * Standard Deviation (σ\sigma): 4040 days.     * Range of interest: 260260 to 340340 days.     * This range represents μσ\mu - \sigma to μ+σ\mu + \sigma.     * According to the normal distribution percentage, 68%68\% of data falls within one standard deviation.     * Calculation: 0.68×10,000=6,8000.68 \times 10,000 = 6,800 batteries.

  • Probability and Percentages (Q51-Q53):     * Scenario 1: μ=2\mu = 2, σ=1\sigma = 1. To find the probability that x > 3:         * 3=μ+1σ3 = \mu + 1\sigma.         * The percentage above 1σ1\sigma is calculated by subtracting the values below it (50%+34%=84%50\% + 34\% = 84\%) from 100%100\%.         * Result: 16%16\%.     * Scenario 2: μ=25\mu = 25, σ=2\sigma = 2. To find the probability that x < 27:         * 27=μ+1σ27 = \mu + 1\sigma.         * The area to the left of +1σ+1\sigma includes the mean and below (50%50\%) plus the region between the mean and +1σ+1\sigma (34%34\%).         * Result: 84%84\%.     * Scenario 3: μ=12\mu = 12, σ=2\sigma = 2. To find P(10 < x < 16).         * Lower bound 10=μ1σ10 = \mu - 1\sigma; Upper bound 16=μ+2σ16 = \mu + 2\sigma.         * Area breakdown: Between μ1σ and μ=34%\text{Between } \mu-1\sigma \text{ and } \mu = 34\%, Between μ and μ+1σ=34%\text{Between } \mu \text{ and } \mu+1\sigma = 34\%, Between μ+1σ and μ+2σ=13.5%\text{Between } \mu+1\sigma \text{ and } \mu+2\sigma = 13.5\%.         * Total probability: 34%+34%+13.5%=81.5%34\% + 34\% + 13.5\% = 81.5\%.

  • Finding Parameters from Range (Q54-Q55):     * Standard Deviation from 99% Range: If 99%99\% of students score between 1313 and 4949.         * In prep materials, 99%99\% is often treated as covering approximately 6σ6\sigma (from μ3σ\mu - 3\sigma to μ+3σ\mu + 3\sigma).         * 4913=3649 - 13 = 36.         * 6σ=36    σ=66\sigma = 36 \implies \sigma = 6.         * Note: The excluded 1%1\% represents 0.5%0.5\% on the right tail and 0.5%0.5\% on the left tail.     * Mean from 95% Range: If 95%95\% of students weigh between 5757 and 6565.         * The mean is the midpoint of the range.         * Calculation: μ=57+652=61\mu = \frac{57 + 65}{2} = 61.

Binomial Distribution and Skewness

  • Visual Analysis of Skewness:     * A graph is described as having a Positive Skew (Skewed Right) when the tail extends toward the positive direction (higher values) on the x-axis.

  • Binomial Mean and Standard Deviation Formulas:     * Mean: μ=n×p\mu = n \times p     * Variance: σ2=n×p×q\sigma^2 = n \times p \times q     * Standard Deviation: σ=n×p×q\sigma = \sqrt{n \times p \times q}     * Where nn is the number of trials, pp is the probability of success, and qq is the probability of failure (1p1-p).

  • Calculating the Mean (Q57):     * Given: p=35%p = 35\% (or 0.350.35), n=4n = 4.     * Calculation: μ=4×0.35=1.4\mu = 4 \times 0.35 = 1.4.

  • Calculating Standard Deviation (Q58):     * Given: n=20n = 20, μ=12\mu = 12.     * First, find pp: 12=20×p    p=0.612 = 20 \times p \implies p = 0.6.     * Then, find qq: q=10.6=0.4q = 1 - 0.6 = 0.4.     * Calculate Standard Deviation: σ=20×0.6×0.4=12×0.4=4.8\sigma = \sqrt{20 \times 0.6 \times 0.4} = \sqrt{12 \times 0.4} = \sqrt{4.8}.

Complex Numbers and De Moivre's Theorem

  • De Moivre's Theorem Formula: For a natural number nn, zn=[r(cos(θ)+isin(θ))]n=rn(cos(nθ)+isin(nθ))z^n = [r(\cos(\theta) + i \sin(\theta))]^n = r^n(\cos(n\theta) + i \sin(n\theta)).

  • Power Calculation Examples:     * Example 1: [2(cos(22.5)+isin(22.5))]4[2(\cos(22.5^\circ) + i \sin(22.5^\circ))]^4         * 24=162^4 = 16         * Angle: 4×22.5=904 \times 22.5^\circ = 90^\circ         * Result: 16(cos(90)+isin(90))=16(0+i×1)=16i16(\cos(90^\circ) + i \sin(90^\circ)) = 16(0 + i \times 1) = 16i.     * Example 2: (cos(15)+icos(75))6(\cos(15^\circ) + i \cos(75^\circ))^6         * Using the identity cos(θ)=sin(90θ)\cos(\theta) = \sin(90^\circ - \theta), we find cos(75)=sin(15)\cos(75^\circ) = \sin(15^\circ).         * The expression simplifies to (cos(15)+isin(15))6(\cos(15^\circ) + i \sin(15^\circ))^6.         * Applying De Moivre's: cos(6×15)+isin(6×15)=cos(90)+isin(90)=i\cos(6 \times 15^\circ) + i \sin(6 \times 15^\circ) = \cos(90^\circ) + i \sin(90^\circ) = i.

  • Division of Complex Numbers: r1(cos(θ1)+isin(θ1))r2(cos(θ2)+isin(θ2))=r1r2(cos(θ1θ2)+isin(θ1θ2))\frac{r_1(\cos(\theta_1) + i \sin(\theta_1))}{r_2(\cos(\theta_2) + i \sin(\theta_2))} = \frac{r_1}{r_2}(\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)).     * Example: 20(cos(80)+isin(80))4(cos(20)+isin(20))=5(cos(60)+isin(60))\frac{20(\cos(80^\circ) + i \sin(80^\circ))}{4(\cos(20^\circ) + i \sin(20^\circ))} = 5(\cos(60^\circ) + i \sin(60^\circ))     * Rectangular conversion: 5(12+i32)=2.5+2.53i5(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2.5 + 2.5\sqrt{3}i.

Roots of Complex Numbers

  • n-th Root Formula: Wk=rn[cos(θ+2kπn)+isin(θ+2kπn)]W_k = \sqrt[n]{r} [\cos(\frac{\theta + 2k\pi}{n}) + i \sin(\frac{\theta + 2k\pi}{n})] for k=0,1,2,,n1k = 0, 1, 2, …, n-1.

  • Key Properties:     * The Modulus of all roots is the same: rn\sqrt[n]{r}.     * Example (Q18/Q20): Finding the third roots of 8(cos(θ)+isin(θ))8(\cos(\theta) + i \sin(\theta)). The modulus of any root (including the second root) is 83=2\sqrt[3]{8} = 2.     * Roots of Unity: The modulus of any $n$-th root of 11 is always 11.

  • Finding the Argument (Amplitude) of Roots:     * Example: Finding the fifth roots of 3(cos(π3)+isin(π3))\sqrt{3}(\cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3})).     * The argument of the first root (k=0k=0) is θn=π/35=π15\frac{\theta}{n} = \frac{\pi/3}{5} = \frac{\pi}{15}.

Polar Coordinates and Equations

  • Coordinate Representation: A point is given by (r,θ)(r, \theta), where rr is the directed distance from the pole (origin) and θ\theta is the directed angle from the polar axis.

  • Equivalent Representations:     * (r,θ)(r, \theta) is equivalent to (r,θ+360k)(r, \theta + 360^\circ k) or (r,θ+180)(-r, \theta + 180^\circ).

  • Distance in Polar Coordinates: The distance between (r1,θ1)(r_1, \theta_1) and (r2,θ2)(r_2, \theta_2) is determined by the formula: d=r12+r222r1r2cos(θ2θ1)d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_2 - \theta_1)}.     * Example: Distance between (5,120)(5, 120^\circ) and (2,300)(2, 300^\circ). The angle difference is 180180^\circ. Distance is 5+2=7|5| + |2| = 7. (Calculation: 52+222(5)(2)cos(180)=25+420(1)=49=7\sqrt{5^2 + 2^2 - 2(5)(2)\cos(180^\circ)} = \sqrt{25 + 4 - 20(-1)} = \sqrt{49} = 7).

  • Polar Curves Identification:     * r=ar = a: Represents a circle centered at the pole with radius a|a|.     * θ=α\theta = \alpha: Represents a straight line passing through the pole with a slope of tan(α)\tan(\alpha).

  • Conversions (Polar to Rectangular):     * x=rcos(θ)x = r \cos(\theta), y=rsin(θ)y = r \sin(\theta).     * Example Equation: y=x2y = x^2 transformed to polar form.         * rsin(θ)=(rcos(θ))2r \sin(\theta) = (r \cos(\theta))^2         * rsin(θ)=r2cos2(θ)r \sin(\theta) = r^2 \cos^2(\theta)         * r=sin(θ)cos2(θ)=tan(θ)sec(θ)r = \frac{\sin(\theta)}{\cos^2(\theta)} = \tan(\theta) \sec(\theta).

Vocabulary and Principles

  • Probability Distribution: A function that connects each outcome of the sample space to its probability of occurring.

  • Correlation: A relationship between two events or variables.

  • Bias: A design in a survey or study that intentionally favors certain results over others.

  • Control Group: A group in an experimental study that receives a formal treatment (or no treatment) that has no actual effect on the result, used for comparison.

  • Margin of Error: Defines the interval representing the potential difference in response between the sample and the population.

  • Argand Plane: A coordinate system where the x-axis represents the real parts and the y-axis represents the imaginary parts of complex numbers.

  • Modulus: The absolute value or distance from the origin for a complex number.

  • Amplitude (Symmetry): The angle θ\theta in the polar form of a complex number.

  • Pole: The fixed point (origin) in the polar coordinate system.

  • Polar Axis: The horizontal ray extending to the right from the pole.