15.4 Normal Distribution, Statistical Sampling, Correlation, and Regression

15.4 Normal Distribution, Statistical Sampling, Correlation, and Regression

0. Descriptive Statistics from Raw Sample Data

Before Normal models and regression, Gaokao statistics questions often ask you to organize a raw sample and compute its basic features. Treat this as the mandatory descriptive layer of 统计.

Frequency tables, histograms, and stem-and-leaf plots

Given a sample of size $n$, group values into class intervals and record frequencies $f_i$ (counts) and relative frequencies $f_i/n$.

Worked Example (raw sample). Scores: $72, 85, 78, 90, 66, 85, 73, 88, 79, 84$.

Stem-and-leaf (tens | ones):

6 | 6
7 | 2 3 8 9
8 | 4 5 5 8
9 | 0

The sample is unimodal and slightly left-concentrated in the 70s–80s; a histogram with width-10 classes would show the same pattern.

Sample mean, variance, and quartiles

For raw data $x_1,\dots,x_n$:

• Sample mean: $\bar{x} = \dfrac{1}{n}\sum_{i=1}^n x_i$
• Sample variance (population-style form often used in high-school contexts): $s^2 = \dfrac{1}{n}\sum_{i=1}^n (x_i-\bar{x})^2$ (some problems use $n-1$; follow the formula stated in the question)
• Sample standard deviation: $s = \sqrt{s^2}$
• Ordered sample $x_{(1)} \le \cdots \le x_{(n)}$; median is the middle value (or average of two middle values). Quartiles $Q_1,Q_3$ split the lower/upper halves; the interquartile range is $IQR = Q_3 - Q_1$. A box plot marks min, $Q_1$, median, $Q_3$, max.

Worked Example (continued). For the ten scores above, $\sum x_i = 800$, so $\bar{x} = 80$. Using the $1/n$ variance form: $s^2 = \frac{1}{10}\left[(72-80)^2+(85-80)^2+\cdots+(84-80)^2\right]=\frac{544}{10}=54.4,\quad s=\sqrt{54.4}\approx 7.38.$ Ordered list: $66,72,73,78,79,84,85,85,88,90$. Median $= \frac{79+84}{2} = 81.5$. Lower half median $Q_1 = 73$, upper half median $Q_3 = 85$, so $IQR = 12$.

These descriptive summaries are what sampling later estimates for a population, and they feed the correlation/regression displays that follow.

1. Normal Distribution

The Normal Distribution (or Gaussian distribution) is the most fundamental continuous probability distribution in statistics, describing phenomena affected by numerous small, independent random forces.

1.1 Probability Density Function

A continuous random variable $X$ follows a Normal Distribution with mean $\mu$ and standard deviation $\sigma > 0$, denoted $X \sim N(\mu, \sigma^2)$, if its probability density function (PDF) is: $f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \quad -\infty < x < \infty$

1.2 Key Geometric Properties of the Normal Curve

1. Symmetry: The curve is bell-shaped and symmetric about the vertical line $x = \mu$.
2. Maximum Value: The maximum value occurs at $x = \mu$, where $f(\mu) = \frac{1}{\sqrt{2\pi}\sigma}$. As $\sigma$ decreases, the peak becomes taller and narrower; as $\sigma$ increases, the curve flattens.
3. Asymptotes: The x-axis ($y = 0$) is a horizontal asymptote as $x \to \pm\infty$.
4. Inflection Points: The curve has inflection points at $x = \mu - \sigma$ and $x = \mu + \sigma$.
5. Standard Normal Distribution: When $\mu = 0$ and $\sigma = 1$, $Z \sim N(0, 1)$ is the Standard Normal Distribution. Any $X \sim N(\mu, \sigma^2)$ can be standardized via $Z = \frac{X - \mu}{\sigma}$.

1.3 The $3\sigma$ Empirical Rule

For $X \sim N(\mu, \sigma^2)$, the area under the PDF curve yields standard probabilities:

• $P(\mu - \sigma \le X \le \mu + \sigma) \approx 0.6827$
• $P(\mu - 2\sigma \le X \le \mu + 2\sigma) \approx 0.9545$
• $P(\mu - 3\sigma \le X \le \mu + 3\sigma) \approx 0.9973$

Due to symmetry about $x = \mu$: $P(X \le \mu) = P(X \ge \mu) = 0.5$ $P(X \le \mu - a) = P(X \ge \mu + a) = 1 - P(X \le \mu + a)$

---

2. Statistical Sampling Methods

Statistical inference relies on selecting representative samples from a target population of size $N$.

2.1 Simple Random Sampling

Every combination of $n$ units has an equal probability of selection.

• Lottery Method: Draw numbered tokens from an urn (suitable for small $N$).
• Random Number Table: Select units matching pseudo-random digits.

2.2 Stratified Random Sampling

When the population consists of distinct sub-groups (strata) with high between-stratum variability and low within-stratum variability:

1. Divide population $N$ into $k$ mutually exclusive strata of sizes $N_1, N_2, \dots, N_k$ ($\sum N_i = N$).
2. Sample $n_i$ items from stratum $i$ proportionally to its size: $\frac{n_i}{N_i} = \frac{n}{N} \implies n_i = n \times \frac{N_i}{N}$
3. Sample Mean Formula: $\bar{x} = \sum_{i=1}^k \frac{N_i}{N} \bar{x}_i$ where $\bar{x}_i$ is the sample mean of stratum $i$.

---

3. Bivariate Correlation and Linear Regression Analysis

3.1 Scatter Plots and Pearson Correlation Coefficient $r$

For paired sample data $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$, the sample correlation coefficient $r$ measures the strength and direction of linear association: $r = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^n (x_i - \bar{x})^2} \sqrt{\sum_{i=1}^n (y_i - \bar{y})^2}} = \frac{\sum_{i=1}^n x_i y_i - n \bar{x} \bar{y}}{\sqrt{(\sum x_i^2 - n \bar{x}^2)(\sum y_i^2 - n \bar{y}^2)}}$

• $-1 \le r \le 1$.
• $r > 0$: Positive linear correlation; $r < 0$: Negative linear correlation.
• $|r| \to 1$: Strong linear relationship; $|r| \to 0$: Weak linear relationship.

3.2 Least Squares Linear Regression

When a strong linear relationship exists, we fit the linear model $\hat{y} = \hat{b}x + \hat{a}$ by minimizing the sum of squared residual errors $\sum_{i=1}^n (y_i - \hat{y}_i)^2$.

Estimation Formulas:

$\hat{b} = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^n (x_i - \bar{x})^2} = \frac{\sum_{i=1}^n x_i y_i - n \bar{x} \bar{y}}{\sum_{i=1}^n x_i^2 - n \bar{x}^2}$ $\hat{a} = \bar{y} - \hat{b} \bar{x}$

Critical Property of Linear Regression:

The regression line always passes through the sample centroid $(\bar{x}, \bar{y})$, satisfying $\bar{y} = \hat{b} \bar{x} + \hat{a}$.

---

4. $2 \times 2$ Contingency Table and $\chi^2$ Independence Test

To test whether two categorical variables $A$ and $B$ are statistically independent, we construct a $2 \times 2$ contingency table:

4.1 Test Statistic $K^2$ ($\chi^2$) Formula

$K^2 = \frac{n(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}$

4.2 Decision Rule Table (Critical Values)

Compare calculated $K^2$ against standard critical thresholds $x_\alpha$:

• If $K^2 \ge 3.841$, we have at least $95%$ confidence ($p < 0.05$) to reject independence and conclude $A$ and $B$ are associated.
• If $K^2 < 3.841$, there is insufficient evidence at the $5%$ significance level to claim association.

---

5. Worked Gaokao Exam Examples

Example 1: Normal Distribution Probability Calculation

Problem: The weight $X$ of apples produced by an orchard follows a normal distribution $N(200, 100)$ (in grams, so $\mu = 200$, $\sigma^2 = 100 \implies \sigma = 10$).

1. Find $P(190 \le X \le 220)$.
2. If 10,000 apples are harvested, estimate the number of apples weighing more than $220\text{ g}$.

Detailed Solution:

1. Express intervals in terms of $\mu$ and $\sigma$:

   • $\mu - \sigma = 200 - 10 = 190$
   • $\mu + \sigma = 200 + 10 = 210$
   • $\mu + 2\sigma = 200 + 20 = 220$

   We know from the empirical rule:

   • $P(190 \le X \le 210) = P(\mu-\sigma \le X \le \mu+\sigma) \approx 0.6827$
   • $P(180 \le X \le 220) = P(\mu-2\sigma \le X \le \mu+2\sigma) \approx 0.9545$

   By symmetry:

   • $P(X \ge 220) = \frac{1 - P(180 \le X \le 220)}{2} \approx \frac{1 - 0.9545}{2} = \frac{0.0455}{2} = 0.02275$
   • $P(X \le 190) = \frac{1 - P(190 \le X \le 210)}{2} \approx \frac{1 - 0.6827}{2} = \frac{0.3173}{2} = 0.15865$

   Therefore: $P(190 \le X \le 220) = 1 - P(X < 190) - P(X > 220) = 1 - 0.15865 - 0.02275 = 0.8186$

2. Expected count weighing $> 220\text{ g}$: $N_{>220} = 10000 \times P(X > 220) = 10000 \times 0.02275 \approx 227.5 \approx 228 \text{ apples}$

Example 2: Comprehensive Statistics (Regression + $\chi^2$ Test)

Problem: A study investigated the relationship between weekly study hours ($x$) and exam score ($y$) for 5 students:

1. Calculate the linear regression equation $\hat{y} = \hat{b}x + \hat{a}$.
2. Predict the score of a student who studies 7 hours per week.
3. In a survey of 100 students on whether studying $> 5$ hours improves passing rates, a $2 \times 2$ table gave $a=40, b=10, c=20, d=30$. Test if studying $>5$ hours is associated with passing at the $99%$ confidence level ($x_{0.01} = 6.635$).

Detailed Solution:

1. Regression Calculations:

   • $\bar{x} = \frac{2 + 4 + 5 + 6 + 8}{5} = \frac{25}{5} = 5$
   • $\bar{y} = \frac{60 + 70 + 75 + 80 + 90}{5} = \frac{375}{5} = 75$
   • $\sum x_i^2 = 2^2 + 4^2 + 5^2 + 6^2 + 8^2 = 4 + 16 + 25 + 36 + 64 = 145$
   • $\sum x_i y_i = (2 \times 60) + (4 \times 70) + (5 \times 75) + (6 \times 80) + (8 \times 90) = 120 + 280 + 375 + 480 + 720 = 1975$

   Now compute slope $\hat{b}$: $\hat{b} = \frac{\sum x_i y_i - 5 \bar{x} \bar{y}}{\sum x_i^2 - 5 \bar{x}^2} = \frac{1975 - 5(5)(75)}{145 - 5(25)} = \frac{1975 - 1875}{145 - 125} = \frac{100}{20} = 5$

   Compute intercept $\hat{a}$: $\hat{a} = \bar{y} - \hat{b} \bar{x} = 75 - 5(5) = 50$

   Thus, the regression equation is: $\hat{y} = 5x + 50$

2. Prediction for $x = 7$: $\hat{y} = 5(7) + 50 = 35 + 50 = 85 \text{ points}$

3. $\chi^2$ Independence Test: $n = 40 + 10 + 20 + 30 = 100$. $ad - bc = (40 \times 30) - (10 \times 20) = 1200 - 200 = 1000$. $(ad - bc)^2 = 1000^2 = 1,000,000$. Row totals: $a+b = 50$, $c+d = 50$. Column totals: $a+c = 60$, $b+d = 40$. $K^2 = \frac{100 \times 1,000,000}{50 \times 50 \times 60 \times 40} = \frac{100,000,000}{6,000,000} = \frac{100}{6} \approx 16.667$

   Since $K^2 \approx 16.667 > 6.635 = x_{0.01}$ (and even $> 10.828 = x_{0.001}$), we have over $99.9%$ confidence to reject independence and conclude that studying $>5$ hours per week is significantly associated with passing.