Comprehensive Notes on the Normal Distribution and Standard Normal Curve

The Standard Normal Distribution $Z \sim N(0, 1)$

• Definition of the Standard Normal Distribution: The standard normal distribution is a specific case of the normal distribution where the mean ($\mu$) is $0$ and the standard deviation ($\sigma$) is $1$. It is denoted as $Z \sim N(0, 1)$.

• Exercise Requirements for Full Credit: According to Exercise 14H, to provide a complete solution for probability problems involving the standard normal distribution, one must:     * Draw the distribution (the bell curve).     * Shade or mark the area under the curve that is being summed.     * State the command entered into the Graphical Display Calculator (GDC).     * State the final probability value.

• Calculating Probabilities with GDC: For a standard normal variable $Z$, probabilities are found using the cumulative distribution function (CDF). For any interval $[a, b]$, the probability is $P(a < Z < b)$. Typical GDC commands follow the syntax: normalcdf(lower_bound, upper_bound, 0, 1).

• Specific Probability Examples:     * To find $P(-2 < Z < 1)$, the GDC command is normalcdf(-2, 1, 0, 1).     * To find $P(Z < 1)$, the lower bound is effectively negative infinity (often entered as $-10^{99}$), resulting in normalcdf(-1E99, 1, 0, 1).     * To find $P(Z > -1.5)$, the upper bound is effectively positive infinity, resulting in normalcdf(-1.5, 1E99, 0, 1).     * To find $P(Z < 0)$, which represents exactly half the distribution, the value is $0.5$.     * For absolute values, such as $P(|Z| > 0.8)$, the probability is the sum of the tails: $P(Z < -0.8) + P(Z > 0.8)$.     * For $P(|Z| < 0.4)$, the probability is the central area between $-0.4$ and $0.4$.

Standard Deviation Intervals and Areas

• Intervals Based on Standard Deviations: For a standard normal distribution, the units on the horizontal axis correspond directly to standard deviations from the mean.     * Between 1 and 2 standard deviations: This refers to the area $P(1 < Z < 2)$.     * Between 0.5 and 1.5 standard deviations: This refers to the area $P(0.5 < Z < 1.5)$.     * More than 1 standard deviation above the mean: This is represented as $P(Z > 1)$.     * More than 2.4 standard deviations above the mean: This is represented as $P(Z > 2.4)$.     * Less than 1 standard deviation below the mean: This is represented as $P(Z < -1)$.     * Less than 1.75 standard deviations below the mean: This is represented as $P(Z < -1.75)$.

General Normal Distributions $X \sim N(\mu, \sigma^2)$

• Transformation Logic: A general normal distribution is defined by its mean $\mu$ and its variance $\sigma^2$. Probability calculations on GDC use the mean and the standard deviation $\sigma$.

• Scenario 1: $X \sim N(14, 5^2)$:     * Mean $\mu = 14$, Standard Deviation $\sigma = 5$.     * Probability of $X < 16$: normalcdf(-1E99, 16, 14, 5).     * Probability of $X \ge 9$: normalcdf(9, 1E99, 14, 5).     * Probability of $9 \le X < 12$: normalcdf(9, 12, 14, 5).     * Probability of $X < 14$: Since $14$ is the mean, this is exactly $0.5$.

• Scenario 2: $X \sim N(48, 81)$:     * Mean $\mu = 48$, Variance $\sigma^2 = 81$, so Standard Deviation $\sigma = 9$.     * Probability of $X < 52$: normalcdf(-1E99, 52, 48, 9).     * Probability of $X \ge 42$: normalcdf(42, 1E99, 48, 9).     * Probability of $37 < X < 47$: normalcdf(37, 47, 48, 9).

• Scenario 3: $X \sim N(3.15, 0.02^2)$:     * Mean $\mu = 3.15$, Standard Deviation $\sigma = 0.02$.     * Probability of $X < 3.2$: normalcdf(-1E99, 3.2, 3.15, 0.02).     * Probability of $X \ge 3.11$: normalcdf(3.11, 1E99, 3.15, 0.02).     * Probability of $3.1 < X < 3.15$: normalcdf(3.1, 3.15, 3.15, 0.02).

Real-World Modeling with Normal Distributions

• Household Expenditure in Portugal:     * Mean ($\mu$) = $\in 100$ per week.     * Standard Deviation ($\sigma$) = $\in 20$ per week.     * Spending less than $\in 130$ per week: $P(X < 130)$.     * Spending more than $\in 90$ per week: $P(X > 90)$.     * Spending between $\in 80$ and $\in 125$ per week: $P(80 < X < 125)$.

• Manufacturing Quality Control (Bolts):     * Bolt diameters: Mean $\mu = 4\,\text{mm}$, Standard Deviation $\sigma = 0.25\,\text{mm}$.     * Rejection criteria: Smaller than $3.5\,\text{mm}$ or bigger than $4.5\,\text{mm}$.     * Acceptance criteria: Between $3.5\,\text{mm}$ and $4.5\,\text{mm}$. In other words, $P(3.5 \le X \le 4.5)$.     * Expected Value Calculation: For a batch of $500$ bolts, the number of accepted bolts is $500 \times P(3.5 \le X \le 4.5)$.

• Medical Waiting Times:     * Mean $\mu = 14\,\text{minutes}$, Standard Deviation $\sigma = 4\,\text{minutes}$.     * Find the probability of waiting more than $20\,\text{minutes}$.     * Find the percentage of patients waiting less than $10\,\text{minutes}$ by calculating $P(X < 10) \times 100$.

The Inverse Normal Distribution

• The Inverse Normal Concept: This function is used to find a threshold value $a$ when the probability (area under the curve) is already known. The GDC command is typically invNorm(area, mean, standard_deviation).

• Standard Normal Inverse Operations ($Z \sim N(0, 1)$):     * To find $a$ such that $P(Z < a) = 0.922$: use invNorm(0.922, 0, 1).     * To find $a$ such that $P(Z < a) = 0.342$: use invNorm(0.342, 0, 1).     * To find $a$ such that $P(Z > a) = 0.005$: First calculate the left-tail area, which is $1 - 0.005 = 0.995$. Then use invNorm(0.995, 0, 1).     * To find $a$ such that $P(a < Z < 1.6) = 0.787$: Find the area below $1.6$ using normalcdf, subtract $0.787$, and use that result in invNorm.     * Symmetric Intervals: For $P(-a < Z < a) = 0.3$, the central area is $0.3$, meaning the two tails combined are $0.7$. One tail is $0.35$. Thus, $P(Z < -a) = 0.35$. Therefore, $-a = invNorm(0.35, 0, 1)$.

Advanced Problems and Combined Probability Models

• Conditional Probability in Normal Distributions:     * Problem: Given $P(X > 32 | X > 28)$ for $X \sim N(28, 5)$.     * Formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$.     * Calculation: $P(X > 32 \cap X > 28)$ is simply $P(X > 32)$ because $32$ is greater than $28$. The result is $\frac{P(X > 32)}{P(X > 28)}$.

• Linking Normal and Binomial Distributions:     * Example: Heights of women are $N(163, 8^2)$. If three women are selected, find the probability at least one is taller than $172\,\text{cm}$.     * Step 1: Calculate the success probability $p = P(Height > 172)$. From the normal model, $p \approx 0.130$.     * Step 2: Model the number of women taller than $172$ as a Binomial distribution: $Y \sim B(3, 0.130)$.     * Step 3: Calculate $P(Y \ge 1) = 1 - P(Y = 0) = 1 - (1-0.130)^3 \approx 0.342$.

• Acceptance Percentages and Batches:     * A bottle is acceptable if its height is within $2$ standard deviations of the mean ($\mu \pm 2\sigma$).     * For a normal distribution, the percentage within $2$ standard deviations is approximately $95.4\%$.     * If $8$ bottles are selected, finding the probability that at least $6$ are acceptable requires a binomial calculation: $X \sim B(8, 0.95449)$. Then calculate $P(X \ge 6) = 1 - P(X \le 5) \approx 0.996$.

• Finding Unknown Parameters ($\mu$ and $\sigma$):     * If parameters are unknown, we use z-scores: $z = \frac{x - \mu}{\sigma}$.     * Finding $\sigma$ given $\mu$: For test scores with mean $22$ and $95\%$ of scores below $30$: $P(Z < z) = 0.95 \Rightarrow z = 1.64485$. Solution: $1.64485 = \frac{30 - 22}{\sigma} \Rightarrow \sigma \approx 4.86$.     * Simultaneous Equations for $\mu$ and $\sigma$: If $15\%$ of fish are $> 54\,\text{cm}$ ($z_1 \approx 1.036$) and $10\%$ are $< 25\,\text{cm}$ ($z_2 \approx -1.282$):         1. $\frac{54 - \mu}{\sigma} = 1.03643$         2. $\frac{25 - \mu}{\sigma} = -1.28155$         Solving these reveals $\mu \approx 41.0$ and $\sigma \approx 12.5$.