Chapter 5: The Normal Distribution and Other Continuous Distributions

Chapter 5 Learning Objectives

Compute probabilities from the normal distribution and understand its characteristics.
Utilize the normal distribution to solve complex business problems.
Determine if a dataset is approximately normally distributed using normal probability plots.
Compute probabilities from the uniform distribution.
Compute probabilities from the exponential distribution.

Continuous Probability Distributions

Definition: A continuous random variable can assume any value within a defined interval or on a continuum (uncountable number of values).
Real-World Examples: * Thickness of a manufactured item. * Time required to complete a specific task. * Temperature of a chemical solution. * Height measured in inches.
Variable values depend solely on the ability to measure precisely and accurately.
Cumulative Distribution Function (CDF): Let $F(x)$ be the CDF for a continuous random variable $X$ . It expresses the probability that $X$ does not exceed a value $x$ : * $F(x) = P(X \leq x)$ * For two possible values $a$ and $b$ where a < b, the probability that $X$ lies between them is: P(a < X < b) = F(b) - F(a).
Probability Density Function (PDF): The density function $f(x)$ for random variable $X$ has the following core properties: 1. $f(x) \geq 0$ for all values of $x$ . 2. The total area under the probability density function $f(x)$ over all possible values of $X$ equals $1.0$ . 3. The probability that $X$ lies between two values is the area under the density graph between those values. 4. The cumulative density function $F(x_0)$ is defined as the area under the PDF from the minimum value up to $x_0$ : * $F(x_0) = \int_{x_{m}}^{x_0} f(x)\,dx$ * Where $x_{m}$ is the minimum value of $X$ . 5. The probability of any individual point is always zero: $P(X = a) = 0$ . Therefore: P(a \leq X \leq b) = P(a < X < b).

The Normal Distribution

Characteristics: * Bell-shaped and perfectly symmetrical around the mean. * The Mean, Median, and Mode are all equal. * Infinite theoretical range spanning from $-\infty$ to $+\infty$ .
Parameters: * Mean ( $\mu$ ): Determines the location (center) of the distribution. Shifting $\mu$ moves the distribution left or right. * Standard Deviation ( $\sigma$ ): Determines the spread (width). Increasing $\sigma$ increases the spread and flattens the curve.
Normal Density Function Formula: * $f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$ * $e \approx 2.71828$ (mathematical constant). * $\pi \approx 3.14159$ (mathematical constant). * $x$ = any value of the continuous variable.
Standardized Normal Distribution (Z): * Any normal distribution can be transformed into the standardized normal distribution. * Mean $\mu = 0$ . * Standard Deviation $\sigma = 1$ . * Translation formula (Z-score): $Z = \frac{X - \mu}{\sigma}$ . * Z-values specify the number of standard deviations a value is from the mean. Positive Z-values are above the mean, and negative Z-values are below the mean.
Z-score Example: If $X \sim N(100, 50^2)$ , for $X = 200$ : * $Z = \frac{200 - 100}{50} = 2.0$ * This indicates the value is $2$ standard deviations (increments of $50$ ) above the mean of $100$ .

Probability Calculations Using the Z-Table

Cumulative Table: The Standardized Normal Table (e.g., Appendix E.2) provides probabilities for the area to the left of a specific Z-score (from $\infty$ to $Z$ ). * Example: P(Z < 2.00) = 0.9772.
General Procedure: 1. Draw the normal curve for the problem in terms of $X$ . 2. Translate $X$ -values to $Z$ -values. 3. Use the Z-table to find the required area.
Download Time Example: * Scenario: Mean download time $\mu = 18.0\,\text{s}$ , $\sigma = 5.0\,\text{s}$ . * Problem: Find P(X < 18.6). * Step: $Z = \frac{18.6 - 18.0}{5.0} = 0.12$ . * Result: P(Z < 0.12) = 0.5478.
Upper Tail Probabilities: To find P(X > 18.6), use the complement rule: * P(Z > 0.12) = 1.0 - P(Z \leq 0.12) = 1.0 - 0.5478 = 0.4522.
Probabilities Between Two Values: To find P(18 < X < 18.6), calculate Z-scores for both points: * $Z_1 = \frac{18-18}{5} = 0$ ; $Z_2 = \frac{18.6-18}{5} = 0.12$ . * P(0 < Z < 0.12) = P(Z < 0.12) - P(Z < 0) = 0.5478 - 0.5000 = 0.0478.
Finding X for a Known Probability: Given a probability, find the corresponding $X$ value. 1. Find the Z-value for the known probability in the table. 2. Convert to units of $X$ using the formula: $X = \mu + Z\sigma$ . 3. Example: Find $X$ such that $20\%$ of download times are less than $X$ . * P(Z < ?) = 0.20 \rightarrow Z = -0.84. * $X = 18.0 + (-0.84)(5.0) = 18.0 - 4.2 = 13.8\,\text{s}$ .

Evaluating Normality

Theoretical Properties: Normal data should be bell-shaped (symmetrical), follow the empirical rule, have an Interquartile Range (IQR) $\approx 1.33\sigma$ , and a range $\approx 6\sigma$ .
Visual Assessment: * Small data: Use stem-and-leaf displays or boxplots to check symmetry. * Large data: Check histograms or polygons for bell shape.
Descriptive Measures: Compare mean, median, and mode for similarity.
Normal Probability Plot (Q-Q Plot): * Data is arranged into an ordered array. * Standardized normal quantile values (Z) are calculated. * Observed data ( $X$ ) is plotted on the vertical axis against quantile values ( $Z$ ) on the horizontal axis. * Linearity indicates a normal distribution. Nonlinear/curved plots indicate deviations like left-skew, right-skew, or rectangular distributions.

The Uniform Distribution

Definition: Outcomes are equally likely over a given range (rectangular distribution).
Density Function: * $f(x) = \begin{cases} \frac{1}{b-a} & a \leq X \leq b \ 0 & \text{otherwise} \end{cases}$
Summary Measures: * Mean: $\mu = \frac{a+b}{2}$ * Standard Deviation: $\sigma = \sqrt{\frac{(b-a)^2}{12}}$
Example (Range 2 to 6): * $f(x) = \frac{1}{6-2} = 0.25$ * $\mu = \frac{2+6}{2} = 4$ * $\sigma = \sqrt{\frac{16}{12}} \approx 1.1547$ * Probability calculation $P(3 \leq X \leq 5) = (\text{Base}) \times (\text{Height}) = (5-3) \times 0.25 = 0.5$ .

The Exponential Distribution

Definition: Often used to model time between occurrences/arrivals.
Probability Density Function: * $f(x) = \lambda e^{-\lambda X}$ for X > 0 * $\lambda$ = mean number of arrivals per unit time.
Summary Measures: * Mean time between arrivals ( $\mu$ ) = $\frac{1}{\lambda}$ . * Standard deviation ( $\sigma$ ) = $\frac{1}{\lambda}$ .
Cumulative Probability: Probability that arrival time is less than specified time $x$ : * P(\text{arrival time} < x) = 1 - e^{-\lambda x}.
Example: Arrivals at $15$ per hour ( $\lambda = 15$ ). Probability between customers is less than $3$ minutes ( $0.05$ hours): * P(X < 0.05) = 1 - e^{-(15)(0.05)} = 1 - e^{-0.75} \approx 0.5276.

Normal Approximation of the Binomial Distribution

Rationale: Binomial calculations become tedious as $n$ grows large. A normal distribution with the same mean and standard deviation can approximate it.
Requirements: Approximation is valid if $n\pi \geq 5$ and $n(1-\pi) \geq 5$ .
Parameters: * $\mu = np$ * $\sigma = \sqrt{np(1-p)}$
Continuity Adjustment: Since binomial is discrete and normal is continuous, adjust points into intervals. * $P(X = k)$ becomes P(k - 0.5 < W < k + 0.5). * $P(X \geq 25)$ becomes P(W > 24.5).
Example: tire production ( $n = 1600$ , defect rate $p = 0.08$ ). Probability of $150$ or fewer defects: * $\mu = 1600 \times 0.08 = 128$ * $\sigma = \sqrt{1600 \times 0.08 \times 0.92} = 10.85$ * $P(X \leq 150.5) \rightarrow Z = \frac{150.5 - 128}{10.85} \approx 2.07$ * P(Z < 2.07) = 0.9808.

Joint Distributions and Sums of Variables

Joint CDF: Defines the probability that variables $X_1, X_2 \dots X_k$ are simultaneously less than specified values: F(x_1, \dots, x_k) = P(X_1 < x_1 \cap \dots \cap X_k < x_k).
Independence: Random variables are independent if and only if $F(x_1, \dots, x_k) = F(x_1)F(x_2)\dots F(x_k)$ .
Covariance and Correlation: * $Cov(X, Y) = E[(X - \mu_x)(Y - \mu_y)] = E(XY) - \mu_x\mu_y$ . * If $X, Y$ are independent, $Cov(X, Y) = 0$ . * $p = Corr(X, Y) = \frac{Cov(X, Y)}{\sigma_x\sigma_y}$ .
Sums and Differences: * $E(X \pm Y) = \mu_x \pm \mu_y$ . * $Var(X \pm Y) = \sigma_x^2 + \sigma_y^2 \pm 2Cov(X, Y)$ .
Linear Combinations: For $W = aX + bY$ : * $\mu_w = a\mu_x + b\mu_y$ * $\sigma_w^2 = a^2\sigma_x^2 + b^2\sigma_y^2 + 2abCov(X, Y)$ .

Applied Problems and Exercises

Coffee Shop Staffing: $\mu = 150$ , $\sigma = 20$ . 1. Quiet Day (X < 130): $Z = -1.0 \rightarrow P \approx 0.1587$ . 2. Busy Day (X > 180): Z = 1.5 \rightarrow P(Z > 1.5) = 1 - 0.9332 = 0.0668. 3. Average Day (150 < X < 170): $Z_1 = 0, Z_2 = 1.0 \rightarrow 0.8413 - 0.5000 = 0.3413$ .
Mathematics Test: $\mu = 65, \sigma = 12$ . * Percentage below score $83$ : $Z = 1.5 \rightarrow 93.32\%$ . * Top $10\%$ distinction: $Z = 1.28 \rightarrow X = 65 + 1.28(12) = 80.36$ .
Cholesterol Classification: $\mu = 190, \sigma = 35$ . * High cholesterol (X > 240): $Z = 1.43 \rightarrow P = 0.0764$ . * Conditional: $P(X \geq 240 | X \geq 200) = \frac{0.0764}{1 - 0.6126} \approx 0.197$ .
Battery Lifetime: $\mu = 500, \sigma = 40\ cycles$ . * Warranty life for failure rate < 2\%: P(Z < ?) = 0.02 \rightarrow Z \approx -2.05 \rightarrow X \approx 418\ cycles.