Study Notes: The Normal Distribution and Other Continuous Distributions

Fundamentals of Continuous Probability Distributions

Continuous Random Variables:
- A continuous variable is defined as a variable that can assume any numerical value along a continuum, encompassing an uncountable number of possible values.
- Values are not restricted to discrete, distinct points; instead, they can take on any real value within an interval depending exclusively on the precision and accuracy of the measuring instrument.
- Typical real-world examples include:
  - The thickness of an manufactured item.
  - The time required to complete an assigned task.
  - The temperature of a chemical solution.
  - Human height measured in inches.
Variation by Shape in Continuous Distributions:
- Continuous probability distributions exhibit distinctive shapes characterized by symmetry, skewness, and theoretical ranges.
- Normal Distribution: Symmetrical, bell-shaped distribution ranging theoretically from negative infinity ( $-\infty$ ) to positive infinity ( $+\infty$ ).
- Uniform Distribution: Symmetrical, rectangular distribution where every value between the smallest value and the largest value is equally likely to occur.
- Exponential Distribution: Right-skewed distribution characterized by a mean strictly greater than its median ( $\text{Mean} > \text{Median}$ ), with values ranging from zero ( $0$ ) to positive infinity ( $+\infty$ ).

Continuous probability distribution shapes: Normal, Uniform, and Exponential

The Normal Distribution: Properties and Mathematical Formulation

Core Characteristics:
- The graphical representation is symmetrical and bell-shaped.
- The central tendency measures coincide precisely: $\text{Mean} = \text{Median} = \text{Mode}$ .
- The location of the distribution along the horizontal axis is determined entirely by the population mean, $\mu$ .
- The dispersion or spread of the distribution is determined entirely by the population standard deviation, $\sigma$ .
- Theoretical range extends across the entire real number line: from $-\infty$ to $+\infty$ .
Parameter Sensitivity:
- Varying the parameters $\mu$ and $\sigma$ produces an infinite family of distinct normal distributions.
- Altering $\mu$ while keeping $\sigma$ constant shifts the distribution bodily to the left or right without modifying its spread.
- Altering $\sigma$ while keeping $\mu$ constant increases or decreases the spread (flattening or narrowing the bell shape) without moving the central location.
- Two distributions can share identical means while possessing different standard deviations, or possess different means and different standard deviations simultaneously.

Normal distributions with varying means and standard deviations

Normal Probability Density Function:
- The height of the curve $, f(X),$ at any given value $X$ is defined by the formula:

$f(X) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{X - \mu}{\sigma}\right)^2}$

*   Mathematical constants and parameters include:
    *    $e$ : Mathematical constant approximated by  $2.71828$ .
    *    $\pi$ : Mathematical constant approximated by  $3.14159$ .
    *    $\mu$ : Population mean.
    *    $\sigma$ : Population standard deviation.
    *    $X$ : Any specific value of the continuous random variable ( $-\infty < X < \infty$ ).

The Standardized Normal Distribution (Z-Distribution)

Standardization Principle:
- Any normal distribution, regardless of its specific parameter combination of $\mu$ and $\sigma$ , can be transformed directly into the standardized normal distribution, denoted as $Z$ .
- To calculate normal probabilities, original units ( $X$ ) are converted into standardized units ( $Z$ ).
- The standardized normal distribution always has a mean of zero and a standard deviation of one:

$\mu = 0$

$\sigma = 1$

Transformation Formula:
- Conversion from an original variable $X$ to the standardized variable $Z$ is performed by subtracting the mean and dividing by the standard deviation:

$Z = \frac{X - \mu}{\sigma}$

*   Values of  $X$  located above the population mean ( $X > \mu$ ) yield positive  $Z$ -values.
*   Values of  $X$  located below the population mean ( $X < \mu$ ) yield negative  $Z$ -values.
*   A  $Z$ -value represents the exact number of standard deviation increments that a given value  $X$  lies away from the mean  $\mu$ .

Standardized Normal Probability Density Function:
- Substituting $\mu = 0$ and $\sigma = 1$ into the general density function yields:

$f(Z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}Z^2}$

Transformation Example:
- Assume $X$ is normally distributed with a mean of $\$100$ and a standard deviation of $\$50$ .
- To evaluate $X = \$200$ in standardized units:

$Z = \frac{\$200 - \$100}{\$50} = \frac{100}{50} = 2.0$

*   This indicates that an observation of  $\$200$  sits exactly  $2.0$  standard deviations ( $2$  increments of  $\$50$ ) above the mean of  $\$100$ .
*   The fundamental shape of the curve does not change upon transformation; only the horizontal scale is altered from dollars ( $X$ ) to standard deviations ( $Z$ ).

Calculating Probabilities from the Normal Distribution

Probability as Area Under the Curve:
- Probability for a continuous distribution corresponds to the area bounded by the curve, the horizontal axis, and the vertical interval limits.
- The probability that $X$ assumes a value between limits $a$ and $b$ is written as:

$P(a \le X \le b) = P(a < X < b)$

*   The probability of any single, exact individual value is strictly zero ( $P(X = c) = 0$ ) because a single point has zero width under the continuous curve.
*   The total cumulative area under the probability density curve across all real numbers equals one:

$P(-\infty < X < \infty) = 1.0$

*   Because the distribution is perfectly symmetrical about the mean, the area is divided evenly:

$P(-\infty < X < \mu) = 0.5$

$P(\mu < X < \infty) = 0.5$

Cumulative Standardized Normal Distribution Table:
- The standard normal probability table (e.g., Appendix Table E.2) provides cumulative probabilities corresponding to $P(Z < z)$ from $-\infty$ up to a specific positive or negative $Z$ .
- The left column of the table lists the value of $Z$ to one decimal place (e.g., $0.0, 0.1, \dots, 2.0$ ).
- The top row provides the second decimal place (the hundredths digit: $.00, .01, .02, \dots$ ).
- The intersection of row and column gives the cumulative area.
- For example, locating $Z = 2.00$ ( $2.0$ in the row, $.00$ in the column) yields a cumulative probability of $0.9772$ :

$P(Z < 2.00) = 0.9772$

General Three-Step Procedure:
1. Sketch the normal curve for the problem in terms of the original variable $X$ , shading the target region.
2. Translate the bounding $X$ -values into standardized $Z$ -values via $Z = \frac{X - \mu}{\sigma}$ .
3. Look up the corresponding probabilities in the Cumulative Standardized Normal Table.
Worked Application: Image Download Times:
- Let continuous random variable $X$ represent the download time of an image file from the Internet in seconds.
- Assume $X$ follows a normal distribution with parameters $\mu = 18.0\,\text{s}$ and $\sigma = 5.0\,\text{s}$ .
- Case 1: Finding a Lower Cumulative Probability ( $P(X < 18.6)$ ):
  - Calculate the standardized $Z$ -value:

$Z = \frac{18.6 - 18.0}{5.0} = \frac{0.6}{5.0} = 0.12$

    *   Look up  $Z = 0.12$  in the table (row  $0.1$ , column  $.02$ ), which gives  $0.5478$ .
    *   Therefore:

$P(X < 18.6) = P(Z < 0.12) = 0.5478$

*   *Case 2: Finding an Upper Tail Probability* ( $P(X > 18.6)$ ):
    *   Utilize the complement rule:

$P(X > 18.6) = P(Z > 0.12) = 1.0 - P(Z \le 0.12)$

$P(X > 18.6) = 1.0 - 0.5478 = 0.4522$

*   *Case 3: Finding Probability Between Two Values* ( $P(18.0 < X < 18.6)$ ):
    *   Standardize both bounds:

$Z_1 = \frac{18.0 - 18.0}{5.0} = 0.00$

$Z_2 = \frac{18.6 - 18.0}{5.0} = 0.12$

    *   Compute the difference between cumulative areas:

$P(18.0 < X < 18.6) = P(0 < Z < 0.12) = P(Z < 0.12) - P(Z \le 0.00)$

$P(18.0 < X < 18.6) = 0.5478 - 0.5000 = 0.0478$

*   *Case 4: Finding Probability in a Symmetric Lower Interval* ( $P(17.4 < X < 18.0)$ ):
    *   Standardize the lower bound:

$Z_1 = \frac{17.4 - 18.0}{5.0} = \frac{-0.6}{5.0} = -0.12$

    *   Evaluate using cumulative table probabilities:

$P(17.4 < X < 18.0) = P(-0.12 < Z < 0) = P(Z < 0) - P(Z \le -0.12)$

$P(17.4 < X < 18.0) = 0.5000 - 0.4522 = 0.0478$

    *   Due to the symmetrical nature of the bell curve,  $P(-0.12 < Z < 0)$  equals  $P(0 < Z < 0.12)$ .

The Empirical Rule (68-95-99.7 Rule)

Theoretical Coverage Percentages:
- For any normal distribution, the spread around the mean conforms to precise proportions:
  - $\mu \pm 1\sigma$ contains approximately $68.26\%$ of all observations ( $P(\mu - 1\sigma < X < \mu + 1\sigma) \approx 0.6826$ ).
  - $\mu \pm 2\sigma$ contains approximately $95.44\%$ of all observations ( $P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 0.9544$ ).
  - $\mu \pm 3\sigma$ contains approximately $99.73\%$ of all observations ( $P(\mu - 3\sigma < X < \mu + 3\sigma) \approx 0.9973$ ).

Inverse Normal Calculations: Finding X Values for Known Probabilities

Two-Step Inversion Procedure:
1. Locate the standardized score $Z$ that corresponds to the given cumulative tail probability using the body of the normal table.
2. Algebraically convert the standardized score $Z$ back into the original scale $X$ using:

$X = \mu + Z\sigma$

Worked Application: Determining the 20th Percentile:
- Using the previous image download scenario ( $\mu = 18.0\,\text{s}$ , $\sigma = 5.0\,\text{s}$ ), find the threshold value $X$ such that $20\%$ ( $0.2000$ ) of download times are less than $X$ .
- Step 1: Scan the interior values of the Standardized Normal Probability Table for cumulative area $0.2000$ .
  - At row $-0.8$ and column $.04$ , the table gives $0.2005$ .
  - At row $-0.8$ and column $.05$ , the table gives $0.1977$ .
  - The closest value to $0.2000$ is $0.2005$ , corresponding to a $Z$ -score of $-0.84$ .
- Step 2: Substitute $Z = -0.84$ , $\mu = 18.0$ , and $\sigma = 5.0$ into the conversion equation:

$X = \mu + Z\sigma$

$X = 18.0 + (-0.84)(5.0)$

$X = 18.0 - 4.20 = 13.80\,\text{s}$

*   *Conclusion:* Approximately  $20\%$  of download times fall below  $13.80$  seconds.

Implementing Normal Distribution Calculations in Microsoft Excel

Standardization Function:
- =STANDARDIZE(X, Mean, Standard_dev)
- Calculates the $Z$ -score for any given observation.
Cumulative Distribution Function:
- =NORM.DIST(X, Mean, Standard_dev, TRUE)
- Calculates the cumulative probability $P(X \le x)$ from $-\infty$ up to $X$ .
- Setting the final cumulative argument to TRUE returns the distribution function; FALSE returns the probability density height.
Standard Normal Inverse Function:
- =NORM.S.INV(probability)
- Converts a cumulative probability directly into its corresponding standardized normal score ( $Z$ ).
General Normal Inverse Function:
- =NORM.INV(probability, Mean, Standard_dev)
- Converts a cumulative probability directly into the original variable scale ( $X$ ).
Excel Implementation Architecture (Parameters: Mean in cell B4 = 7, Std Dev in cell B5 = 2):
- Lower Cumulative Tail: For $X = 7$ (cell B8):
  - $Z$ -value formula (cell B9): =STANDARDIZE(B8, B4, B5) yields $0$ .
  - Cumulative probability (cell B10): =NORM.DIST(B8, B4, B5, TRUE) yields $0.5000$ .
- Upper Tail Probability: For $X = 9$ (cell B13):
  - $Z$ -value formula (cell B14): =STANDARDIZE(B13, B4, B5) yields $1$ .
  - Upper probability (cell B15): =1 - NORM.DIST(B13, B4, B5, TRUE) yields $0.1587$ .
- Extreme Tail Combination: For $P(X < 7 \text{ or } X > 9)$ (cell B18):
  - Formula: =B10 + B15 yields $0.6587$ .
- Probability for an Interval Range ( $P(5 \le X \le 9)$ ):
  - Lower limit $X_1 = 5$ (cell B21): $Z$ -value (cell B23) =STANDARDIZE(B21, B4, B5) yields $-1$ ; cumulative area (cell B25) =NORM.DIST(B21, B4, B5, TRUE) yields $0.1587$ .
  - Upper limit $X_2 = 9$ (cell B22): $Z$ -value (cell B24) =STANDARDIZE(B22, B4, B5) yields $1$ ; cumulative area (cell B26) =NORM.DIST(B22, B4, B5, TRUE) yields $0.8413$ .
  - Range probability (cell B27): =ABS(B26 - B25) yields $0.6827$ .
- Finding $X$ and $Z$ Given a Cumulative Percentage ( $10.00\%$ in cell B30):
  - $Z$ -value (cell B31): =NORM.S.INV(B30) yields $-1.28$ .
  - $X$ -value (cell B32): =NORM.INV(B30, B4, B5) yields $4.44$ .
- Finding Interval Limits for a Central Percentage ( $95.00\%$ in cell B35):
  - Two-tailed outer boundary $Z$ -value (cell B36): =NORM.S.INV((1 - B35)/2) yields $-1.96$ .
  - Lower $X$ -value (cell B37): =B4 + (B36 * B5) yields $3.08$ .
  - Upper $X$ -value (cell B38): =B4 - (B36 * B5) yields $10.92$ .

Methods for Evaluating Normality

Importance of Normality Assessment:
- Not all continuous empirical distributions are normally distributed.
- Statistical techniques reliant on normality assumptions require thorough evaluation of sample data to check alignment with theoretical normal properties.
Theoretical Benchmarks of a Normal Distribution:
- Symmetry and Modality: Perfectly bell-shaped, single-peaked, with identical mean and median.
- Interquartile Range ( $IQR$ ): Exactly equal to $1.33$ standard deviations ( $IQR \approx 1.33\sigma$ ).
- Effective Range: Bounded effectively within $6$ standard deviations (Range $\approx 6\sigma$ ).
- Skewness and Kurtosis: Skewness statistic equals $0$ ; kurtosis statistic equals $0$ .
- Empirical Rule Proportions:
  - Approximately $68.26\%$ (about two-thirds) of data fall within $\mu \pm 1\sigma$ .
  - Approximately $80\%$ of data fall within $\mu \pm 1.28\sigma$ .
  - Approximately $95.44\%$ (about $95\%$ ) of data fall within $\mu \pm 2\sigma$ .
Diagnostic Evaluation Protocol:
- Graphical Displays:
  - For small- to moderate-sized datasets, construct a stem-and-leaf display or boxplot to inspect for balance and symmetry.
  - For large datasets, construct a histogram or polygon to examine if the profile displays a symmetrical bell shape.
- Descriptive Summary Comparison:
  - Compare mean, median, and mode to see if values cluster tightly together.
  - Calculate $IQR$ and evaluate the ratio $\frac{IQR}{\sigma}$ against $1.33$ .
  - Evaluate the ratio $\frac{\text{Range}}{\sigma}$ against $6.0$ .
- Empirical Spread Checks:
  - Determine the percentage of observations residing in intervals $\bar{X} \pm 1s$ , $\bar{X} \pm 1.28s$ , and $\bar{X} \pm 2s$ .
- Normal Probability Plot (Q-Q Plot):
  - Arrange sample observations into an ordered array from smallest to largest.
  - Determine the corresponding standardized normal quantile values ( $Z$ ).
  - Plot ordered sample observations ( $X$ ) on the vertical axis against theoretical standardized quantile scores ( $Z$ ) on the horizontal axis.
  - Inspect the resulting scatter for positive linearity: data from an approximately normal population form a straight line.
Interpreting Non-Linear Patterns on Normal Probability Plots:
- Normal: Points adhere strictly to a linear trajectory.
- Left-Skewed: The plot curves downward, displaying downward/convex curvature with deceleration.
- Right-Skewed: The plot curves upward, displaying upward/concave curvature with acceleration.

Shapes of normal probability plots for left-skewed, normal, and right-skewed data

Comprehensive Case Study: Evaluating Normality of Mutual Fund Returns

Empirical Dataset Summary ( $n = 318$ Mutual Funds):
- Sample size (Count): $318$
- Mean: $21.84$
- Median: $21.65$
- Mode: $21.74$
- Standard Error: $0.3604$
- Standard Deviation ( $s$ ): $6.4263$
- Sample Variance ( $s^2$ ): $41.2968$
- Coefficient of Variation: $29.43\%$
- Skewness: $1.6976$
- Kurtosis: $8.4670$
- Minimum: $3.39$
- First Quartile ( $Q_1$ ): $17.76$
- Third Quartile ( $Q_3$ ): $24.74$
- Maximum: $62.91$
- Range: $62.91 - 3.39 = 59.52$
- Interquartile Range ( $IQR$ ): $24.74 - 17.76 = 6.98$
Graphical Inspection:
- The boxplot exhibits an elongated whisker extending far to the right, showing right-skewness.

Boxplot for three-year mutual fund return percentages displaying right skewness

Comparison to Theoretical Properties:
- Mean vs. Median: The sample mean ( $21.84$ ) is close to the median ( $21.65$ ), which resembles a normal profile.
- Interquartile Range: The sample $IQR$ of $6.98$ equals $1.09$ standard deviations ( $\frac{6.98}{6.4263} \approx 1.09$ ), substantially smaller than the theoretical benchmark of $1.33\sigma$ .
- Range: The sample range of $59.52$ equals $9.26$ standard deviations ( $\frac{59.52}{6.4263} \approx 9.26$ ), greatly exceeding the theoretical benchmark of $6\sigma$ .
- Interval Coverages:
  - Percentage within $\bar{X} \pm 1s$ : $77.04\%$ (theoretical normal: $68.26\%$ ).
  - Percentage within $\bar{X} \pm 1.28s$ : $86.79\%$ (theoretical normal: $80.00\%$ ).
  - Percentage within $\bar{X} \pm 2s$ : $96.86\%$ (theoretical normal: $95.44\%$ ).
- Skewness and Kurtosis: Skewness is $1.698$ and kurtosis is $8.467$ , substantially departing from the theoretical normal value of $0$ .
- Normal Probability Plot Analysis: The Excel Q-Q plot bends noticeably upward, deviating sharply from a straight line at upper quantiles.

Normal probability plot for three-year return percentages exhibiting right-skewed curvature

Conclusion:
- The mutual fund returns are right-skewed.
- Observations are more densely clustered around the mean than expected in a normal distribution.
- The sample range is substantially wider than expected due to extreme outliers on the high end.
- The dataset fails the normality criteria and cannot be modeled by a theoretical normal distribution.

The Continuous Uniform Distribution

Definition and Properties:
- The continuous uniform distribution (or rectangular distribution) has equal probability density across every possible subinterval within a bounded range.
- Every point between the lower bound $a$ and upper bound $b$ is equally likely.
Probability Density Function:

$f(X) = \begin{cases} \frac{1}{b - a} & \text{if } a \le X \le b \\ 0 & \text{otherwise} \end{cases}$

*    $a$ : Minimum value of  $X$ .
*    $b$ : Maximum value of  $X$ .

Summary Parameters:
- Expected Value (Mean):

$\mu = E(X) = \frac{a + b}{2}$

*   *Variance:*

$\sigma^2 = \text{Var}(X) = \frac{(b - a)^2}{12}$

*   *Standard Deviation:*

$\sigma = \sqrt{\frac{(b - a)^2}{12}}$

General Computational Example:
- Given a continuous uniform distribution over the interval $2 \le X \le 6$ :
  - $a = 2$ , $b = 6$ .
  - Height of the density function:

$f(X) = \frac{1}{6 - 2} = \frac{1}{4} = 0.25$

    *   Mean:

$\mu = \frac{2 + 6}{2} = \frac{8}{2} = 4$

    *   Standard deviation:

$\sigma = \sqrt{\frac{(6 - 2)^2}{12}} = \sqrt{\frac{4^2}{12}} = \sqrt{\frac{16}{12}} = \sqrt{1.3333} \approx 1.1547$

    *   Probability across the subinterval  $3 \le X \le 5$ :

$P(3 \le X \le 5) = (\text{Base}) \times (\text{Height}) = (5 - 3) \times 0.25 = 2 \times 0.25 = 0.5$

Applied Uniform Distribution Case: Slater's Buffet

Scenario Description:
- Slater's Buffet charges customers based on the weight of salad taken from the salad bar.
- Empirical sampling demonstrates that salad plate filling weight, denoted by continuous variable $x$ , follows a uniform distribution between $5\,\text{ounces}$ and $15\,\text{ounces}$ .
Mathematical Specifications:
- Interval limits: $a = 5$ , $b = 15$ .
- Probability density function:

$f(x) = \begin{cases} \frac{1}{15 - 5} = \frac{1}{10} = 0.10 & \text{for } 5 < x < 15 \\ 0 & \text{elsewhere} \end{cases}$

Expected Value and Dispersion Measures:
- Expected Value (Mean):

$E(x) = \frac{a + b}{2} = \frac{5 + 15}{2} = \frac{20}{2} = 10\,\text{oz}$

*   *Variance:*

$\text{Var}(x) = \frac{(b - a)^2}{12} = \frac{(15 - 5)^2}{12} = \frac{10^2}{12} = \frac{100}{12} \approx 8.33\,\text{oz}^2$

Subinterval Probability Computation:
- To find the probability that an individual customer takes between $12$ and $15\,\text{ounces}$ of salad ( $P(12 < x < 15)$ ):

$P(12 < x < 15) = (\text{Base}) \times (\text{Height}) = (15 - 12) \times \frac{1}{10} = 3 \times 0.10 = 0.30$

*   There is a  $30\%$  chance that a customer takes between  $12$  and  $15$  ounces of salad.