Notes on Density Curves and Normal Distributions

Computational Limitations and Data Transformations

Calculators and computers possess inherent limitations when handling large sets of digits. This is evidenced when calculating the standard deviation for specific sets of observations. For instance, given the values $20,001$, $20,002$, and $20,003$, the mean is calculated as $\\bar{x} = 20,002$ and the standard deviation as $s = 1$. As zeros are added to the center of these numbers (e.g., $200,001$, $200,002$, $200,003$), the mathematical standard deviation remains $s = 1$. However, hardware and software will eventually return incorrect answers as the number of digits increases, demonstrating that they cannot manage an arbitrary degree of numerical precision.

In statistical analysis, the shape of a distribution dictates the appropriate numerical summary. Data sets such as the survival times of $72$ guinea pigs in medical studies, cancer patient survival rates, or car battery lifespans typically exhibit right-skewed distributions. For skewed data, measures of center and spread must be selected carefully; the five-number summary (minimum, $Q_1$, median, $Q_3$, maximum) is often more appropriate than the mean and standard deviation. Furthermore, while quartiles divide data into four equal parts, large data sets may benefit from quintiles (the $20\text{th}$, $40\text{th}$, $60\text{th}$, and $80\text{th}$ percentiles) or deciles to provide a more granular view of the distribution.

Linear transformations affect measures of center and spread differently. Multiplying every observation by a value (such as converting inches to centimeters using the factor $2.54$) multiplies the mean, median, quartiles, interquartile range (IQR), and standard deviation ($s$) by that same factor. However, the variance is multiplied by the square of that factor (e.g., $2.54^2$). In the context of weight, a kilogram is equal to $2.2$ pounds. For flower lengths, if millimeters are converted to inches, one must consider that $1\text{ mm} = \frac{1}{1000}\text{ m}$ and $1\text{ m} = 39.37\text{ inches}$. Specifically, for Henry Cavendish's measurements of the density of the earth, recorded as multiples of the density of water ($1\text{ g/cm}^3$), conversion to American units requires using the density of water as $62.43\text{ lb/ft}^3$.

Advanced Measures of Center and Refined Density Estimation

The trimmed mean serves as a measure of center that is more resistant than the arithmetic mean but utilizes more information than the median. A $10\%$ trimmed mean is computed by discarding the highest $10\%$ and lowest $10\%$ of observations and averaging the middle $80\%$. A $20\%$ trimmed mean follows the same logic, removing the outer $20\%$ from each tail. This method effectively neutralizes the impact of a small number of outliers.

Modern statistical software can also employ density estimators to describe the overall shape of data without assuming a specific pre-defined form like the Normal distribution. These estimators create smooth curves that reflect the overall pattern while ignoring minor irregularities and outliers. For example, software might apply a density curve to rainwater pH values (roughly symmetric) or guinea pig survival times (strongly skewed). The choice of classes in a histogram can sometimes obscure the true mode of a distribution, whereas a density curve can more accurately pinpoint peaks.

Principles and Properties of Density Curves

A density curve acts as a smooth, idealized mathematical model for a distribution of data. It serves to describe the overall pattern by "smoothing over" minor irregularities. For a curve to be defined as a density curve, it must satisfy two conditions: it must always be located on or above the horizontal axis, and the total area underneath the curve must be exactly $1$. Proportions of observations are represented by the area under the curve within a specific range of values. For example, in a study of vocabulary scores for $947$ seventh-graders in Gary, Indiana, the proportion of students scoring $6.0$ or lower was $0.303$ based on the actual data, while the density curve approximated this area as $0.293$.

Measures of center can be identified visually on a density curve. The median is the "equal-areas point," where exactly half the area lies on either side. The mean is the "balance point," defined as the location where the curve would balance if made of solid material. In a perfectly symmetric density curve, the mean and median are identical and located at the center. However, in a skewed distribution, the mean is pulled away from the median in the direction of the long tail. Standard deviation ($\\sigma$) and mean ($\\mu$) for these idealized distributions utilize Greek letters to distinguish them from the sample statistics $\\bar{x}$ and $s$ calculated from actual data.

The Normal Distribution and the 68–95–99.7 Rule

Normal distributions represent a specific class of symmetric, unimodal, bell-shaped density curves denoted as $N(\\mu, \\sigma)$. These distributions are entirely characterized by their mean ($\\mu$) and standard deviation ($\\sigma$). The mean is situated at the center of symmetry, while the standard deviation controls the spread. On a Normal curve, $\\sigma$ can be located by identifying the points where the curvature changes from "falling steeply" to "falling less steeply" (the inflection points), which are exactly distance $\\sigma$ from the mean $\\mu$. The specific mathematical equation for the height of a Normal density curve at any point $x$ is:

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

The 68–95–99.7 rule (also known as the Empirical Rule) describes the proportion of data within specific standard deviations of the mean in any Normal distribution:

• Approximately $68\%$ of observations lie within $\\pm 1\\sigma$ of $\\mu$.
• Approximately $95\%$ of observations lie within $\\pm 2\\sigma$ of $\\mu$.
• Approximately $99.7\%$ of observations lie within $\\pm 3\\sigma$ of $\\mu$.

For example, the height of young women (aged $18$ to $24$) is approximately $N(64.5, 2.5)$ in inches. This implies $95\%$ of these women are between $59.5$ and $69.5$ inches tall. Normal distributions are common in biological populations, test scores (like the ISTEP, SAT, or ACT), and measurements of physical quantities, and they serve as an excellent approximation for chance outcomes.

Standardizing and $z$-Scores

Standardizing is the process of converting an observation to a common scale that measures how many standard deviations it lies from the mean. The standardized value, or $z$-score, for an observation $x$ from a distribution with mean $\\mu$ and standard deviation $\\sigma$ is calculated as:

$z = \\frac{x - \\mu}{\\sigma}$

A positive $z$-score indicates the value is above the mean, while a negative $z$-score indicates it is below. Standardizing a variable $X$ that follows any Normal distribution $N(\\mu, \\sigma)$ results in a new variable $Z$ that follows the Standard Normal distribution, which is $N(0, 1)$. This allows for the comparison of values from different distributions, such as comparing an SAT score to an ACT score. For instance, in the class of $2003$, SAT scores followed $N(1026, 209)$ while ACT scores followed $N(20.8, 4.8)$.

Normal Distribution Calculations and Table A

Proportions of observations in a Normal distribution correspond to the area under the curve. A cumulative proportion is the area to the left of a specific value $x$. These proportions are found using software or the Standard Normal Table (Table A). When using Table A, one must first standardize $x$ to $z$ (rounding to two decimal places). The table provides the area for $Z < z$.

Calculations often include:

1. Area to the left: Directly from Table A or software.
2. Area to the right: $1 - \\text{Area to the left}$.
3. Area between two values: $\\text{Area left of upper value} - \\text{Area left of lower value}$.

For example, if the NCAA requires an SAT score of $820$ for Division I athletes ($N(1026, 209)$), we find the $z$-score is $z = \\frac{820 - 1026}{209} \\approx -0.99$. The area to the left is $0.1611$, so the proportion of students who qualify is $1 - 0.1611 = 0.8389$ (approximately $84\%$). For partial qualifiers needing a score of $720$, the $z$-score is $-1.46$, and the area between $720$ and $820$ is $0.1611 - 0.0721 = 0.0890$.

Inverse Normal Calculations

Inverse Normal calculations are used to find the observed value $x$ that corresponds to a given cumulative proportion. If finding the value from a table, one must locate the cumulative proportion in the body of Table A, identify the corresponding $z$-score, and then "unstandardize" it using the formula:

$x = \\mu + z\\sigma$

For instance, to find the SAT Verbal score ($N(505, 110)$) required for the top $10\%$, we seek the score with a cumulative proportion of $0.90$. The closest entry in Table A is $0.8997$, which corresponds to $z = 1.28$. Solving gives $x = 505 + (1.28)(110) = 645.8$. Thus, a student needs a score of at least $646$.

Assessing Normality with Normal Quantile Plots

A Normal quantile plot is the primary tool for assessing whether a data set follows a Normal distribution. The procedure involves:

1. Arranging the observed data from smallest to largest and assigning each a percentile rank.
2. Calculating the Normal scores (the $z$-scores for those same percentiles in $N(0,1)$).
3. Plotting the actual data values on one axis and the Normal scores on the other.

If the data are Normal, the points will fall roughly along a straight line. Outliers appear as points far from the overall linear pattern. Systematic curves indicate skewness; for example, a right-skewed distribution will show the largest observations falling systematically above a line drawn through the main body of points. Granularity, shown as short horizontal runs of points, is caused by repeated values due to limited measurement precision (e.g., repeating wire breaking strengths of $1150\\,lb$).

Clinical and Social Case Studies in Distribution

Various standardized tests and biological measurements illustrate these principles:

• WAIS (IQ Tests): Scores are approximately $N(100, 15)$. Mentally retarded status (Social Security benefits) applies to scores below $70$ (the bottom $2.3\%$). MENSA membership requires a score of $130$ or higher (the top $2.3\%$).
• Cholesterol Levels: For women aged $20$ to $34$, levels follow $N(185, 39)\\text{ mg/dl}$. Medical attention is needed above $240\\text{ mg/dl}$. For men aged $55$ to $64$, the distribution is $N(222, 37)\\text{ mg/dl}$.
• Osteoporosis Diagnosis: Based on Bone Mineral Density (BMD). The WHO criterion defines osteoporosis as a BMD $2.5$ standard deviations below the mean ($z = -2.5$).
• Human Pregnancy: Approximately $N(266, 16)\\text{ days}$. Horse pregnancies are longer at approximately $N(336, 3)\\text{ days}$.