10.1 The Normal Distribution, Empirical Rule, and Z-Scores

10.1 The Normal Distribution, Empirical Rule, and Z-Scores

Quick Answer: The normal distribution is a symmetric, unimodal, bell-shaped continuous probability distribution where $\text{mean} = \text{median} = \text{mode}$ at the center $\mu$. The total area under the density curve equals $1$ ($100%$). The Empirical Rule dictates that approximately $68.2%$ of values fall within $\mu \pm 1\sigma$, $95.4%$ within $\mu \pm 2\sigma$, and $99.7%$ within $\mu \pm 3\sigma$. The $z$-score formula $z = \frac{x - \mu}{\sigma}$ standardizes raw scores into directed standard deviation units; memorize it, because it is not printed on the reference sheet. On graphing calculators, use $\text{normalcdf}(\text{lower}, \text{upper}, \mu, \sigma)$ to calculate cumulative probabilities and $\text{invNorm}(\text{area}, \mu, \sigma)$ to calculate cutoff scores from left-tail percentiles.

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1. Characteristics and Geometry of Normal Distributions (AII-S.ID.4a, AII-S.ID.4b)

[!NOTE] Standard note. NYSED split S-ID.4 into two assessed components for Algebra II: AII-S.ID.4a - recognize whether or not a normal curve is appropriate for a given data set - and AII-S.ID.4b - if appropriate, determine population percentages using a graphing calculator for an appropriate normal curve. Section 10.4 covers the 4a judgment call; this section covers the 4b machinery.

In statistics, quantitative data can follow many different distributions. Among continuous probability distributions, the normal distribution (often called the Gaussian distribution or bell curve) is the most fundamental model in Algebra II and inferential statistics.

A normal distribution is completely determined by two parameters:

1. Population Mean ($\mu$): The center of the distribution, locating the line of symmetry.
2. Population Standard Deviation ($\sigma$): The measure of spread, determining the dispersion and height of the curve.

Essential Geometric Properties

• Unimodal and Symmetric: The curve exhibits a single central peak and is perfectly symmetric about the vertical line $x = \mu$. If the curve is folded along $x = \mu$, the two halves coincide.
• Central Equivalence: Because of perfect symmetry, the mean, median, and mode are all identical and located at the exact center: $\mu = \text{median} = \text{mode}$.
• Total Area Under the Curve: The total area bounded between the normal curve and the horizontal axis equals exactly $1$ (representing $100%$ of the population).
• Asymptotic Tails: As $x$ moves away from the mean in either direction ($x \to \infty$ or $x \to -\infty$), the tails approach the horizontal axis asymptotically, getting closer to $0$ without ever touching or crossing it.
• Inflection Points at $\mu \pm \sigma$: The curve transitions from concave down (near the peak) to concave up (in the tails) at points of inflection located at exactly one standard deviation from the mean: $x = \mu - \sigma$ and $x = \mu + \sigma$.

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2. The Empirical Rule (68-95-99.7% Rule)

When a dataset is approximately normally distributed, the Empirical Rule provides precise percentage approximations for data falling within integer multiples of the standard deviation from the mean.

                           Mean
                            μ
                            |
                     34.1%  |  34.1%
                   +--------+--------+
            13.6%  |                 |  13.6%
          +--------+                 +--------+
   2.15%  |                                   |  2.15%
  +-------+                                   +-------+
  | 0.15% |                                   | 0.15% |
--+-------+--------+--------+--------+--------+-------+-->
 μ-3σ    μ-2σ     μ-1σ      μ       μ+1σ     μ+2σ    μ+3σ

Detailed Interval Breakdown

Because the normal distribution is symmetric about $\mu$, each symmetric interval can be partitioned into constituent sub-regions:

[!NOTE] Summing Percentages on Regents Problems: To find the percentage between asymmetric boundaries such as $\mu - 2\sigma$ and $\mu + 1\sigma$, sum the specific partitions: $\text{Area} = 13.6\% + 34.1\% + 34.1\% = 81.8\%$

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3. Standardization and the Z-Score Formula

When comparing values from different distributions or evaluating values not located at integer standard deviations, data must be standardized.

The Formula

[!CAUTION] This formula is not on the reference sheet. The Next Generation Algebra II reference sheet prints the quadratic formula, trigonometric identities, cubic factorizations, probability rules, sequence and series formulas, and exponential growth/decay models - and nothing statistical. The $z$-score formula and the Empirical Rule percentages must both be memorized.

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

Where:

• $x$ is the raw data value.
• $\mu$ is the population mean.
• $\sigma$ is the population standard deviation ($\sigma > 0$).
• $z$ is the standard score ($z$-score).

Mathematical Interpretation of the Z-Score

A $z$-score represents the directed distance between a raw score $x$ and the mean $\mu$, measured in units of standard deviation:

• If $z > 0$, the raw score lies above the mean.
• If $z < 0$, the raw score lies below the mean.
• If $z = 0$, the raw score is equal to the mean.
• For example, $z = -1.75$ indicates that the observation lies $1.75$ standard deviations below the distribution mean.

Converting from Z-Score Back to Raw Score

Multiplying both sides by $\sigma$ and adding $\mu$ yields the inverse algebraic formula:

$x = \mu + z\sigma$

The transformation maps any normal distribution $X \sim N(\mu, \sigma)$ onto the Standard Normal Distribution $Z \sim N(0, 1)$, which has a mean of $0$ and a standard deviation of $1$.

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4. Graphing Calculator Technology: normalcdf & invNorm

The graphing calculator is required on the Regents examination for computing non-integer cumulative probabilities and inverse percentiles.

Practical Execution Details

1. Accessing Distribution Functions: Press 2nd $\to$ VARS (DISTR).
2. Infinite Tail Approximations:
   • For left-tail probabilities ($P(X < b)$), set lower to $-1\text{E}99$ (entered as (-) 1 [2nd] [,] 99).
   • For right-tail probabilities ($P(X > a)$), set upper to $1\text{E}99$ (entered as 1 [2nd] [,] 99).
3. Left-Area Rule for invNorm: The area argument in invNorm must always represent the cumulative area from the far left tail. If an exam question asks for the "top $15%$," the area to the left is $1 - 0.15 = 0.85$.

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5. Worked Examples

Worked Problem 1: Empirical Rule Interval Calculation

Problem: The packaging weights of organic flour bags are normally distributed with a mean of $5.00\text{ lb}$ and a standard deviation of $0.08\text{ lb}$. Determine the percentage of bags weighing between $4.84\text{ lb}$ and $5.08\text{ lb}$.

• Step 1: Express the interval boundaries in terms of standard deviations from the mean.
  • Lower boundary: $4.84 = 5.00 - 0.16 = 5.00 - 2(0.08) = \mu - 2\sigma$.
  • Upper boundary: $5.08 = 5.00 + 0.08 = 5.00 + 1(0.08) = \mu + 1\sigma$.
• Step 2: Partition the interval into Empirical Rule sections.
  • From $\mu - 2\sigma$ to $\mu - 1\sigma$: $13.6%$
  • From $\mu - 1\sigma$ to $\mu$: $34.1%$
  • From $\mu$ to $\mu + 1\sigma$: $34.1%$
• Step 3: Sum the individual percentages. $\text{Percentage} = 13.6\% + 34.1\% + 34.1\% = 81.8\%$ Approximately $81.8%$ (or $81.85%$ with unrounded decimals) of flour bags fall within this range.

Worked Problem 2: Comparing Relative Standing via Z-Scores

Problem: Sofia scored $86$ on a chemistry midterm with $\mu = 74$ and $\sigma = 8$. On her history midterm, she scored $88$ where $\mu = 79$ and $\sigma = 6$. On which exam did Sofia perform better relative to her peers?

• Step 1: Calculate the $z$-score for chemistry. $z_{\text{chem}} = \frac{x - \mu}{\sigma} = \frac{86 - 74}{8} = \frac{12}{8} = 1.50$
• Step 2: Calculate the $z$-score for history. $z_{\text{hist}} = \frac{x - \mu}{\sigma} = \frac{88 - 79}{6} = \frac{9}{6} = 1.50$
• Step 3: Compare standard scores. Both $z$-scores are equal ($1.50$), meaning Sofia scored exactly $1.5$ standard deviations above the class mean on both examinations. Relative to her peers, her performance was identical.

Worked Problem 3: Finding Percentiles and Cutoff Scores with Calculator Functions

Problem: A manufacturing plant produces metal rods whose lengths are normally distributed with $\mu = 120.0\text{ cm}$ and $\sigma = 2.5\text{ cm}$. Rods in the longest $8%$ must be trimmed. Determine the minimum cutoff length for a rod to require trimming, rounded to the nearest tenth of a centimeter.

• Step 1: Identify the cumulative left-tail area. Because trimming applies to the top $8%$ (the right tail), the proportion of rods below this cutoff is: $\text{Cumulative Left Area} = 1 - 0.08 = 0.92$
• Step 2: Apply the inverse normal function. Using the calculator syntax invNorm(area, μ, σ): $\text{Cutoff} = \text{invNorm}(0.92, 120.0, 2.5)$
• Step 3: Compute the raw score.
  • Standard normal critical value: $z = \text{invNorm}(0.92, 0, 1) \approx 1.40507$
  • Raw score: $x = 120.0 + 1.40507(2.5) = 120.0 + 3.5127 \approx 123.51\text{ cm}$ Rounded to the nearest tenth, the cutoff length is $123.5\text{ cm}$.

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6. Common Regents Pitfalls & Exam Strategies

• Pitfall 1: Inverting the Z-Score Numerator. Students often write $z = \frac{\mu - x}{\sigma}$, which reverses the algebraic sign and misplaces the score on the opposite side of the mean.
• Pitfall 2: Inputting Right-Tail Area into invNorm. When a problem specifies the "highest $10%$" or "top $5%$," entering $0.10$ or $0.05$ computes the bottom cutoff instead. Always enter $1 - \text{right-tail area}$.
• Pitfall 3: Conflating Z-Scores with Probabilities. A $z$-score is a coordinate position along the horizontal axis, not a probability. A $z$-score of $-1.5$ is valid; a negative probability is mathematically impossible.