Last modified: August 05, 2018

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The Normal Curve and Z-Scores

The Normal Curve

A normal distribution (often referred to as the normal curve or Gaussian distribution) is a continuous probability distribution that is symmetric about the mean, where most of the observations cluster around the central peak and taper off symmetrically towards both ends. Many real-world datasets such as human heights, IQ scores, and measurement errors exhibit this kind of distribution.

The probability density function (PDF) of the normal distribution is given by:

$$ f(x | \mu, \sigma^2) = \frac{1}{\sigma \sqrt{2 \pi}} e^{ -\frac{(x - \mu)^2}{2\sigma^2} } $$

where:

• $\mu$ is the mean,
• $\sigma^2$ is the variance (with $\sigma$ being the standard deviation),
• $e$ is Euler's number,
• $\pi$ is the constant Pi.

This formula describes the shape of the curve mathematically. The mean $\mu$ determines the center of the distribution, and the standard deviation $\sigma$ determines the width of the bell curve.

Examples

• The distribution of adult heights usually follows a normal pattern, where most individuals cluster around the average height, with fewer people being much shorter or taller.
• When it comes to IQ scores, they are designed to follow a normal distribution, with the average score set at 100 and a standard deviation of 15, meaning most people score close to this mean.
• Not all data sets are normally distributed; for example, incomes tend to be skewed because a small proportion of people earn extremely high salaries, which significantly raises the overall average.
• Similarly, house prices are often skewed, as a small number of very expensive properties can disproportionately affect the overall data, making the mean house price higher than what most people actually pay.

The Empirical Rule (68-95-99.7 Rule)

The Empirical Rule provides a rough estimate for the spread of data in a normal distribution. It applies to any dataset that approximately follows the normal distribution.

• 68% of the data falls within 1 standard deviation of the mean, i.e., between $\mu - \sigma$ and $\mu + \sigma$.
• 95% of the data falls within 2 standard deviations of the mean, i.e., between $\mu - 2\sigma$ and $\mu + 2\sigma$.
• 99.7% of the data falls within 3 standard deviations of the mean, i.e., between $\mu - 3\sigma$ and $\mu + 3\sigma$.

Example

Suppose we have a dataset where the heights of fathers are normally distributed with:

• Mean height $\mu = 68.3$ inches,
• Standard deviation $\sigma = 1.8$ inches.

According to the Empirical Rule:

• About 68% of the heights will lie between $68.3 - 1.8 = 66.5$ inches and $68.3 + 1.8 = 70.1$ inches.
• About 95% of the heights will lie between $68.3 - 2(1.8) = 64.7$ inches and $68.3 + 2(1.8) = 71.9$ inches.
• About 99.7% of the heights will lie between $68.3 - 3(1.8) = 62.9$ inches and $68.3 + 3(1.8) = 73.7$ inches.

These intervals can be visualized as follows:

$$ \text{Interval} \quad \mu \pm n\sigma \quad \text{Proportion of Data Contained} $$

$$ \mu \pm \sigma \quad (66.5 \text{ to } 70.1) \quad \approx 68\% $$

$$ \mu \pm 2\sigma \quad (64.7 \text{ to } 71.9) \quad \approx 95\% $$

$$ \mu \pm 3\sigma \quad (62.9 \text{ to } 73.7) \quad \approx 99.7\% $$

Standardizing Data and Z-Scores

To compare values from different normal distributions or to work with a standardized form of a dataset, we can convert raw data values to z-scores.

Definition of Z-Score

The z-score is a way of describing a value in terms of how many standard deviations it is away from the mean. The formula for the z-score is:

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

Where:

• $x$ is the raw score (the value you are standardizing),
• $\mu$ is the mean of the dataset,
• $\sigma$ is the standard deviation.

A z-score tells us:

• $z = 0$: The value is exactly at the mean.
• $z > 0$: The value is above the mean.
• $z < 0$: The value is below the mean.

Example

Suppose a father is 71.9 inches tall. We want to find his z-score given that the mean height is 68.3 inches and the standard deviation is 1.8 inches.

$$ z = \frac{71.9 - 68.3}{1.8} = \frac{3.6}{1.8} = 2 $$

This means that a height of 71.9 inches is 2 standard deviations above the mean.

Conversely, if a father is 67.4 inches tall:

$$ z = \frac{67.4 - 68.3}{1.8} = \frac{-0.9}{1.8} = -0.5 $$

This means that a height of 67.4 inches is 0.5 standard deviations below the mean.

Standard Normal Distribution

After converting all values in a normal distribution to z-scores, we obtain the standard normal distribution, which has:

• A mean $\mu = 0$,
• A standard deviation $\sigma = 1$.

The standard normal distribution is often used in statistics because it allows for easy computation of probabilities and comparison across different datasets.

Finding Areas Under the Normal Curve

To find the proportion of data within a certain range, we can use z-scores to convert the raw data points and then look up the corresponding probabilities using a z-table or statistical software. The area under the normal curve between two z-scores represents the proportion of data that lies between those values.

Example

To find the proportion of fathers with heights between 67.4 inches and 71.9 inches, we first compute the z-scores for these heights:

For 67.4 inches:

$$ z_{67.4} = \frac{67.4 - 68.3}{1.8} = -0.5 $$

For 71.9 inches:

$$ z_{71.9} = \frac{71.9 - 68.3}{1.8} = 2 $$

Next, using a z-table (or software), we find the area to the left of these z-scores:

• The area to the left of $z = -0.5$ is approximately 0.3085 (30.85%).
• The area to the left of $z = 2$ is approximately 0.9772 (97.72%).

Thus, the proportion of fathers with heights between 67.4 and 71.9 inches is:

$$ P(67.4 \leq \text{height} \leq 71.9) = 0.9772 - 0.3085 = 0.6687 \approx 66.87\% $$

Computing Percentiles

The percentile of a value in a normal distribution tells us the percentage of the data that is less than or equal to that value. To compute percentiles, we:

1. Find the z-score corresponding to the desired percentile using a z-table or statistical software.
2. Convert the z-score back into the raw data value.

Example

Suppose we want to compute the 30th percentile of fathers' heights. Using a z-table, we find that the z-score corresponding to the 30th percentile is approximately $z = -0.52$.

To find the corresponding height, we use the z-score formula in reverse:

$$ \text{height} = \mu + z\sigma = 68.3 + (-0.52)(1.8) = 68.3 - 0.936 = 67.364 \text{ inches}. $$

Thus, the 30th percentile corresponds to a height of approximately 67.36 inches.

This process can be applied to any percentile by finding the appropriate z-score and converting it back to the original scale using the mean and standard deviation of the dataset.