Standardization, Z-Scores, and the Normal Distribution

Z-Scores and Standardization

Understanding Z-Scores

Z-scores are a way to standardize scores from different distributions, allowing for direct comparison. A z-score indicates how many standard deviations a value is from the mean of its distribution.

• Definition: The z-score of a value $X$ is calculated as:

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

• Interpretation: A positive z-score means the value is above the mean, while a negative z-score means it is below the mean.

• Application: Z-scores allow comparison of scores from different distributions by converting them to a common scale.

Example: Comparing Scores Across Distributions

Scenario: Suppose you received a score of X=60 on a statistics exam and X=56 on a research methods test. The exams have the following means and standard deviations:

• Research Methods: $\mu = 48$, $\sigma = 4$

• Statistics: $\mu = 50$, $\sigma = 10$

Question: For which course should you expect a better grade?

• Solution: Calculate the z-scores for each exam:

Research Methods: $z = \frac{56 - 48}{4} = 2$ Statistics: $z = \frac{60 - 50}{10} = 1$

• Conclusion: The z-score is higher for Research Methods, so you performed better relative to the class in that course.

Standardizing Distributions

Transforming Scores to a New Distribution

Standardization can be used to transform scores from one distribution to another with a different mean and standard deviation.

• Process: To convert a score $X$ from an original distribution ($\mu_{old}$, $\sigma_{old}$) to a new standardized distribution ($\mu_{new}$, $\sigma_{new}$):

1. Find the z-score in the original distribution: $z = \frac{X - \mu_{old}}{\sigma_{old}}$ 2. Convert to the new score: $X_{new} = z \cdot \sigma_{new} + \mu_{new}$

• Example: A population with mean $\mu = 44$ and standard deviation $\sigma = 6$ is standardized to a new distribution with $\mu = 50$ and $\sigma = 10$.

• Question a: What is the new standardized value for a score of $X = 47$ from the original distribution?

$z = \frac{47 - 44}{6} = 0.5$ $X_{new} = 0.5 \times 10 + 50 = 55$

• Question b: One individual has a new standardized score of $X_{new} = 65$. What was this person’s score in the original distribution?

$z = \frac{65 - 50}{10} = 1.5$ $X_{old} = 1.5 \times 6 + 44 = 53$

The Normal Distribution

Properties and Applications

The normal distribution is a continuous, symmetric, bell-shaped distribution characterized by its mean ($\mu$) and standard deviation ($\sigma$).

• Key Properties:

  • Mean, median, and mode are equal.

  • Approximately 68% of data falls within 1 standard deviation of the mean, 95% within 2, and 99.7% within 3 (the Empirical Rule).

• Standard Normal Distribution: A normal distribution with $\mu = 0$ and $\sigma = 1$.

Example: Calculating Probabilities

Scenario: A client has an investment portfolio with $\mu = $500,000 and$\sigma = $15,000. Determine the probability that the value of her portfolio is between $485,000 and $530,000.

• Step 1: Convert the values to z-scores:

$z_1 = \frac{485,000 - 500,000}{15,000} = -1$ $z_2 = \frac{530,000 - 500,000}{15,000} = 2$

• Step 2: Use the standard normal table to find probabilities:

• P($z$ between -1 and 2) = P($z < 2$) - P($z < -1$)

• From the z-table: P($z < 2$) ≈ 0.9772, P($z < -1$) ≈ 0.1587

• Probability = 0.9772 - 0.1587 = 0.8185

Conclusion: There is approximately an 81.85% chance that the portfolio value is between $485,000 and $530,000.

Summary Table: Z-Score and Standardization Formulas

Concept	Formula	Description
Z-score	$z = \frac{X - \mu}{\sigma}$	Standardizes a value relative to its distribution
Score from Z-score (new distribution)	$X_{new} = z \cdot \sigma_{new} + \mu_{new}$	Converts a z-score to a value in a new distribution
Probability (Normal Distribution)	P($a < X < b$) = P($z_a < z < z_b$)	Find area under the normal curve between two values