Lecture on Z-Scores and Standardization
Introduction
- Understanding z-scores and standardization.
- Calculating exact proportions using the standard normal distribution.
Standard Normal Distribution
- Definition: A type of normal distribution with a mean of 0 and a standard deviation of 1.
- Characteristics:
- Centered at 0.
- Intervals increase by 1.
- Numbers on the horizontal axis correspond to z-scores.
Z-Scores
- Definition: Indicates how many standard deviations an observation is from the mean (mu).
- Examples:
- Z-score of -2: 2 standard deviations to the left of the mean.
- Z-score of 1.5: 1.5 standard deviations to the right of the mean.
- Purpose: Allows calculation of the area associated with a specific z-score using a z-score table (standard normal table).
Z-Score Table (Standard Normal Table)
- Usage: Provides the total area to the left of any z-value.
- Top row and first column for z-values.
- Middle numbers for areas.
- Examples:
- Z-score of -1.95: Area of 0.0256 to the left.
- To find area to the right: 1 minus the area to the left.
- Example: Z-score of 0.57 has area 0.7157 to the left; right area = 1 - 0.7157 = 0.2843.
Reverse Look-Up
- Determine z-score based on a given area.
- Example: Area 0.8461 corresponds to z-score of 1.02.
Standardization
- Concept: Converting any normal distribution to a standard normal distribution.
- Benefits: Allows use of the z-score table for any normal distribution.
- Formula:
- Z = (X - mu) / sigma
- X = observation, mu = population mean, sigma = population standard deviation.
Example Problems
-
Chemistry Exam Scores
- Normal distribution: Mean = 60, SD = 10.
- Convert to standard normal distribution:
- Mean = 0, SD = 1.
- Example conversion: X = 50 gives Z = -1.
- Finding proportion of students scoring less than 49:
- Standardize: Z = (49 - 60) / 10 = -1.1.
- Use table: Area to left of Z = -1.1 is 0.1357.
-
Heights of Students
- Normal distribution: Mean = 5.5 ft, SD = 0.5 ft.
- Find proportion between 5.81 ft and 6.3 ft:
- Standardize: Z for 5.81 = 0.62, Z for 6.3 = 1.6.
- Use table: Area for Z = 0.62 is 0.7324, for Z = 1.6 is 0.9452.
- Calculate proportion: 0.9452 - 0.7324 = 0.2128.
Conclusion
- Understanding z-scores and standardization helps in interpreting and calculating areas under the standard normal distribution.
- Practice using the z-score table for various statistical problems.
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