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

1. 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.

2. 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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