Mon, Sep 19
- Normal Curves
- Exam 1
- Part 1 Instructions
- Review for Exam 1 on Mon, Sep 26
- Exam 1 Part 2 in-class exam on Wed, Sep 28
- Standard Normal Distribution
- Mean = 0
- Standard Deviation = 1
- (x – mean)/std dev
- Example (p. 61)
- Young women heights normally distributed
- Mean = 64.5 in
- Standard Deviation = 2.5 in
- Z-Score for woman 68 inches tall
- Z = (68 – 64.5)/2.5 = 1.4
- Z-Score for woman 60 inches tall
- Z = (60 – 64.5)/2.5 = -1.8
- Normal Curve Calculations with Z-Scores
- % Greater than
- z > 0
- z > 1
- % Less than
- z < 0
- z < 1
- % Between
- 0 < z < 1
- 1 < z < 2
- % Greater than
- Normal Calculations video
1. The distribution of heights of young women (18 – 24 years old) is approximately normal with mean = 64.5 inches and standard deviation = 2.5 inches.
- Draw a normal curve with a horizontal axis. Label mean.
- What is the z-score for a 20-year-old woman who is 6 feet tall? Label point.
- Between what two values do the heights of the central 95% of young women lie? Shade this area.
- What % of young women are more than two standard deviations taller than the mean?
2. There are two national college-entrance examinations, the SAT and the American College Testing program (ACT). Scores on individual SAT exams are approximately normal with mean = 500 and standard deviation = 100. Scores on the ACT exams are approximately normal with mean = 18 and standard deviation = 6.
- Julie’s SAT Math score is 630. John’s ACT Math score is 22. Calculate the standardized Z-scores for both.
- Assuming that both tests measure the same kind of ability, who has the higher score?
- What percent of all SAT scores are above 600?
3. Find the percentage of observations from a standard normal distribution that satisfy the following.
- z < −1
- z > 2
- −1 < z < 2
- Read pp. 61-64
- Start working on Exam 1: Part 1 Take Home