Lesson 5: Normal Curve and Z-Scores
September 9, 2025
Review:
- 8675309
- Standard Deviation
- Exam 1 on Thu Sep 18
Presentation:
- Normal Curves
- Normal Curve explained in 1 min
- Normal Curves explained on 1 page
- Density Curves
- Height of curve indicates proportion of values
- Area under the curve = 1.0
- Any sub-area under the curve is then a proportion (% of values)
- Normal Curves
- A special case of density curves
- Bell shaped and symmetrical
- Mean and Standard deviation
- 68 – 95 – 99.7 Empirical Rule
- Against All Odds – Video Unit 7
- Demonstrate with student height data
- Standard Normal Distribution
- Mean = 0
- Standard Deviation = 1
- Calculating Z-Scores
- (x – mean)/std dev
- Example
- 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
Activity:
Problem 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-yr-old woman who is 6 ft 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?
Problem 2. There are two national college-entrance examinations, the SAT and the 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?
Problem 3. Find the percentage of observations from a standard normal distribution that satisfy the following. Also, draw a normal curve and shade the corresponding area under the curve.
- z < −1
- z < 1
- z > -2
- z > 2
- −1 < z < 2
- -2 < z < 0
Problem 4. The Indiana Statewide Testing for Educational Progress (ISTEP) is a program for assessing the skills of students in various grades. In a recent year, 76,531 tenth grade Indiana students took the English language arts exam. The mean score was 572 and the standard deviation was 51. Assume the ISTEP scores are approximately Normally distributed to answer the following questions.
- Use the 68-95-99.7 rule to give a range of scores that includes 95% of students.
- Use the 68-95-99.7 rule to give a range of scores that includes 99.7% of students.
- Calculate the Z-score for a student who scored 600.
- Calculate the Z-score for a student who scored 500.
