1. Honors Statistics The Standard Deviation as a Ruler and the Normal Model
Chapter 6 Part 3

2. Learning Goals
Recognize when standardization can be used to compare values.
Be able to use Normal models and the Rule to estimate the percentage of observations falling within 1, 2, or 3 standard deviations of the mean.
Know how to find the percentage of observations falling below any value in a Normal model using a Normal table or appropriate technology.
Know how to check whether a variable satisfies the Nearly Normal Condition by making a Normal Probability plot or histogram.

3. Procedure for Finding Normal Percentiles
State the problem in terms of the observed variable y.
Example : y > 24.8
Standardize y to restate the problem in terms of a z-score.
Example: z > ( μ)/σ, therefore z > ?
Draw a picture to show the area under the standard normal curve to be calculated.
Find the required area using Table Z or the TI-83/84 calculator.

4. Example 1:
The heights of men are approximately normally distributed with a mean of 70 and a standard deviation of 3. What proportion of men are more than 6 foot tall?

5. Answer: State the problem in terms of y. (6’=72”)
Standardize and state in terms of z.
Draw a picture of the area under the curve to be calculated.
Calculate the area under the curve.

6. Example 2:
Suppose family incomes in a town are normally distributed with a mean of $1,200 and a standard deviation of $600 per month. What are the percentage of families that have income between $1,400 and $2,250 per month?

7. Answer: State the problem in terms of y.
Standardize and state in terms of z.
Draw a picture.
Calculate the area.

8. Your Turn:
The Chapin Social Insight (CSI) Test evaluates how accurately the subject appraises other people. In the reference population used to develop the test, scores are approximately normally distributed with mean 25 and standard deviation 5. The range of possible scores is 0 to 41.
What percent of subjects score above a 32 on the CSI Test?
What percent of subjects score at or below a 13 on the CSI Test?
What percent of subjects score between 16 and 34 on the CSI Test?

9. From Percentiles to Scores: z in Reverse
Sometimes we start with areas and need to find the corresponding z-score or even the original data value.
Example: What z-score represents the first quartile in a Normal model?

10. z in Reverse
Given a normal distribution proportion (area under the standard normal curve), find the corresponding observation value.
Table Z – find the area in the table nearest the given proportion and read off the corresponding z-score.
TI-83/84 Calculator – Use the DISTR menu, 3rd entry invNorm. Syntax for invNorm(area,[μ,σ]) is the area to the left of the z-score (or Observation y) wanted (left-tail area).

11. From Percentiles to Scores: z in Reverse (cont.)
Look in Table Z for an area of
The exact area is not there, but is pretty close.
This figure is associated with z = –0.67, so the first quartile is 0.67 standard deviations below the mean.

12. Inverse Normal Practice
Proportion (area under curve, left tail)
Using Table Z
.3409
.7835
.9268
.0552
Using TI-83/84

13. Procedure for Inverse Normal Proportions
Draw a picture showing the given proportion (area under the curve).
Find the z-score corresponding to the given area under the curve.
Unstandardize the z-score.
Solve for the observational value y and answer the question.

14. Example 1: SAT VERBAL SCORES
SAT Verbal scores are approximately normal with a mean of 505 and a standard deviation of 110
How high must a student score in order to place in the top 10% of all students taking the verbal section of the SAT.

15. Analyze the Problem and Picture It.
The problem wants to know the SAT score y with the area 0.10 to its right under the normal curve with a mean of 505 and a standard deviation of 110. Well, isn't that the same as finding the SAT score y with the area 0.9 to its left? Let's draw the distribution to get a better look at it.

16. 1. Draw a picture showing the given proportion (area under the curve).
y=505
y = ?

17. 2. Find Your Z-Score
Using Table Z - Find the entry closest to It is This is the entry corresponding to z = So z = 1.28 is the standardized value with area 0.90 to its left.
Using TI-83/84 – DISTR/invNorm(.9). It is

18. 3. Unstandardize
Now, you will need to unstandardize to transform the solution from the z, back to the original y scale. We know that the standardized value of the unknown y is z = So y itself satisfies:

19. 4. Solve for y and Summarize
Solve the equation for y:
The equation finds the y that lies 1.28 standard deviations above the mean on this particular normal curve. That is the "unstandardized" meaning of z = 1.28.
Answer: A student must score at least 646 to place in the highest 10%

20. Example 2:
A four-year college will accept any student ranked in the top 60 percent on a national examination. If the test score is normally distributed with a mean of 500 and a standard deviation of 100, what is the cutoff score for acceptance?

21. Answer: Draw picture of given proportion.
Find the z-score. From TI-83/84, invNorm(.4) is z = -.25.
Unstandardize:
Solve for y and answer the question.
y = 475, therefore the minimum score the college will accept is 475.

22. Your Turn:
Intelligence Quotients are normally distributed with a mean of 100 and a standard deviation of 16. Find the 90th percentile for IQ’s.

23. Are You Normal? How Can You Tell?
When you actually have your own data, you must check to see whether a Normal model is reasonable.
Looking at a histogram of the data is a good way to check that the underlying distribution is roughly unimodal and symmetric.

24. Are You Normal? How Can You Tell? (cont.)
A more specialized graphical display that can help you decide whether a Normal model is appropriate is the Normal probability plot.
If the distribution of the data is roughly Normal, the Normal probability plot approximates a diagonal straight line. Deviations from a straight line indicate that the distribution is not Normal.

25. The Normal Probability Plot
A normal probability plot for data from a normal distribution will be approximately linear: X 90 60 30 Z -2 -1 1 2

26. The Normal Probability Plot
Left-Skewed
Right-Skewed X X 90 90 60 60 30 30 Z Z -2 -1 1 2 -2 -1 1 2
Rectangular
Nonlinear plots indicate a deviation from normality 30 60 90 -2 -1 1 2 Z X

27. Are You Normal? How Can You Tell? (cont.)
Nearly Normal data have a histogram and a Normal probability plot that look somewhat like this example:

28. Are You Normal? How Can You Tell? (cont.)
A skewed distribution might have a histogram and Normal probability plot like this:

29. Summary Assessing Normality (Is The Distribution Approximately Normal)
Construct a Histogram or Stemplot. See if the shape of the graph is approximately normal.
Construct a Normal Probability Plot (TI-83/84). A normal Distribution will be a straight line. Conversely, non-normal data will show a nonlinear trend.

30. Assess the Normality of the Following Data
9.7, 93.1, 33.0, 21.2, 81.4, 51.1, 43.5, 10.6, 12.8, 7.8, 18.1, 12.7
Histogram – skewed right
Normal Probability Plot – clearly not linear