1. Applying the Standard Normal Distribution to Problems
Real stuff and real numbers
in the world of x, z, and Area

2. Review Following: a quick review of the Standard Normal Distribution
Recall
We can find area to the left of some z
We can find area to the right of some z
We can find area between two z values
What’s going to be new here:
Take x problems (real-life) and change them to z problems (Standard Normal Distribution)

3. The Standard Normal Distribution is our favorite
This one is the most special of all of them
The mean: 𝜇=0
The standard deviation: 𝜎=1
Horizontal 𝑧=…,−3,−2,−1, 0, 1, 2, 3,⋯
Total area between curve and axis = 1.000

4. Normal Distribution Properties
Bluman, Chapter 6

5. z value (Standard Value)
The z value is the number of standard deviations that a particular X value is away from the mean. The formula for finding the z value is:
Bluman, Chapter 6

6. x value Going the other way: If you know the z value
and you need to find the x value,
𝑥=𝑧∙𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛+𝑚𝑒𝑎𝑛
𝑥=𝑧∙𝜎+𝜇 6

7. 6.2 Applications of the Normal Distributions
The standard normal distribution curve can be used to solve a wide variety of practical problems. The only requirement is that the variable be normally or approximately normally distributed.
For all the problems presented in this chapter, you can assume that the variable is normally or approximately normally distributed.
Bluman, Chapter 6

8. “You can assume that…”
Not every measurement follows the pattern of a bell-shaped a Normal Distribution!
A lot do, so let’s see what we can do with z.
But some are skewed – not normal distribut’n
Some will be uniform – not normal distribut’n
And various others that aren’t normal distrib’n
For those others, today’s methods are Invalid!

9. Applications of the Normal Distributions
To solve problems by using the standard normal distribution, transform the original variable to a standard normal distribution variable by using the z value formula.
This formula transforms the values of the variable into standard units or z values. Once the variable is transformed, then the Procedure Table and Table E in Appendix C can be used to solve problems.
Bluman, Chapter 6

10. What did that last slide say?
We are given an “x” problem from real life.
The measurements are everyday kinds of units
The values happen to fall in a normal distrib’n.
We change the x problem into a z problem
So it fits a Standard Normal Distribution
Area is Probability and Probability is Area

11. Example 6-6: Holiday Spending
A survey by the National Retail Federation found that women spend on average $ for the Christmas holidays. Assume the standard deviation is $ Find the percentage of women who spend less than $ Assume the variable is normally distributed.
Step 1: Draw the normal distribution curve.
Bluman, Chapter 6

12. Example 6-6: Holiday Spending
Step 2: Find the z value corresponding to $
Step 3: Find the area to the left of z = 0.47.
Table E gives us an area of
68% of women spend less than $160.
Bluman, Chapter 6

13. The Holiday Spending Example
What did they do?
Conclusions and Extensions
They took an “x” problem
Measurements in dollars
Situation follows a normal distribution (given)
They changed it into a z problem.
Area represented portion of spenders
x = $160
Converted to z = 0.47
Area to left is
So 68% spend less than $160
Area to right is _______
_____% spend more than $160

14. The Holiday Spending Example
Analysis and Printed Table
Analysis and TI-84 solution
Assumption: Normal distrib
𝜇=146.21,𝜎=29.44
“What % less than $160?”
Converted 𝑥=160 to 𝑧=0.47
What is area to left of 𝑧=0.47?
Lookup in printed table
State conclusion: “68%...”
Same assumption
Same particular normal dist
Same question asked
TI-84 can do it all in one step
More info next slide > > >

15. Low x (use -1E99 for “everything to the left”)
Approximates “negative infinity” by −1× 10 99
2ND COMMA (EE) for the “E”, × 10 𝑝𝑜𝑤𝑒𝑟
-99 worked for z problems but need better here
High x (use 1E99 for “everything to the right”)
Approximate “positive infinity” by 1× 10 99
Mean of the population
Standard deviation of the population

16. We can input it as an x problem Under the hood
The TI-84 converts it to a z problem
But it needed to know the mean and standard deviation
When we omit the mean and standard deviation, the TI-84 assumes we are talking z language
You still need to draw pictures and understand concepts and be able to do it “primitively”

17. Example 6-8: Emergency Response
The American Automobile Association reports that the average time it takes to respond to an emergency call is 25 minutes. Assume the variable is approximately normally distributed and the standard deviation is 4.5 minutes. If 80 calls are randomly selected, approximately how many will be responded to in less than 15 minutes?
(Added non-Bluman content)
FIRST find the proportion (or percent) that are under 15 minutes response time
THEN multiply the 80 calls by that proportion.
Bluman, Chapter 6

18. Example 6-8: Emergency Response
The American Automobile Association reports that the average time it takes to respond to an emergency call is 25 minutes. Assume the variable is approximately normally distributed and the standard deviation is 4.5 minutes. If 80 calls are randomly selected, approximately how many will be responded to in less than 15 minutes?
Step 1: Draw the normal distribution curve.
Bluman, Chapter 6

19. Example 6-8: Emergency Response
Step 2: Find the z value for 15.
Step 3: Find the area to the left of z = It is
Step 4: To find how many calls will be made in less than 15 minutes, multiply the sample size 80 by to get Hence, approximately 1 call will be responded to in under 15 minutes.
Bluman, Chapter 6

20. TI-84 for Emergency Response
Find the proportion using normalcdf(low x, high x, mean, stdev)
Since this is an area, or probability, the result is between and
To scale it up to n=80, multiply by 80
2ND ANS recalls the normalcdf() result

21. More Emergency Response Practice
What percent of calls have response time >25 minutes?
What percent have response time >30 minutes?
What percent have response time between 15 and 30 minutes?
If 3185 calls occur in a month, how many take more than 30 minutes to get a response?

22. Example 6-7a: Newspaper Recycling
Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds. If a household is selected at random, find the probability of its generating between 27 and 31 pounds per month. Assume the variable is approximately normally distributed.
Step 1: Draw the normal distribution curve.
Bluman, Chapter 6

23. Example 6-7a: Newspaper Recycling
Step 2: Find z values corresponding to 27 and 31.
Step 3: Find the area between z = -0.5 and z = 1.5.
Table E gives us an area of =
The probability is 62%.
Bluman, Chapter 6

24. Newspaper Recycling, TI-84
Low x = 27
High x = 31
Mean x = 28
Standard deviation of x = 2

25. Backwards x problems They give a proportion or percent of interest
“What x value separates out the low/middle/high %?”
Think: proportions are areas
DRAW A PICTURE
Find the z value and then convert it to an x value.

26. Example 6-9: Police Academy
To qualify for a police academy, candidates must score in the top 10% on a general abilities test. The test has a mean of 200 and a standard deviation of 20. Find the lowest possible score to qualify. Assume the test scores are normally distributed.
Step 1: Draw the normal distribution curve.
Bluman, Chapter 6

27. Police Academy (Added guidance not given by Bluman)
You have an area and a nice picture.
Solve the z problem first. What z score separates the top 10%
This is same as “What z separates the bottom 90% (0.9000)?”
When you find the z, convert it to an x answer.

28. Example 6-9: Police Academy
Step 2: Subtract to find area to the left, Look for the closest value to that in Table E.
Step 3: Find X.
The cutoff, the lowest possible score to qualify, is 226.
Bluman, Chapter 6

29. TI-84 for Police Academy
Use 2nd DISTR invNorm(area to left, x mean, x stdev)
Area to left = the area (proportion) to the left
Mean of the distribution in x language
Standard deviation of the distribution in x

30. Example 6-10: Systolic Blood Pressure
For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure. If the mean systolic blood pressure is 120 and the standard deviation is 8, find the upper and lower readings that would qualify people to participate in the study.
Step 1: Draw the normal distribution curve.
Bluman, Chapter 6

31. Example 6-10: Systolic Blood Pressure
Area to the left of the positive z: =
Using Table E, z  0.84.
Area to the left of the negative z: – =
Using Table E, z 
The middle 60% of readings are between 113 and 127.
X = (8) =
X = (8) =
Bluman, Chapter 6

32. TI-84 for Blood Pressure
You still have to draw the picture!!!!!!!!!!!!!!!!
Left edge has of the area to its left
Right edge has of the area to its left
(the on the left end between)
Use invNorm twice!
2ND ENTER shortcut recalls first entry and just edit a little

33. More Blood Pressure Practice
What blood pressure readings define the middle 90%?
The middle 95%?
Separate the bottom 1% from everybody else?
Separate the top 1% from everybody else?
The quartiles (25th and 75th percentiles)?

34. “Is this a normal distribution?”
A dangerous assumption has been used for this entire lesson.
The question is an important question for statisticians in general.
What follows is a glimpse of one little corner of the specialized task of determining whether a distribution is a normal distribution.
It’s kind of advanced, a fringe topic in this lesson, presented mostly for awareness.

35. Normal Distributions
A normally shaped or bell-shaped distribution is only one of many shapes that a distribution can assume; however, it is very important since many statistical methods require that the distribution of values (shown in subsequent chapters) be normally or approximately normally shaped.
There are a number of ways statisticians check for normality. We will focus on three of them.
Bluman, Chapter 6

36. Checking for Normality
Histogram
Pearson’s Index PI of Skewness
Outliers
Other Tests
Normal Quantile Plot
Chi-Square Goodness-of-Fit Test
Kolmogorov-Smikirov Test
Lilliefors Test
Bluman, Chapter 6

37. Example 6-11: Technology Inventories
A survey of 18 high-technology firms showed the number of days’ inventory they had on hand. Determine if the data are approximately normally distributed.
Method 1: Construct a Histogram.
The histogram is approximately bell-shaped.
Bluman, Chapter 6

38. Example 6-11: Technology Inventories
Method 2: Check for Skewness.
The PI is not greater than 1 or less than 1, so it can be concluded that the distribution is not significantly skewed.
Method 3: Check for Outliers.
Five-Number Summary:
Q1 – 1.5(IQR) = 45 – 1.5(53) = -34.5
Q3 – 1.5(IQR) = (53) = 177.5
No data below or above 177.5, so no outliers.
Bluman, Chapter 6

39. Example 6-11: Technology Inventories
A survey of 18 high-technology firms showed the number of days’ inventory they had on hand. Determine if the data are approximately normally distributed.
Conclusion:
The histogram is approximately bell-shaped.
The data are not significantly skewed.
There are no outliers.
Thus, it can be concluded that the distribution is approximately normally distributed.
Bluman, Chapter 6