Applying the Standard Normal
Distribution to Problems
Real stuff and real numbers
in the world of x, z, and Area
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)
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
3
Normal Distribution Properties
Bluman, Chapter 6
4
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:
value - mean
z
standard deviation
z
X 

Bluman, Chapter 6
5
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
6
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
7
“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!
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
9
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
Example 6-6: Holiday Spending
A survey by the National Retail Federation found that
women spend on average $146.21 for the Christmas
holidays. Assume the standard deviation is $29.44. Find
the percentage of women who spend less than $160.00.
Assume the variable is normally distributed.
Step 1: Draw the normal distribution curve.
Bluman, Chapter 6
11
Example 6-6: Holiday Spending
Step 2: Find the z value corresponding to $160.00.
z
X 

160.00  146.21

 0.47
29.44
Step 3: Find the area to the left of z = 0.47.
Table E gives us an area of .6808.
68% of women spend less than $160.
Bluman, Chapter 6
12
The Holiday Spending Example
What did they do?
• 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
Conclusions and Extensions
• x = $160
• Converted to z = 0.47
• Area to left is 0.6808
• So 68% spend less than
$160
• Area to right is _______
• _____% spend more than
$160
The Holiday Spending Example
Analysis and Printed Table
• 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%...”
Analysis and TI-84 solution
• Same assumption
• Same particular normal dist
• Same question asked
• TI-84 can do it all in one
step
• More info next slide > > >
1. Low x (use -1E99 for “everything to the left”)
– Approximates “negative infinity” by −1 × 1099
– 2ND COMMA (EE) for the “E”, × 10𝑝𝑜𝑤𝑒𝑟
– -99 worked for z problems but need better here
2. High x (use 1E99 for “everything to the right”)
– Approximate “positive infinity” by 1 × 1099
3. Mean of the population
4. Standard deviation of the population
• 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”
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
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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
18
Example 6-8: Emergency Response
Step 2: Find the z value for 15.
15  25
z
 2.22
4.5
Step 3: Find the area to the left of z = -2.22. It is 0.0132.
Step 4: To find how many calls will be made in less than
15 minutes, multiply the sample size 80 by
0.0132 to get 1.056. Hence, approximately 1 call
will be responded to in under 15 minutes.
Bluman, Chapter 6
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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 0.0000 and 1.0000
• To scale it up to n=80,
multiply by 80
• 2ND ANS recalls the
normalcdf() result
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?
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
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Example 6-7a: Newspaper Recycling
Step 2: Find z values corresponding to 27 and 31.
27  28
z
 0.5
2
31  28
z
 1.5
2
Step 3: Find the area between z = -0.5 and z = 1.5.
Table E gives us an area of .9332 - .3085 = .6247.
The probability is 62%.
Bluman, Chapter 6
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Newspaper Recycling, TI-84
•
•
•
•
Low x = 27
High x = 31
Mean x = 28
Standard deviation of x = 2
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.
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
26
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.
Example 6-9: Police Academy
Step 2: Subtract 1 - 0.1000 to find area to the left, 0.9000.
Look for the closest value to that in Table E.
Step 3: Find X.
X    z  200 1.28  20  225.60
The cutoff, the lowest possible score to qualify, is 226.
Bluman, Chapter 6
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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
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
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Example 6-10: Systolic Blood Pressure
Area to the left of the positive z: 0.5000 + 0.3000 = 0.8000.
Using Table E, z  0.84. X = 120 + 0.84(8) = 126.72
Area to the left of the negative z: 0.5000 – 0.3000 = 0.2000.
Using Table E, z  - 0.84. X = 120 - 0.84(8) = 113.28
The middle 60% of readings are between 113 and 127.
Bluman, Chapter 6
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TI-84 for Blood Pressure
• You still have to draw the picture!!!!!!!!!!!!!!!!
• Left edge has 0.2000 of the area to its left
• Right edge has 0.8000 of the area to its left
– (the 0.2000 on the left end + 0.6000 between)
• Use invNorm twice!
• 2ND ENTER shortcut
recalls first entry
and just edit a little
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)?
“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.
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
35
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
36
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.
5 29 34 44 45 63 68 74 74
81 88 91 97 98 113 118 151 158
Method 1: Construct a Histogram.
The histogram is approximately bell-shaped.
Bluman, Chapter 6
37
Example 6-11: Technology Inventories
Method 2: Check for Skewness.
X  79.5, MD  77.5, s  40.5
3( X  MD) 3  79.5  77.5 
PI 

 0.148
s
40.5
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: 5 - 45 - 77.5 - 98 - 158
Q1 – 1.5(IQR) = 45 – 1.5(53) = -34.5
Q3 – 1.5(IQR) = 98 + 1.5(53) = 177.5
No data below -34.5 or above 177.5, so no outliers.
Bluman, Chapter 6
38
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.
5 29 34 44 45 63 68 74 74
81 88 91 97 98 113 118 151 158
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
39