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

Applying the Standard Normal Distribution to Problems

Real stuff and real numbersin 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: – The standard deviation:

• Horizontal • 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:

Bluman, Chapter 6 5

value - mean

standard deviationz

X

z

x valueGoing the other way:

If you know the z value

and you need to find the x value,

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 SpendingA 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 SpendingStep 2: Find the z value corresponding to $160.00.


Table E gives us an area of .6808.

68% of women spend less than $160.

160.00 146.210.47

29.44

X

z

Step 3: Find the area to the left of z = 0.47.

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

• “What % less than $160?”• Converted to • What is area to left of ?• 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 – 2ND COMMA (EE) for the “E”, – -99 worked for z problems but need better here

2. High x (use 1E99 for “everything to the right”)– Approximate “positive infinity” by

3. Mean of the population4. 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”

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.

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?


Example 6-8: Emergency Response


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.

Example 6-8: Emergency ResponseStep 2: Find the z value for 15.


15 252.22

4.5

z

Step 3: Find the area to the left of z = -2.22. It is 0.0132.

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?

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.


Example 6-7a: Newspaper Recycling


Example 6-7a: Newspaper RecyclingStep 2: Find z values corresponding to 27 and 31.


Table E gives us an area of .9332 - .3085 = .6247.

The probability is 62%.

27 280.5

2

z

Step 3: Find the area between z = -0.5 and z = 1.5.

31 281.5

2

z

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.

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.


Example 6-9: Police Academy


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.

The cutoff, the lowest possible score to qualify, is 226.

Example 6-9: Police AcademyStep 2: Subtract 1 - 0.1000 to find area to the left, 0.9000.

Look for the closest value to that in Table E.


200 1.28 20 225.60 X z Step 3: Find X.

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

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.


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.

Area to the left of the negative z: 0.5000 – 0.3000 = 0.2000.

Using Table E, z - 0.84.

The middle 60% of readings are between 113 and 127.

Example 6-10: Systolic Blood Pressure


X = 120 + 0.84(8) = 126.72

X = 120 - 0.84(8) = 113.28

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 entryand 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.


Checking for Normality Histogram Pearson’s Index PI of Skewness Outliers Other Tests

Normal Quantile PlotChi-Square Goodness-of-Fit TestKolmogorov-Smikirov TestLilliefors Test


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.

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: 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.



3 79.5 77.53( )PI 0.148

40.5

X MD

s

79.5, 77.5, 40.5 X MD s

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.



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

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