Statistics -- Normal Distribution

5/27/2018 Statistics -- Normal Distribution

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Chapter 6: Whats Normal?

Stats in Your World

BOCK & MARIANO


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Whats Normal?

After this chapter, you should be able to

MEASURE position using percentiles

INTERPRET cumulative relative frequency graphs

MEASURE position using z-scores

TRANSFORM data

DEFINE and DESCRIBE density curves

Learning Objectives


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Whats Normal

DESCRIBE and APPLY the 68-95-99.7 Rule

DESCRIBE the standard Normal Distribution

PERFORM Normal distribution calculations

ASSESS Normality

Learning Objectives (contd)


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DescribingLoca

tioninaDistr

ibution

Measuring Position: Percentiles

One way to describe the location of a value in a distribution

is to tell what percent of observations are less than it.

Definition:

The pthpercentileof a distribution is the value

withppercent of the observations less than it.

6 7

7 2334

7 5777899

8 00123334

8 569

9 03

Deja earned a score of 86 on her test. How did she perform

relative to the rest of the class?

Example

Her score was greater than 21 of the 25observations. Since 21 of the 25, or 84%, of the

scores are below hers, Deja is at the 84th

percentile in the classs test score distribution.

6 7

7 2334

7 5777899

8 00123334

8 569

9 03

What percentile is the person who earned a 72?What is the percentile is the person who earned

a 93?

What is the percentile of the two students who

earned an 80?

If two observations have the same value, they will be at the same

percentile. To find the percentile, calculate the percent of the values in the

distribution that are below bothvalues.

Percentiles should be whole numbers, so if you get a decimal, you should

round to the nearest integer.


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Just knowing that Deja is 6 points above average doesnt tell

you much about her location in the distribution. Depending on

the spread of the distribution, Deja might be just barely above

average or really far above average. This is why we need toincorporate a measure of spread (standard deviation) to have

a really good understanding of how far above the mean she is.


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The stemplot below shows the number of wins for each of

the 30 Major League Baseball teams in 2009.

Find the percentiles for the following teams:

1) The Colorado Rockies, who won 92 games.

2) The New York Yankees, who won 103 games.

3) The Kansas City Royals and Cleveland Indians, who both

won 65 games.


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Measuring Position: z-Scores

A z-score tells us how many standard deviations from the

mean an observation falls, and in what direction.

Definition:

Ifxis an observation from a distribution that has known mean

and standard deviation, the standardized valueofxis:

A standardized value is often called a z-score.

z x

meanstandard deviation

Deja earned a score of 86 on her test. The class mean is 80and the standard deviation is 6.07. What is her standardized

score?

z x mean

standard deviation

86 80

6.07

0.99


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scribingLoc

ationinaDistribution

Using z-scores for Comparison

We can use z-scores to compare the position of individuals in

different distributions.

Deja earned a score of 86 on her statistics test. The class mean was

80 and the standard deviation was 6.07. She earned a score of 82

on her chemistry test. The chemistry scores had a fairly symmetric

distribution with a mean 76 and standard deviation of 4. On whichtest did Deja perform better relative to the rest of her class?

Example

zstats

86 80

6.07

zstats

0.99

zchem

82 76

4

zchem

1.5


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Using z-scores for Comparison

We can use z-scores to compare the position of individuals in

different distributions.

Now go back to Slide 6 and find and interpret the z-scores for the

following teams:

a) The New York Yankees with 103 wins.

b) The New York Mets, with 70 wins.


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CHECK YOUR UNDERSTANDING

Ms. Raskins Statistics class recorded their heights on a

dotplot.

1) Find and interpret the z-score of the student who is 65 inches tall.

2) Find and interpret the z-score of the student who is 74 inches tall.

3) The student who is 76 inches tall is on the basketball team. His

height translates to az-score of -0.85 in the teams heightdistribution. The standard deviation of basketball team members

is 3.5 inches. What is the mean of the team members heights?


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DescribingLoc



On one test your class achieved an average grade of 80 with

a standard deviation of 8 points.

1) If you got a 96, what is your z-score?

2) Your best friend got a 76. What is her z-score?3) What test grade has a z-score of +1.5?

4) Ms. Raskin calls home whenever a students z-score is worse

than -2.0. What grade earns that phone call?


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DescribingLoc



In the last chapter, we made boxplots for the number of

calories and the fiber content in 23 kinds of Kelloggs cereals.Now think about the sugar content. Those cereals average

7.6 grams of sugar per serving with a standard deviation of

4.5 grams.

1) Find the z-scores for the following cereals and describe what the

z-score tells you about that cereal:

1) Frosted Flakes: 11g of sugar

2) Apple Jacks: 14g of sugar

3) Crispix: 3g of sugar2) The z-score for Honey Smacks sugar content is a very high 3.87!

How many grams of sugar are in one serving?

3) Product 19 is very low in sugar, with a z-score of -0.8. How many

grams of sugar are in a serving of this cereal?


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DescribingLoc



The calorie content for 23 varieties of Kelloggs cereals

averages 109 calories per serving with a standard deviation of

22.2 calories. The mean fiber content for these cereals is 2.7

grams per serving with a standard deviation of 3.2 grams. A

serving of Kelloggs All-Bran with Extra Fiber has a very low50 calories and a very high 14 grams of fiber. Which is more

remarkablethe calorie content or the fiber content?

Explain.


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DescribingLoc



Ms. Raskin has just announced that the lower of your two test

scores will be dropped (JK!) You got a 90 on Test 1 and an 80on Test 2. Youre all set to drop the 80 until she announces

that she grades on a curve. She standardizes the scores in

order to decide which is the lower one. If the mean on the

first test was 88 with a standard deviation of 4 and the mean

on the second was 75 with a standard deviation of 5, whichone will be dropped?

Is this fair?


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DescribingLoc



The mens combined skiing event in the winter Olympics

consists of two races: a downhill and a slalom. Times for thetwo events are added together, and the skier with the lowest

total time wins. In the 2006 Winter Olympics, the mean

slalom time was 94.2714 seconds with a standard deviation of

1.8356 seconds. Ted Ligety of the United States, who won

the gold medal with a combined time of 189.35 seconds,skied the slalom in 87.93 seconds and the downhill in 101.42

seconds. On which race did he do better compared with the

competition?


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DescribingLoc



A single-season home run record for Major League Baseball

has been set just 3 times since Babe Ruth hit 60 home runs in1927. In an absolute sense, Barry Bonds had the best

performance of these four players, since he hit the most home

runs in a single season. However, in a relative sense, this

may not be true. Baseball historians suggest that hitting a

home un has been easier in some eras than others. This isdue to many factors, including quality of batters, quality of

pitchers, hardness of the baseball, dimensions of the parks,

and possible use of performance-enhancing drugs. To make

a fair comparison, we should see how these performances

rate relative to those of other hitters during the same year.


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DescribingLoc



Compute the standardized scores for each performance.Which player had the most outstanding performance

relative to his peers?


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Nor

malDistributions

That z-score jont:

A z-score gives us an indication of how unusual a value is

because it tells us how far it is from the mean.

A data value that sits right at the mean has a z-score of 0.

A z-score of 1 means that the data value is 1 standard

deviation above the mean.

A z-score of -1 means that the data value is 1 standarddeviation below the mean.


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What happens when a z-score is REALLY BIG?

How far from 0 does a z-score have to be to be interesting or

unusual? There is no universal standard, but the larger (+/-)

a z-score is, the more unusual it is.

There is no universal standard for z-scores, but there is a

model that shows up over and over in Statistics. This modelis called the Normal Model.

All models are wrong,but some are useful!

-George Box, statistician


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Normal Distributions

You may have heard of bell-shaped curves. Statisticians

call them Normal models. Normal modelsare appropriate

for distributions whose shapes are unimodel and roughlysymmetric

All Normal curves are symmetric, single-peaked, and bell-

shaped

A Specific Normal curve is described by giving its mean

and standard deviation .

Two Normal curves, showing the mean and standard deviation .


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Normal Distributions Nor

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Definition:

A Normal distributionis described by a Normal density curve. Anyparticular Normal distribution is completely specified by two numbers: its

mean and standard deviation .

The mean of a Normal distribution is the center of the symmetric

Normal curve.

The standard deviation is the distance from the center to thechange-of-curvature points on either side.

We abbreviate the Normal distribution with mean and standard

deviation as N(,).

Normal distributions are good descriptions for some distributions of real data(test scores, characteristics of biological populations, etc.).

Normal distributions are good approximations of the results of many kinds of

chance outcomes.

Many statistical inferenceprocedures are based on Normal distributions.


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NormalDistribu

tions

Although there are many Normal curves, they all have properties

in common.

The 68-95-99.7 Rule

Definition: The 68-95-99.7 Rule

In the Normal distribution with meanand standard deviation :

Approximately 68% of the observations fall within of .

Approximately 95% of the observations fall within 2 of .

Approximately 99.7% of the observations fall within 3 of .


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NormalDistribu

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The distribution of Iowa Test of Basic Skills (ITBS) vocabulary

scores for 7thgrade students in Gary, Indiana, is close to

Normal. Suppose the distribution is N(6.84, 1.55).

a) Sketch the Normal density curve for this distribution.

b) What percent of ITBS vocabulary scores are less than 3.74?

c) What percent of the scores are between 5.29 and 9.94?

Example


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The distribution of heights of young women aged 1824 is

approximately Normal: N(64.5, 2.5).

a) Sketch the Normal density curve for this distribution.

b) What percent of young women have heights greater than 67

inches?

c) What percent of young women have heights between 67 and72 inches?

Example


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The average clean-up time for a crew of a medium-sized firm

is 84.0 hours and the standard deviation is 6.8 hours.

Assuming a Normal distribution, within what time interval will

95% of the clean-up times fall?

Over a long period of time, a farmer notes that the eggs

produced by his chickens have a mean weight of 60g and astandard deviation of 15g. If eggs are classified by weight and

small eggs are those having a weight of less than 45g, what

percent of the farmers eggs will be classified as small?

Eggs are classified as jumbo if their weight is 90g or more.What percent of the farmers eggs will be jumbo?



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NormalDistribu

tions

As a group, the Dutch are among the tallest people in the

world. The average Dutch man is 184cm talljust over 6 feet!

The standard deviation of mens heights is about 8cm.

Assuming the distribution is approximately Normal, use the 68-

95-99.7 rule to sketch a model for the heights of Dutch men.

Label the axis clearly and indicate appropriate percentages.


Based on this model, what percentage of all Dutch men shouldbe over 2 meters (66) tall?


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Lets say it takes you 20 minutes, on average, to drive to

school, with a standard deviation of 2 minutes. Suppose a

Normal model is appropriate for the distributions of driving

times. Based on the 68-95-99.7 Rule:

About how often will it take you between 18 and 22 minutes toget to school?

How often will you arrive at school in less than 22 minutes?

How often will it take you more than 24 minutes?



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In the 2006 Winter Olympics mens combined event, Jean

Baptiste Grange of France skied the slalom in 88.46 seconds

about 1 standard deviation faster than the mean. If a Normal

model is useful in describing slalom times, about how many of

the 35 skiers finishing the event would you expect skied the

slalom fasterthan Jean-Baptiste?



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NormalDistribu

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Scores on the Wechsler Adult Intelligence Scale for the 20-34

year old age group are normally distributed with a mean of 110

and standard deviation of 25. Use the 68-95-99.7 rule to

determine what percent of people score

Above 110

Above 150 Below 85

Below 185



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EPA fuel economy estimates for automobiles tested recently

predicted a mean of 24.8mpg and a standard deviation of 6.2

for highway driving. Assume a Normal model to determine

The range of gas mileage for the central 68% of cars

The percent of autos getting more than 31 mpg

The percent of autos getting between 31 and 37.2mpg Why you shouldnt use these numbers to predict driving in-town

gas mileage.



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NormalDistributions

The Standard Normal Distribution

All Normal distributions are the same if we measure in units

of size from the mean as center.

We can standardize these data by changing to z-scores:

z= (x- )/ . If the variable we standardize has a Normal

distribution, then so does the new variable z.

This new distribution is called the Standard Normal

Distribution.


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NormalDistributions

The Standard Normal Distribution

All Normal distributions are the same if we measure in units

of size from the mean as center.

Definition:

The standard Normal distributionis the Normal distribution

with mean 0 and standard deviation 1.

If a variablexhas any Normal distribution N(,) with mean

and standard deviation , then the standardized variable

has the standard Normal distribution, N(0,1).

zx -


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The Standard Normal Table

Because all Normal distributions are the same when we

standardize, we can find areas under any Normal curve froma single table.

Definition: The Standard Normal Table

Table Ais a table of areas under the standard Normal curve. The table

entry for each value zis the area under the curve to the left of z.

Z .00 .01 .02

0.7 .7580 .7611 .7642

0.8 .7881 .7910 .7939

0.9 .8159 .8186 .8212

P(z < 0.81) = .7910

Suppose we want to find the

proportion of observations from the

standard Normal distribution that are

less than 0.81.

We can use Table A:


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Finding Areas Under the Standard Normal Curve

Find the proportion of observations from the standard Normal distribution that

are between -1.25 and 0.81.

Example

Can you find the same proportion using a different approach?

1 - (0.1056+0.2090) = 10.3146

= 0.6854


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NormalDistribu

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are greater than -1.78.

Example


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Find the proportion of observations in a Normal distribution that are more than

1.53 standard deviations above the mean.

Example

The table entry0.9370 is for the

area to the left of z

= 1.53.

The area to the

right of z= 1.53 is 1

0.9370 = 0.0630


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Find the proportion of observations in a Normal distribution that are between -

0.58 and 1.79.

Example

=

Area to the left

of z= 1.79 is

0.9633

Area to the left

of z= -0.58 is

0.2810

Area between z= -

0.58 and z= 1.79 is

0.6823


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Remember those tall Dutch men? Their mean height was

184cm and the standard deviation was 8cm. Answer each of

these questions by sketching a Normal model, shading the

appropriate area, finding the z-scores, and using the table to

determine the percentage.

What percent of Dutch men should be less than 190cm tall?

What percent of Dutch men should be between 170 and

180cm tall?

What fraction of Dutch men should be over 198cm tall?



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Working Backward: Finding z-scores from Percentiles

Sometimes we start with areas and need to find the corresponding z-score or

even the original data value.

What z-score represents the first quartile in a Normal Model?


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For each, sketch the standard Normal distribution, shade the area

described, and find the z-score cutpoints

1. The lowest 40% of the distribution

2. The highest 30% of the distribution

3. The highest 2% of the distribution

4. The middle 30% of the distribution


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To Find a z-score from a percentile:

1.Look for the percentile (as a decimal) in the middleof the z-table.

2.Plug the z-score, the mean, and the standard deviation into the

formula: z= (x)/

3.Solve forx


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Going Back to those Dutch men

Lets think about the Normal model for the heights of Dutch men one

more time. Remember that their mean height was 184cm and the

standard deviation was 8cm. Answer each of these questions by

sketching a Normal model, shading the appropriate area, finding the

cutpoint z-score, and then determining the cutpoint height.

1.How tall are the tallest 10% of all Dutch men?

2.How tall are the shortest 20% of Dutch men?

3.How tall are the middle 50% of Dutch men?


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Normal Distribution Calculations

State:Express the problem in terms of the observed variablex.

Plan:Draw a picture of the distribution and shade the area ofinterest under the curve.

Do:Perform calculations.

Standardizexto restate the problem in terms of a standard

Normal variable z.

Use Table A and the fact that the total area under the curve

is 1 to find the required area under the standard Normal curve.

Conclude:Write your conclusion in the context of the problem.

How to Solve Problems Involving Normal Distributions


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When Tiger Woods hits his driver, the distance the ball travels can be

described by N(304, 8). What percent of Tigers drives travel between 305

and 325 yards?

Whenx = 305, z =305- 304

8 0.13

Whenx = 325, z =325- 304

8 2.63

Using Table A, we can find the area to the left of z=2.63 and the area to the left of z=0.13.

0.99570.5517 = 0.4440. About 44%of Tigers drives travel between 305 and 325 yards.


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In the 2008 Wimbledon tennis tournament, Rafael Nadal averaged 115 miles

per hour on his first serves: N(115, 6). About what proportion of his first

serves would you expect to exceed 120mph?

Do: Whenx =120, z =120-115

6 0.83

Looking up a z-score of 0.83 shows us that the area less than z= -.83 is 0.7967. This means

that the area to the right of z= 0.83 is 10.7967 = 0.2033.

Conclude: About 20% of Nadals first serves will travel more than 120mph.

State: Letx= the speed of Nadals first serve. The variablexhas a Normal distribution

with = 115 and = 6. We want the proportion of first serves withx 120.

x= 120

z

= 0.83

Plan:


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In the 2008 Wimbledon tennis tournament, Rafael Nadal averaged 115 miles

per hour on his first serves: N(115, 6). About what proportion of his first

serves are between 100 and 110mph?

Do: Whenx =100, z= 100-115

6 2.5

When x = 110, z110 115

6 0.83

Looking up a z-score of -2.50 shows us that the area less than z= -2.50 is 0.0062. Looking

up a z-score of -0.83 shows us that the area less than z= -0.83 is 0.2033. Thus, the area

between z= -2.50 and z= 0.83 is 0.20330.0062 = 0.1971.

Conclude: About 20% of Nadals first serves will travel between 100 and 110mph.

State: Letx= the speed of Nadals first serve. The variablexhas a Normal distribution

with = 115 and = 6. We want the proportion of first serves withx 120.

x= 100

z

= -2.50

x= 110

z

= -0.83


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Normal Distribution Calculations: Cholesterol in

Teenage Boys

High levels of cholesterol in the blood increase the risk of

heart disease. For 14-year-old boys, the distribution of blood

cholesterol is approximately Normal: N(170, 30). What is thefirst quartile of the distribution of blood cholesterol?


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High levels of cholesterol in the blood increase the risk of

heart disease. For 14-year-old boys, the distribution of blood

cholesterol is approximately Normal: N(170, 30). Cholesterollevels above 240mg/dL may require medical attention. What

percent of 14-year-old boys have more than 240mg/dL of

cholesterol? What percent of 14-year-old boys have blood

cholesterol between 200 and 240 mg/dL?



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Assessing Normality

The Normal distributions provide good models for some

distributions of real data. Many statistical inference proceduresare based on the assumption that the population is

approximately Normally distributed. Consequently, we need a

strategy for assessing Normality.

Plot the data.

Make a dotplot, stemplot, or histogram and see if the graph is

approximately symmetric and bell-shaped.

Check whether the data follow th e 68-95-99.7 rule.

Count how many observations fall within one, two, and three

standard deviations of the mean and check to see if these

percents are close to the 68%, 95%, and 99.7% targets for a

Normal distribution.


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Normal Probability Plots

Most software packages can construct Normal probability plots.

These plots are constructed by plotting each observation in a data set

against its corresponding percentiles z-score.

If the points on a Normal probability plotlie close to a straight line,

the plot indicates that the data are Normal. Systematic deviations from

a straight line indicate a non-Normal distribution. Outliers appear aspoints that are far away from the overall pattern of the plot.

Interpreting Normal Probability Plots

E l


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Assessing Normality


are between -1.25 and 0.81.

Example

Can you find the same proportion using a different approach?

1 - (0.1056+0.2090) = 10.3146

= 0.6854


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Assessing Normality: Is Unemployment

Normal?

At right are the data onunemployment rates in the 50

states in November, 2009.

Below is a histogram showing

the data.

1 - (0.1056+0.2090) = 10.3146

= 0.6854

4 1 5

5 0

6 3 3 4 4 6 7 7 7 9

7 0 0 2 4 4 4 8

8 0 0 2 2 4 5 5 6 7 8 9

9 1 2 5 6 6 7

10 2 3 5 6 6 8 9

11 1 5

12 3 3 3 7

13

14 7

Mean = 8.682 StDev = 2.225

Key: 4|1 = 4.1%


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Assessing Normality: Is Unemployment

Normal?

Refer to the tableat right:

Mean = 8.682 StDev = 2.225

Mean 1SD 6.457 to 10.907 36/50 = 72%

Mean2SD 4.232 to 13.132 48/50 = 96%

Mean3SD 2.007 to 15.357 50/50 = 100%

These percents are quite close to the68%, 95%, and 99.7% targets for a

Normal distribution.


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Assessing Normality: Were last years Seniors

GPAs Normal?0.78 2 2.31 2.79 2.93 3.23 3.46 3.89

0.81 2 2.39 2.8 2.93 3.23 3.48 3.92

0.92 2.04 2.53 2.83 3 3.27 3.5 3.98

1.28 2.2 2.55 2.84 3.11 3.34 3.79 4.15

1.28 2.28 2.63 2.86 3.12 3.36 3.79

1.78 2.29 2.77 2.88 3.13 3.38 3.79

Mean1SD

Mean2SD

Mean3SD


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Assessing Normality: Are your GPAs Normal?

1.22 1.36 1.39 1.45 1.49 1.60 1.63 1.95

1.95 2 2 2.03 2.06 2.26 2.41 2.44

2.45 2.46 2.49 2.51 2.53 2.64 2.75 2.80

2.83 2.83 2.94 2.94 2.99 2.99 3.11 3.14

3.16 3.23 3.27 3.31 3.31 3.37 3.453.48

3.55 3.75 4.04 4.06 4.33 4.35

Mean1SD

Mean2SD

Mean3SD

Mean = 2.70 St.Dev. = 0.81


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Assessing Normality: No Space in the Fridge?

The measurements listed below describe the useable capacity

(in cubic feet) of a sample of 36 side-by-side refrigerators.(source: Consumer Reports, May 2010)Are the data close to Normal?

Mean1SD

Mean2SD

Mean3SD

12.9 13.7 14.1 14.2 14.5 14.5 14.6

14.7 15.1 15.2 15.3 15.3

15.3 15.3 15.5 15.6 15.6 15.8 16.016.0 16.2 16.2 16.3 16.4

16.5 16.6 16.6 16.6 16.8 17.0 17.0

17.2 17.4 17.4 17.9 18.4


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Assessing Normality: Are Twizzlers Normal?

Ms. Raskin will give you a Twizzler. Measure its length to the

nearest cm. Are Twizzlers Normal?

Mean1SD

Mean2SD

Mean3SD


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What Can Go Wrong?

Dont use a Normal model when the distribution is

not unimodal and symmetric.


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What Can Go Wrong?

Dont use the mean and standard deviation when

outliers are present. Both mean and standard

deviation can be distorted by outliers.

Dont round off too soon.

Dont round your results in the middle of a calculation.

Dont worry about minor differences in results.

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In this chapter, we learned that

The Normal Distributionsare described by a special family of bell-shaped, symmetric density curves called Normal curves. The mean

and standard deviation completely specify a Normal distribution

N(,). The mean is the center of the curve, and is the distance

from to the change-of-curvature points on either side.

All Normal distributions obey the 68-95-99.7 Rule, which describes

what percent of observations lie within one, two, and three standard

deviations of the mean.

Summary

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All Normal distributions are the same when measurements are

standardized. The standard Normal distributionhas mean =0

and standard deviation =1.

Table Agives percentiles for the standard Normal curve. By

standardizing, we can use Table A to determine the percentile for a

given z-score or the z-score corresponding to a given percentile in

any Normal distribution.

To assess Normality for a given set of data, we first observe its

shape. We then check how well the data fits the 68-95-99.7 rule.We

can also construct and interpret a Normal probability plot.

Summary (contd)

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+Ch. 6

Describing Location in a Distribution

There are two ways of describing an individuals location within a

distributionthe percentileand z-score.

A cumulative relative frequency graphallows us to examine

location within a distribution.

It is common to transform data, especially when changing units of

measurement. Transforming data can affect the shape, center, and

spread of a distribution.

We can sometimes describe the overall pattern of a distribution by a

density curve(an idealized description of a distribution that smooths

out the irregularities in the actual data).

Summary (contd)

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+Looking Ahead

Well learn how to describe relationships between two

quantitative variables

Well study

Scatterplots and correlation

Least-squares regression

In the next Chapter

Statistics -- Normal Distribution

Documents

Transcript of Statistics -- Normal Distribution