Transcript 6.1 Normal Distributions

Chapter 6
The Normal Distribution
(The original slide set from the Bluman text has been trimmed
and augmented with extra explanations and TI-84 examples.
The Bluman slides have the footer and rose background.)
© McGraw-Hill, Bluman, 5th ed., Chapter 6
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6.1 Normal Distributions
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Many continuous variables have distributions
that are bell-shaped and are called
approximately normally distributed
variables.
The theoretical curve, called the bell curve or
the Gaussian distribution, can be used to
study many variables that are not normally
distributed but are approximately normal.
Bluman, Chapter 6
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Normal Distributions
The mathematical equation for the a normal
distribution is:
y
e
( X   )2 (2 2 )
 2
where
e  2.718
  3.14
  population mean
  population standard deviation
Bluman, Chapter 6
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Normal Distributions
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The shape and position of the normal
distribution curve depend on two parameters,
the mean and the standard deviation.
Each normally distributed variable has its own
normal distribution curve, which depends on the
values of the variable’s mean and standard
deviation.
Bluman, Chapter 6
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Normal Distributions - different ones
for different values of 𝜇 and 𝜎
Bluman, Chapter 6
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Normal Distribution Properties
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The normal distribution curve is bell-shaped.
The mean, median, and mode are equal and
located at the center of the distribution.
The normal distribution curve is unimodal (i.e.,
it has only one mode).
The curve is symmetrical about the mean,
which is equivalent to saying that its shape is
the same on both sides of a vertical line
passing through the center.
Bluman, Chapter 6
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Normal Distribution Properties
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The curve is continuous—i.e., there are no
gaps or holes. For each value of X, here is a
corresponding value of Y.
The curve never touches the x axis.
Theoretically, no matter how far in either
direction the curve extends, it never meets the
x axis—but it gets increasingly closer.
Bluman, Chapter 6
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Normal Distribution Properties
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They said “it never meets the x axis—but it gets
increasingly closer.”
Example: for the standard normal distribution
where 𝜇 = 0, 𝜎 = 1, when 𝑥 = 5, 𝑦 = 0.0000015
And when 𝑥 = 10, 𝑦 = 7.7 × 10−23
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Normal Distribution Properties
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The total area under the normal distribution
curve is equal to 1.00 or 100%.
The area under the normal curve that lies within
 one standard deviation of the mean is
approximately 0.68 (68%).
 two standard deviations of the mean is
approximately 0.95 (95%).
 three standard deviations of the mean is
approximately 0.997 ( 99.7%).
Bluman, Chapter 6
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Normal Distribution Properties
Bluman, Chapter 6
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“The Empirical Rule”
Bluman, Chapter 6
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Standard Normal Distribution
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Since each normally distributed variable has its
own mean and standard deviation, the shape
and location of these curves will vary. In
practical applications, one would have to have
a table of areas under the curve for each
variable. To simplify this, statisticians use the
standard normal distribution.
The standard normal distribution is a normal
distribution with a mean of 0 and a standard
deviation of 1.
Bluman, Chapter 6
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The Standard Normal
Distribution is our favorite

This one is the most special of all of them
mean: 𝜇 = 0
 The standard deviation: 𝜎 = 1
 The
Horizontal 𝑧 = ⋯ , −3, −2, −1, 0, 1, 2, 3, ⋯
 Total area between curve and axis = 1.000

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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
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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
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Area Problems
KEY CONNECTION:
 PROBABILITY is AREA
 AREA is PROBABILITY
HOW THIS AFFECTS YOUR LIFE:
 You want to answer Probability questions
 You find the answers by finding Areas of
regions underneath the Normal Curve.
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Three kinds of area problems
“What is the area to the left of some 𝑧 ?”
 “What is the area to the right of some 𝑧 ?”
 “What is the area between 𝑧1 and 𝑧2 ?”
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How do you find these areas?
1.
2.
3.
By using a printed table that gives areas
to the left of 𝑧 = −#. ## and 𝑧 = #. ##
Or by using the TI-84 normalcdf(z1,z2)
function.
Or by using Excel’s
=NORM.S.DIST(z,TRUE) function
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Area under the Standard Normal
Distribution Curve
1. To the left of any z value:
Look up the z value in the table and use the
area given.
Bluman, Chapter 6
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Area under the Standard Normal
Distribution Curve
2. To the right of any z value:
Look up the z value and subtract the area
from 1.
Bluman, Chapter 6
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Area under the Standard Normal
Distribution Curve
3. Between two z values:
Look up both z values and subtract the
corresponding areas.
Bluman, Chapter 6
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TI-84 Methods
To get normalcdf()
 2ND DISTR

Area between two z vals.
 Normalcdf(zLow, zHigh)
It’s on the VARS key

2:normalcdf(
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Beware! You do NOT
want to use 1:normalpdf(
in most cases in this
course.
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Example: Area between
𝑧 = −1 and 𝑧 = +1
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Agrees with the Empirical
Rule (68/95/99.7) !
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TI-84 Methods
Area to the left of z
 normalcdf(-99,z)
 Example: Area to the left
of z=1.23
Area to the right of z
 normalcdf(z,99)
 Example: Area to the
right of z=1.23
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Observe: area to left +
area to right = 1.0000000
Not a coincidence!!!!
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Example 6-1: Area under the Curve
Find the area to the left of z = 1.99.
The value in the 1.9 row and the .09 column of
Table E is .9767. The area is .9767.
Bluman, Chapter 6
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Example 6-1: Area under the Curve
Find the area to the left of z = 1.99.
They got 0.9767 using the printed table.
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Example 6-2: Area under the Curve
Find the area to right of z = -1.16.
The value in the -1.1 row and the .06 column of
Table E is .1230. The area is 1 - .1230 = .8770.
Bluman, Chapter 6
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Example 6-2: Area under the Curve
Find the area to right of z = -1.16.
They used the printed table, which only gives
areas to the left, so they had to subtract,
1 -0.1230 = 0.8770.
The TI-84 normalcdf() was more direct.
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Example 6-3: Area under the Curve
Find the area between z = 1.68 and z = -1.37.
The values for z = 1.68 is .9535 and for z = -1.37
is .0853. The area is .9535 - .0853 = .8682.
Bluman, Chapter 6
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Example 6-3: Area under the Curve
Find the area between z = 1.68 and z = -1.37.
With the printed tables, they had to do two
lookups and subtract the results to get .8682
The TI-84 normalcdf() was more direct.
Bluman, Chapter 6
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Problems that work backwards
They give you the area
 You have to work backwards to find the z
score.
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Example of a backwards
problem
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What z value divides the area under the
standard normal curve so that the area to
the left of that z is 0.7123? And what is
the area to the right of that z score?
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How to solve it using the printed table
The area to the left is .7123. Then look for that
value inside Table E.
The z value is 0.56.
Bluman, Chapter 6
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Backwards problem using TI-84
invNorm( is the tool
 2ND DISTR again
 3:invNorm(
 Stands for “Inverse
Normal”
 Tell him area to the left
 He responds with the z
score.
Example: area 0.7123
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If they give you area to the right
Example: Find z so area to
the right of z is 0.7500
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But Tables & TI-84 deal
with areas to the left.
COMPLEMENT: If area
to the right is 0.7500, are
to the left is 1 – 0.7500
So we seek the z that has
0.2500 to its left.
How to solve it
In table:
 Lookup 0.2500 in table.
 If it’s not there, take the
closest value.
 Read out to find z again.
 With TI-84 invNorm(
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Backwards Area-Between
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“Find the z scores that delimit the middle
80% of the area” DRAW A PICTURE!!!
0.8000 is in the middle
 So 1.0000 – 0.8000 = 0.2000 in two tails
 0.2000 ÷ 2 = 0.1000 in each tail
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Backwards Area-Between
Using Printed Table
 Look deep in table for
closest match to 0.1000
 Read out to find the
negative z on the left.
 Because of symmetry,
the positive z on the right
is the opposite of that
value.
Using TI-84
 invNorm(.1000)
 Because of symmetry,
the positive z on the right
is the opposite of that
value.
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Answers: 𝑧 = −1.28 and
𝑧 = +1.28
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Confirming this with TI-84
Using normalcdf()
 normalcdf(-1.28,1.28)
should give about area
.8000 they asked for
 More decimals for more
precision
Using DRAW
 2ND DRAW 1:ClrDraw
 2ND DISTR
 Right arrow to DRAW
 1:ShadeNorm(z1,z2)
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Confirming this with TI-84
WINDOW settings
 It probably won’t turn out
well on its own. You may
need to do some thinking.
 What we did for this one:
Using DRAW
 2ND DRAW 1:ClrDraw
 2ND DISTR
 Right arrow to DRAW
 1:ShadeNorm(z1,z2)
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