Chapter 6 (part 2) WHEN IS A Z-SCORE BIG? NORMAL MODELS
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Transcript of Chapter 6 (part 2) WHEN IS A Z-SCORE BIG? NORMAL MODELS
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Chapter 6 (part 2) WHEN IS A Z-SCORE BIG? NORMAL
MODELSA Very Useful Model for Data
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X83 6 9 120
Normal Models: A family of bell-shaped curves that differ only in their means and standard deviations.
µ = the mean
= the standard deviation
µ = 3 and = 1
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Normal Models The mean, denoted ,can be any
number The standard deviation can be any
nonnegative number The total area under every normal
model curve is 1 There are infinitely many normal models Notation: X~N() denotes that data
represented by X is modeled with a normal model with mean and standard deviation
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Total area =1; symmetric around µ
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The effects of The effects of and and
How does the standard deviation affect the shape of the bell curve?
= 2
=3 =4
= 10 = 11 = 12How does the expected value affect the location of the bell curve?
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X3 6 9 120
X3 6 9 120
µ = 3 and = 1
µ = 6 and = 1
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X83 6 9 120
X83 6 9 120
µ = 6 and = 2
µ = 6 and = 1
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area under the density curve between 6 and 8 is a number between 0 and 1
3 6 9 12
µ = 6 and = 2
0 X
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area under the density curve between 6 and 8 is a number between 0 and 1
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Standardizing
Suppose X~N( Form a new normal model by
subtracting the mean from X and dividing by the standard deviation :
(X This process is called standardizing the
normal model.
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Standardizing (cont.)
(X is also a normal model; we will denote it by Z:
Z = (X has mean 0 and standard deviation
1: = 0;
The normal model Z is called the
standard normal model.
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Standardizing (cont.) If X has mean and stand. dev. , standardizing
a particular value of x tells how many standard deviations x is above or below the mean .
Exam 1: =80, =10; exam 1 score: 92
Exam 2: =80, =8; exam 2 score: 90
Which score is better?
1 exam on 92 than better is 2 exam on 90
1.258
10
8
8090z
1.210
12
10
8092z
2
1
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X83 6 9 120
µ = 6 and = 2
Z0 1 2 3-1-2-3
.5.5
µ = 0 and = 1
(X-6)/2
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Z~N(0, 1) denotes the standard normal model
= 0 and = 1
Z0 1 2 3-1-2-3
.5.5
Standard Normal Model
.5.5
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Important Properties of Z
#1. The standard normal curve is symmetric around the mean 0
#2. The total area under the curve is 1;
so (from #1) the area to the left of 0 is 1/2, and the area to the right of 0 is 1/2
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Finding Normal Percentiles by Hand (cont.)
Table Z is the standard Normal table. We have to convert our data to z-scores before using the table.
The figure shows us how to find the area to the left when we have a z-score of 1.80:
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Areas Under the Z Curve: Using the Table
Proportion of area above the interval from 0 to 1 = .8413 - .5 = .3413
0 1Z
.1587.3413
.50
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Standard normal areas have been calculated and are provided in table Z.
The tabulated area correspondto the area between Z= - and some z0
Z = z0
Area between - and z0
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Example – begin with a normal model with mean 60 and stand dev 8
In this example z0 = 1.25
0.89440.8944
0.89440.8944
Proportion of the area to the left of 70
under the original curve is the proportion
70 60of the area to the left of 1.25
8under the standard normal Z curve
= 0.8944
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Example
Area between 0 and 1.27) =
1.270 z
Area=.3980
.8980-.5=.3980
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Example
Area to the right of .55 = A1
= 1 - A2
= 1 - .7088 = .2912
0 .55
A2
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Example
Area between -2.24 and 0 =
Area=.4875
.5 - .0125 = .4875
z-2.24 0Area=.0125
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Example
Area to the left of -1.85 = .0322
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Example
A1 A2
0 2.73z
-1.18
Area between -1.18 and 2.73 = A - A1
= .9968 - .1190
= .8778
.1190
.9968
A1
A
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Area between -1 and +1 = .8413 - .1587 =.6826
.8413.1587
.6826
Example
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Example
Is k positive or negative?
Direction of inequality; magnitude of probability
Look up .2514 in body of table; corresponding entry is -.67
-.67
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Example
Area to the right of 250
under original curve
area to the right of
250 275 25.58
43 43under the standard normal
curve = 1 .2810 .7190
Z
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Example
area between 225 and 375 area under
standard normal curve between z = (225 275) 43
= -1.16 and z = (375 275) 43 = 2.33;
the area is .9901 .1230 .8671
.9901.1230
.8671
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N(275, 43); find k so that areato the left is .9846
.9846 area to the left of k under N(275,43)
curve area to left of z = ( 275) 43 under
275N(0,1) curve 2.1643
(from standard normal table)
2.16(43) 275 367.88
k
k
k
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Area to the left of z = 2.16 = .9846
0 2.16 Z.1587
.4846
Area=.5
.9846
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Example
Regulate blue dye for mixing paint; machine can be set to discharge an average of ml./can of paint.
Amount discharged: N(, .4 ml). If more than 6 ml. discharged into paint can, shade of blue is unacceptable.
Determine the setting so that only 1% of the cans of paint will be unacceptable
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Solution
=amount of dye discharged into can
~N( , .4); determine so that
area to the right of 6 is .01
X
X
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Solution (cont.)
6.4
(6 ) .4
=amount of dye discharged into can
~N( , .4); determine so that
the area to the right of x= 6 is .01.
.01 area to the right of 6
area to the right of z =
2.33(from standard normal tab
X
X
x
le)
= 6-2.33(.4) = 5.068
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Are You Normal? Normal Probability Plots
Checking your data to determine if a normal model is appropriate
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Are You Normal? Normal Probability Plots
When you actually have your own data, you must check to see whether a Normal model is reasonable.
Looking at a histogram of the data is a good way to check that the underlying distribution is roughly unimodal and symmetric.
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A more specialized graphical display that can help you decide whether a Normal model is appropriate is the Normal probability plot.
If the distribution of the data is roughly Normal, the Normal probability plot approximates a diagonal straight line. Deviations from a straight line indicate that the distribution is not Normal.
Are You Normal? Normal Probability Plots (cont)
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Nearly Normal data have a histogram and a Normal probability plot that look somewhat like this example:
Are You Normal? Normal Probability Plots (cont)
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A skewed distribution might have a histogram and Normal probability plot like this:
Are You Normal? Normal Probability Plots (cont)