WHEN IS A Z-SCORE BIG?
NORMAL MODELS
A Very Useful Model for Data
µ = 3 and  = 1

0
3
6
8
9
12
X
Normal Models: A family of bell-shaped
curves that differ only in their means and
standard deviations.
µ = the mean
 = the standard deviation
Normal Models
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 
 The
Total area =1; symmetric around µ
The effects of  and 
How does the standard deviation affect the shape of the bell curve?
= 2
 =3
 =4
How does the expected value affect the location of the bell curve?
 = 10  = 11  = 12
µ = 3 and  = 1

0
3
6
3
12
µ = 6 and  = 1

0
9
X
6
9
12
X

0
3
µ = 6 and  = 2
6
8
3
12
µ = 6 and  = 1

0
9
X
6
8
9
12
X
µ = 6 and  = 2
0
3
6
9
12
X
area under the density curve between 6 and 8
is a number between 0 and 1
area under the density curve between 6 and 8
is a number between 0 and 1
Standardizing
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.
 Suppose
Standardizing (cont.)
 is also a normal model; we will
denote it by Z:
Z = (X 
  has mean 0 and standard deviation 1:
 = 0;  = 1.
  , 1
 The normal model Z is called the
standard normal model.
 (X
µ = 6 and  = 2
0
3
6
8
9
12
X
(X-6)/2
µ = 0 and  = 1
.5
-3
-2
-1
.5
0
1
2
3
Z
Standard Normal Model
.5
-3
-2
-1
.5
0
1
2
3
Z
Z~N(0, 1) denotes the standard normal
model
 = 0 and  = 1
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
Standard normal areas have been
calculated and are provided in table Z.
Area between -
and z0
The tabulated area corresponds
to the area between Z= - and some z0
Z = z0
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0
0.5000
0.5040
0.5080
0.5120
0.5160
0.5199
0.5239
0.5279
0.5319
0.5359
0.1
0.5398
0.5438
0.5478
0.5517
0.5557
0.5596
0.5636
0.5675
0.5714
0.5753
0.2
0.5793
0.5832
0.5871
0.5910
0.5948
0.5987
0.6026
0.6064
0.6103
0.6141
0.8413
0.8438
0.8461
0.8485
0.8508
0.8531
0.8554
0.8577
0.8599
0.8621
0.8849
0.8869
0.8888
0.8907
0.8925
0.8944
0.8962
0.8980
0.8997
0.9015
…
1
…
…
…
…
1.2
…
0.01
…
0.00
…
z
…
…
…
…
Example – begin with a normal model
with mean 60 and stand dev 8
0.8944
Proportion of the area to the left of 70
under the original curve is the proportion
70  60
of the area to the left of
= 1.25
8
under the standard normal Z curve
= 0.8944
z
0.0
0.1
0.2
0.8438
0.8461
0.8849
0.8869
0.8888
0.05
0.5199
0.5596
0.5987
0.8485
0.8508
0.8531
0.8907
0.8925
0.8944
0.06
0.5239
0.5636
0.6026
0.07
0.5279
0.5675
0.6064
0.08
0.5319
0.5714
0.6103
0.09
0.5359
0.5753
0.6141
0.8554
0.8577
0.8599
0.8621
0.8962
0.8980
0.8997
0.9015
…
…
…
1.2
0.04
0.5160
0.5557
0.5948
…
0.8413
…
1
0.03
0.5120
0.5517
0.5910
…
0.02
0.5080
0.5478
0.5871
In this example z0 = 1.25
…
0.01
0.5040
0.5438
0.5832
…
0.00
0.5000
0.5398
0.5793
0.8944
0.8944
0.8944
0.8944
…
…
…
…
Areas Under the Z Curve:
Using the Table
Proportion of area above the interval
from 0 to 1 = .8413 - .5 = .3413
.50
.3413 .1587
0
1
Z
Example
Area=.3980
0
 Area
between 0 and 1.27) =
1.27
z
.8980-.5=.3980
Example
A2
0
.55
Area to the right of .55 = A1
= 1 - A2
= 1 - .7088
= .2912
Example
Area=.4875
Area=.0125 -2.24
 Area
0
between -2.24 and 0 =
z
.5 - .0125 = .4875
Example
Area to the left of -1.85
= .0322
Example
.9968
A1
A1
.1190
A2
A
-1.18
0
2.73
z
Area between -1.18 and 2.73 = A - A1

= .9968 - .1190

= .8778

Example
.6826
.1587
.8413
Area between -1 and +1 = .8413 - .1587 =.6826
Example
Area to the right of 250
under original curve
= area to the right of
250  275 25
Z=
=
= .58
43
43
under the standard normal
curve = 1  .2810 = .7190
Example
.8671
.1230
.9901
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
Example: from
percentiles to
scores, z in
reverse.
-.67
Is k positive or negative?
Direction of inequality; magnitude of probability
Look up .2514 in body of table; corresponding entry is -.67
N(275, 43); find k so that area to the left of k is .9846
.9846 = area to the left of k under
N(275,43) curve = area to left of
z = (k  275) 43 under N(0,1)
curve  k  275 = 2.16 (from
43
standard normal table)
 k = 2.16(43)  275 = 367.88
Area to the left of 2.16 = .9846
0.45
0.4
0.35
0.3
0.25
.9846
0.2
0.15
0.1
0.05
0
X
-3
0
2.16
3
Underweight Cereal Boxes
On a production line the amount of
cereal dispensed into individual cereal
boxes by an automatic filling machine
can be described by a Normal model
with mean 16.3 ounces and standard
deviation 0.2 ounces.
If the label on the cereal box reads
16.0 ounces, what percentage of
the boxes will be underweight?
Denote by X the amount of cereal
dispensed into a box. X~N(16.3, 0.2)
Underweight Cereal Boxes-2
Using Statcrunch
What percentage of the
cereal boxes will be
underweight (less than
16.0)? X~N(16.3, 0.2)
Use StatCrunch:
• Click Stat button;
• In drop-down menu
point to Calculators;
• In continuation menu
click Normal.
Underweight Cereal Boxes-3
Using Statcrunch
What percentage of the
cereal boxes will be
underweight (less than
16.0)? X~N(16.3, 0.2)
Use StatCrunch:
• Input the required
information (yellow);
• Click Compute button
Area to the left of 16.0 =
0.0668
Underweight Cereal Boxes-4
Using Statcrunch
What percentage of the cereal boxes will be
underweight (less than 16.0)?
• Area to the left of 16.0 = 0.0668
• Conclusion: approximately 6.7% of the
boxes will contain less than 16.0 ounces of
cereal.
•
Example:
determine the mean
 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
Solution
X =amount of dye discharged into can
X ~N( , .4); determine  so that
area to the right of 6 is .01
Solution (cont.)
X =amount of dye discharged into can
X ~N( , .4); determine  so that
the area to the right of x= 6 is .01.
.01 = area to the right of x = 6
= area to the right of z = (6   ) .4

 6.4
= 2.33(from standard normal table)
  = 6-2.33(.4) = 5.068
Are You Normal? Normal
Probability Plots
Checking your data to determine if a normal
model is appropriate
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.
Are You Normal? Normal Probability
Plots (cont)
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)
 Nearly
Normal data have a histogram
and a Normal probability plot that look
somewhat like this example:
Are You Normal? Normal Probability
Plots (cont)
A
skewed distribution might have a
histogram and Normal probability plot
like this:
End of Normal Models