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```The Normal Distribution
The Distribution
1200
1000
800
600
400
200
Std. Dev = 1.00
Mean = -.01
N = 10000.00
0
50
3.
00
3.
50
2.
00
2.
50
1.
00
1.
0
.5
00
0.
0
-.5 0
.0
-1
0
.5
-1
0
.0
-2
0
.5
-2
0
.0
-3
0
.5
-3
0
.0
-4
X
The Standard Normal
Distribution
• We simply transform all X values to have
a mean = 0 and a standard deviation = 1
• Call these new values z
• Define the area under the curve to be 1.0
z Scores
• Calculation of z
Xμ
z
σ
 where  is the mean of the population and 
is its standard deviation
 This is a simple linear transformation of X.
Tables of z
• We use tables to find areas under the
distribution
• A sample table is on the next slide
• The following slide illustrates areas under
the distribution
Normal Distribution
Cutoff at +1.645
1200
1000
z=
1.64545
45
800
Area =
.05 .05
600
400
200
0
50
3.
00
3.
50
2.
00
2.
50
1.
00
1.
0
.5
00
0.
0
-.5
0
.0
-1
0
.5
-1
0
.0
-2
0
.5
-2
0
.0
-3
0
.5
-3
0
.0
-4
z
Using the Tables
• Define “larger” versus “smaller” portion
• Distribution is symmetrical, so we don’t
need negative values of z
• Areas between z = +1.5 and z = -1.0
 See next slide
Calculating areas
• Area between mean and +1.5 = 0.4332
• Area between mean and -1.0 = 0.3413
• Sum equals
0.7745
• Therefore about 77% of the observations
would be expected to fall between z =
-1.0 and z = +1.5
Converting Back to X
• Assume  = 30 and  = 5
• 77% of the distribution is expected to lie
between 25 and 37.5
z
X 

Therefore X    z  
X  30  1.0  5  25
X  30  1.5  5  37.5
Probable Limits
• X=+z
• Our last example has  = 30 and  = 5
• We want to cut off 2.5% in each tail, so
 z = + 1.96
X    z 
X  30  1.96  5  39.8
X  30  1.96  5  20.2
Cont.
Probable Limits--cont.
• We have just shown that 95% of the
normal distribution lies between 20.2 and
39.8
• Therefore the probability is .95 that an
observation drawn at random will lie
between those two values
Measures Related to z
• Standard score
 Another name for a z score
• Percentile score
 The point below which a specified percentage of the
observations fall
• T scores
 Scores with a mean of 50 and a standard deviation of
10
Cont.
```
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