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Normal Probability Distributions
x
The normal distribution is a descriptive model
that describes real world situations.
• The mean
• Bell shaped and is symmetric about the mean
• The total area that lies under the curve is one
2
3
The Normal Distribution:
as mathematical function (pdf)
f ( x) 
1
 2
1 x 2
 (
)
2

e
This is a bell shaped
curve with different
centers and spreads
depending on  and 
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The Normal PDF
It’s a probability function, so no matter what the values of
 and , must integrate to 1!



1
2
1 x 2
 (
)
 e 2  dx
1
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Normal distribution is defined by its
mean and standard dev.

E(X)= =  x

1
2

1 x 2
 (
)
2

e
dx

Var(X)=2
=
(


x
2
1
 2
1 x 2
 (
)
2

e
Standard Deviation(X)=
dx)   2
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Means and Standard Deviations
Curves with different means, same standard deviation
10 11
12 13 14
15 16 17 18 19
20
Curves with different means, different standard deviations
9 10 11 12 13 14 15 16 17 18 19 20 21 22
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Empirical Rule
68%
About 68% of the area
lies within 1 standard
deviation of the mean
About 95% of the area
lies within 2 standard
deviations
About 99.7% of the area lies within
3 standard deviations of the mean
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The Standard Normal Distribution
The standard normal distribution has a mean of 0 and a
standard deviation of 1.
–4 –3 –2 –1
0 1
2 3
4
z
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The Standard Normal (Z):
“Universal Currency”
The formula for the standardized normal
probability density function is
1
p( Z ) 
e
(1) 2
1 Z 0 2
 (
)
2 1
1

e
2
1
 ( Z )2
2
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The Standard Normal (Z):
“Universal
Currency”
The Standard
Normal Distribut
All normal distributions can be converted into
the standard normal curve by subtracting the
mean and dividing by the standard deviation:
Z
X 

Somebody calculated all the integrals for the standard
normal and put them in a table! So we never have to
integrate! .
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The Standard Score
The standard score, or z-score, represents the number of
standard deviations a random variable x falls from the
mean.
The test scores for a civil service exam are normally
distributed with a mean of 152 and a standard deviation of
7. Find the standard z-score for a person with a score of:
(a) 161
(b) 148
(c) 152
(a)
(b)
(c)
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The Standard Normal Distribution
The standard normal distribution has a mean of 0 and a
standard deviation of 1.
Using z-scores any normal distribution can be
transformed into the standard normal distribution.
–4 –3 –2 –1
0 1
2 3
4
z
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Probabilities are depicted by areas under the curve
• Total area under the
curve is 1
• The red area is equal to
p(z > 1)
• The blue area equal to
p(-1< z <0)
• Since the properties of
the normal distribution
are known, areas can be
looked up on tables or
calculated on computer.
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Suppose Z has standard normal
distribution Find p(0<Z<1.23)
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16
Find p(-1.57<Z<0)
Find p(Z>.78)
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Z is standard normal
Calculate p(-1.2<Z<.78)
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Probabilities and Normal Distributions
Example
IQ scores are normally distributed with a mean of 100
and a standard deviation of 15. Find the probability that a
person selected at random will have an IQ score less
than 115.
100 115
To find the area in this interval, first find the standard
score equivalent to x = 115.
115  100
z
1
15
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Probabilities and Normal Distributions
Normal Distribution
Standard Normal
Distribution
100 115
SAME
SAME
Find P(x < 115).
Find P(z < 1).
0 1
P(z < 1) = 0.5+0.3413=0.8413, so P(x <115) = 0.8413
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Example
Monthly bills in a certain city are normally distributed
with a mean of $100 and a standard deviation of $12. A
utility bill is randomly selected. Find the probability it is
between $80 and $115.
Normal Distribution
P(80 < x < 115)
P(–1.67 < z < 1.25)
0.4525 + 0.3944 = 0.8469
The probability bill is between $80
and $115 is 0.8469.
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From Areas to z-Scores
Find the z-score corresponding to a cumulative area of 0.9803.
z = 2.06 corresponds
Area=0.4803
0.9803
–4 –3 –2 –1 0
1
2
3
4
z
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Finding z-Scores from Areas
Find the z-score such that 45% of the area under the
curve falls between –z and z.
.225
.225
.45
–z 0
z
Half this area is .45/2 = .225 The closest table area is
.2257 and the z-score is 0.60. The positive z score is 0.60.
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From z-Scores to Raw Scores
To find the data value, x when given a standard score, z:
The test scores for a civil service exam are normally
distributed with a mean of 152 and a standard deviation of 7.
Find the test score for a person with a standard score of:
(a) 2.33
(b) –1.75
(c) 0
(a) x = 152 + (2.33)(7) = 168.31
(b) x = 152 + (–1.75)(7) = 139.75
(c) x = 152 + (0)(7) = 152
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Example
Monthly bills in a certain city are normally distributed with a
mean of $100 and a standard deviation of $12. What is the
smallest bill that can be in the top 10% of the bills?
$115.36 is the smallest
value for the top 10%.
90%
10%
z
Find the area in the table that is closest to 0.90% (the
40%)The area 0.3997 corresponds to a z-score of 1.28.
To find the corresponding x-value, use
x = 100 + 1.28(12) = 115.36.