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Mean,Mean.
A FormulaA Formula: page 6
The Sample
Sample
If X is any random variable, then, as n increases without bound,
x  X
the distribution of its standardized sample mean,
, approaches the
X
n
distribution of the standard normal random variable, Z, whose p.d.f. is
1
0.5z 2
f Z ( z) 
e
.
2 
This is called the Central Limit Theorem.
(material continues)

T
C
I

Normal Distributions. Standard Normal
Normal Distributions, Standard Normal
1. STANDARD NORMAL(Mean 0 & Standard deviation 1)
In The Sample Mean we derived the probability density function
1
0.5z 2
f Z ( z) 
e
2 
for the standard normal random variable Z.
We can use integration and fZ to compute probabilities for Z.
Example 1. Compute P(0.74  Z  1.29). As we saw in
Integration,
1.29
P(  0.74  Z  1.29) 

 0.74
1
 0.5z 2
e
dz
2 
To evaluate the integral open Integrating.xls and enter the function as
=(1/SQRT(2*PI()))*EXP(-0.5*x^2).
Recall that x is the only variable that can be used in Integrating.xls.
Integrating.xls
(material continues)
T
C
I

Normal Distributions. General Normal
Normal, General Normal
2. GENERAL NORMAL(Any Mean & Any Standard deviation/But its
standardization is a standard normal distribution)
The adjective “standard”, used in standard normal distributions,
implies that there are “non-standard” normal distributions. This is indeed the
case.
A random variable, X, is called normal if its standardization,
X  X
S
,
X
has a standard normal distribution.
It can be shown that the probability density function for a normal
random variable, X, with mean X and standard deviation X has the following
form.
2
fX ( x) 
1
 X  2 
(material continues)
 x X
 0.5
 X
e




T
C
I

Normal Distributions• Standard Normal Random Variable (Z)


f
z


e
Z
• p.d.f.
2
1
 0.5 z 2
• Can use Integrating.xls to find probabilities
Normal Distributions• Ex1. Find P 1.34  Z  0.68
• Soln: show ex1 excel file
P 1.34  Z  0.68  
0.68
1
1.34
2
 0.6616
e
 0.5 z 2
dz
Normal Distributions
• Ex. Find a number z 0 so that PZ  z 0   0.9750.
• Soln:show ex2 excel file
z0
1
0.5 z 2
PZ  z 0   
e
dz

2
z 0  1.9600
Normal Distributions
• The previous example tells us that 97.5% of all
data for a standard normal random variable lies in
the interval
. 
 , 1.96
• This means that 2.5% of the data lies above z =
1.96
• Graphically, we have the following:
Normal Distributions
• The shaded region corresponds to 97.5% of all
possible area (note 2.5% is not shaded)
-6
-5
-4
-3
-2
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
-1
0
1
2
3
4
5
6
1.96
Normal Distributions
• Due to symmetry, we get 95% of the area shaded
with 5% not shaded (2.5% on each side)
-6
-1.96
-5
-4
-3
-2
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
-1
0
1
2
3
4
5
6
1.96
Normal Distributions
• This means that a 95% confidence interval for
the standard normal random variable Z is (-1.96,
1.96)
-1.96
-6
-5
-4
-3
-2
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
-1
0
1
2
3
4
5
6
1.96
Normal Distributions
•
A 95% confidence interval tells you how well a
particular value compares to known data or sample data
•
The interval that is constructed tells you that there is a
95% probability that the interval will contain the mean
of X.
•
Another interpretation is that 95% of all values found in
a sample should lie within this 95% confidence interval.
Normal Distributions
•
Possible formulas:
Z Standard Normal random variable
P 1.96  Z  1.96  0.95
Normal Distributions
•
Possible formulas:


x  x
P  1.96 
 1.96   0.95
x




x


x
P  1.96  s  1.96   0.95


n


Important
•
Possible formulas:
S
If is
X
X  X
X

x  x
x

x  x
X
n
unknown
The sample standard deviation,
s ( sm all )
, will be a very good approximation for
X
Normal Distributions
•
Remember, that -1.96 and 1.96 were special values that
apply to a 95% confidence interval
•
You need to find different values for other types of
confidence intervals.
•
Ex. Find a 99% confidence interval for Z. Find a 90%
confidence interval for Z.
Normal Distributions
• Soln:PZ  z1   0.9950
PZ  z 2   0.9500
z1  2.5759
z 2  1.6449
• Ex. What is the confidence interval for Z with 1
standard deviation? 2 standard deviations? 3
standard deviations?
Normal Distributions
• Soln:
P 1  Z  1  0.6827
-6
-5
-4
-3
-2
P 2  Z  2  0.9545
-6
-5
-4
-3
-2
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-1
0
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-1
0
1
1
2
2
3
3
4
4
5
6
5
6
Normal Distributions
• Soln:
P 3  Z  3  0.9973
-6
-5
-4
-3
-2
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-1
0
1
2
3
4
5
6
Normal Distributions
• Since Z is a standard normal random variable, Z
would have standardized some variable X.
•
So,
Z
X  X
X
Normal Distributions
• Ex. Suppose X is a normal random variable with
and  X  15. Find a 95%
 X  120
confidence interval for X if the 95% confidence
interval for Z is (-1.96, 1.96).
Normal Distributions
• Soln:
P 1.96  Z  1.96  0.95


X  X
P  1.96 
 1.96   0.95
X


X  120


P  1.96 
 1.96   0.95
15


P 1.96  15  X  120  1.96  15  0.95
P120  1.96  15  X  120  1.96  15  0.95
P90.6  X  149.4  0.95
So, the 95% confidence interval for X is (90.6, 149.4). We are
95% confident that this interval contains the true mean
Normal Distributions
• Ex. Suppose X is a normal random variable. If a
sample of size 34 was taken with  x  120 and
, find a 95% confidence interval for
s  15
the sample mean-remember this is
x
if the 95% confidence interval for Z is (-1.96, 1.96).
Normal Distributions
P 1.96  Z  1.96   0.95
• Soln:


x  x
P  1.96 
 1.96   0.95
x






x


x
P  1.96 
 1.96   0.95
X




n






x


x
P  1.96 
 1.96   0.95
s




n






x

120
P  1.96 
 1.96   0.95
15




34


Normal Distributions
• Soln:

15
15 
P  1.96 
 x  120  1.96 
  0.95
34
34 


15
15 
P120  1.96 
 x  120  1.96 
  0.95
34
34 

P114.9579  x  125.0421  0.95
So, the 95% confidence interval for x is (114.9579,
125.0421).
We are 95% confident that this interval contains the true mean
Normal Distributions
• General Normal Random Variable
• p.d.f.
f X x  
1
 X  2
e
 x X
 0.5
 X



2
• Probabilities are done similarly to Standard NRV
Normal Distributions
• Ex. If X is a normal random variable representing
exam 1 scores with mean 75 and standard
deviation 10, find P70  X  100 .
• Soln:
P70  X  100  
100
1
10  2
 0.6853
70
e
 x 75 
0.5

 10 
2
dx
Normal Distributions
• NORMDIST function in Excel
• Can calculate p.d.f. and c.d.f. values for a normal
random variable
• Ex. If X is a normal random variable representing
exam 1 scores with mean 75 and standard
deviation 10, find P70  X  100.
Normal Distributions(show excel)
• Soln:
P70  X  100  FX 100  FX 70
 NORMDIST 100, 75, 10, TRUE   NORMDIST 70, 75, 10, TRUE 
 0.6853
100
75
10
TRUE
70
75
10
TRUE
Normal Distributions
• Specific values for the p.d.f. can also be calculated
using NORMDIST
• Ex. Find height of p.d.f. of a normal random
variable X at X = 90 that has a normal distribution
with   71 and  X  12
.
X
Normal Distributions
• Soln: Two ways to solve=0.009
1
(1) Evaluate 12  2
e
 90 71 
 0.5

 12 
2
(2) Evaluate =NORMDIST(90, 71, 12, FALSE)
using Excel
Normal Distributions
• How does the mean and standard deviation affect
the shape of the Normal Random Variable graph?
• Ex. Graph the p.d.f. of a normal random variable
with the following characteristics:
(1)  X  0 and  X  1
X 5


0
X
(2)
and

1

4
X
X
(3)
and
X  4


5
X
(4)
and
Why?
Normal Distributions
Recall-General Normal random variable p.d.f
• Soln: (1)  X  0
Max Height

 X 1
and
0.40
The y-values of the graph
around x = -3 and x = 3
are very small
-6
-5
-4
-3
-2
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
-1 0
1
2
3
4
5
6
Normal Distributions
• General Normal Random Variable
e^(0)=1
• p.d.f.
f X x 
1
 X  2
 x X
0.5

e  X
y int ercept  f X 0 
1
1  2



2
 00
0.5
1
e 



2
~
1
~ 0.4
2
Normal Distributions-sec1-4/13
Why?
• Soln: (2)  X  0
and

Max Height
0.08
(This is 0.40  std. dev.)
Y values very small
Around
x = -15 and
x = 15
-25
(This is 3 standard
deviations from the mean- 3 *
-20
-15
X)
-10
X 5
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00
-5
0
5
10
15
20
25
Normal Distributions
• General Normal Random Variable
e^(0)=1
• p.d.f.
 x X
 0.5
 X



 00
 0.5
 5
2
1
f X x  
e
 X  2
 f X 0 
1
e
5  2




2
~
1
~ 0.4 / 5 ~ 0.08
5 2
Normal Distributions
• Soln: (3)  X  4
Max Height  0.40
(At x = 4)
Y values very small
around x = 1 and x = 7
(This is 3 standard
deviations from
the mean)
 X 1
and
-2
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
-1 0
1
2
3
4
5
6
7
8
9
10
Normal Distributions
• Soln: (4)  X  4
Max Height  0.08
(This is 0.40 std. dev.)

Y values very small
around x = -11 and x = 19
(This is 3 standard
deviations from the
mean)
and
X 5
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00
-26 -21 -16 -11 -6 -1
4
9
14
19
24
29
34
Normal Distributions
• Ex. Find the mean and standard deviation for the
following normal random variables graphed.
(A)
0.14
0.12
0.10
0.08
0.06
0.04
0.02
-9
-6
0.00
-3
0
3
6
9
12
15
18
21
Normal Distributions
0.14
0.12
0.10
0.08
0.06
0.04
0.02
-9
-6
0.00
-3
0
3
6
9
12
15
18
21
• Mean is 6 and standard deviation is 3
Normal Distributions
• (B)
0.25
0.20
0.15
0.10
0.05
-17
-15
-13
-11
-9
-7
-5
0.00
-1
-3
1
3
Normal Distributions
0.25
0.20
0.15
0.10
0.05
-17
-15
-13
-11
-9
-7
-5
0.00
-3
-1
1
3
• Mean is -7 and standard deviation is 2
Normal Distributions
• (C)
0.009
0.008
0.007
0.006
0.005
0.004
0.003
0.002
0.001
0
50
100 150 200 250 300 350 400 450 500 550
Normal Distributions
0.009
0.008
0.007
0.006
0.005
0.004
0.003
0.002
0.001
0
50
100 150 200 250 300 350 400 450 500 550
• Mean is 300 and standard deviation is 50
Normal Distributions
• Relationship between p.d.f. and c.d.f.
• Pa  X  b  FX b  FX a
• Pa  X  b    a f X x  dx
b
• So,
FX b   FX a   
b
a
f X x  dx
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