Transcript Slide 1

The following steps are recommended for computing
Normal probabilities for any normal distribution.
1. Identify the random variable of interest
2. Identify the parameters of the random variable (its mean and
standard deviation).
3. What is the appropriate probability statement
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4. Convert to a standard normal distribution by
x
z

This must be done to all parts of the probability statement.
5. Find the probability in the standard normal table.
Range
x
Proportion
Proportion
within range outside range
µ±σ
0.68
0.32
μ±2σ
0.9544
0.0456
μ± 3σ
0.997
0.003
μ ± 1.96 σ
95%
5%
μ ± 2.58 σ
99%
1%
μ ± 3.29 σ
99.9%
0.1%
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Example: The maximum daily temperature of England in
February following Gaussian distribution has a mean of 10ºC
and standard deviation of 2ºC.
(i) Find the probability of a random day with a maximum
temperature above 12ºC .
(ii) 95% of the days have a maximum temperature lying in a
range being symmetrical around 10ºC. Find out the range.
(i) P( x>12)
p( x  12 )  p(
x
x  10
z


2
x  10 12  10

)
2
2
 p( z  1 )
p( z  1 )

 symmetrical
2
0.32

 0.16
2
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(ii)
x
x  10
z


2
p( z  a )  0.95
a  1.96
x 
x  10

 1.96

2
x  10
 1.96 
 1.96
2
 3.92  x  10  3.92
6.08  x  13.92
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A random telegraph signal has two levels of s(t) = ±1V and is
perturbed by random noise of mean squares voltage 0.1V .
The perturbed signal is simply s(t)+n(t).
n(t)
+
s(t)
s(t)+n(t)
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Note: +1V is used to represent binary “1”
-1V is used to represent binary “0”
An error 1->0 is defined as
if s(t) = +1V but s (t)+ n(t) < 0
An error 0 ->1 is defined as
if s(t) = -1V but s (t)+ n(t) > 0
This is to assume that an instantaneous decision is made at the
centre of each received pulse with decision level zero.
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Example: A random telegraph signal has two levels of ±1V
and is perturbed by random noise of mean squares voltage
0.1V. Find the error probabilities for error 1->0 and 0->1
respectively, assuming that an instantaneous decision is made
at the centre of each received pulse with decision level zero.
1. p(1->0)
s(t) = +1V but s (t)+ n(t) < 0
So want to find p( n(t) < -1)
n(t) is zero mean and mean square of 0.1
  0,  
0.1  0.3162
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n( t )  0
z
0.3162
n( t )  0
1
p( n( t )  1 )  p(

)
0.3162
0.3162
1
 p( z 
)
0.3162
 p( z  3.163 )
 p( z  3.163 )  symmetrical
 1  p( z  3.163 )
 1   3.163  1  0.9992  0.0008
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2. p(0->1)
s(t) = -1V but s (t)+ n(t) > 0
So want to find p( n(t) >1)
  0,  
n( t )  0
z
0.3162
0.1  0.3162
n( t )
1
p( n( t )  1 )  p(

)
0.3162 0.3162
1
 p( z 
)
0.3162
 p( z  3.163 )  0.0008
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Sum of Gaussians:
The distribution of a sum of two Gaussian distributed variables
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X, with mean  x and variance of  x and Y, with mean  y
and variance of  2y is another Gaussian distributed variable
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with mean  x   y and variance of  2
x  y
X  Y ~ N (  x   y , 2x   2y )
Difference of Gaussians:
The distribution of a difference of two Gaussian distributed
variables X, with mean  x and variance of  2x and Y, with
mean  y and variance of  2y is another Gaussian distributed
2


variable with mean  x   y and variance of  2
x
y
X Y ~
2
N (  x   y , x
2
 y
)
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Example: suppose that earning of men are M ~ N (68, 4) per
day, and that of women are W ~ N (65, 1) per day. Select a
man and a woman at random. Find the probability of the
woman earning more than the man.
Note that
W  M ~ N( 65  68, 1  4 )  N( 3, 5 )
then
p( W  M )  p( W  M  0 )
0  ( 3 )
 P( z 
)
5
 1  ( 3 / 5 )
 1  0.91  0.09
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