Central Limit Theorem
The theorem states that the sum of a large number of
independent observations from the same distribution has,
under certain general conditions, an approximate normal
distribution.
Moreover, the approximation steadily improves as the
number of observations increases. The theorem is
considered to be the heart of probability theory .
The central limit theorem is one of the most remarkable
results of the theory of probability.
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(iii) Rayleigh distribution
The envelop (instantaneous amplitude) of a narrowband
noise follows a Rayleigh distribution, given by
x
x
p( x )  2 exp(
)
2

2
2
The distribution is similar to Gaussian but is not symmetrical.
The envelop cannot be less than zero but has no upper limit.
In amplitude modulation, the envelop carries information, but
noise perturbs the envelop.
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3
Revision:
Binomial distribution:
There are two outcomes each trial, with probability for
two outcomes given by p, (1-p)
The probability of r successes in n trials is given by
n!
r ( nr )
p( r ) 
p q
( n  r )! r!
Mean
variance
r  np
  nqp
2
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Example: A multiple choice examination have 100 questions,
each having one correct answer, three incorrect answers.
(i) Find the mean and the standard deviation for the
distribution of the correct answers for one who answers the
questions by random guess.
r  np  100 0.25  25
  100 0.25  0.75  18.75
2
  18.75  4.33
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(ii) What is the probability of getting 50% correct answers by
guessing answers for all of the questions?
n=100, p=1/4, r=50. …
n!
p( r ) 
p r q( nr )
( n  r )! r!
100!

0.2550 ( 1  0.25 )50
100  50!50!
 4.51 108
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If for each question, the exam taker knows that two answers
are wrong, but have to guess one from the two others. What
is the probability of getting 70% correct answers?
n=100, p=1/2, r=70. ….
n!
p( r ) 
p r q( nr )
( n  r )! r!
100!

0.570 ( 1  0.5 )30
100  70!70!
 2.317  10 5
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(iii) If one knows the answers to 90 questions, but have to
purely guess answers for the remaining 10 questions.
What is the probability of getting more than 95 correct
answers?
Solution: He (she) needs to guess correctly more than 5
answers out of 10 questions. P(r>5)
So use n =100-90=10, p=1/4.
and then sum up the probabilities for p(r) for r= 6, r= 7, r= 8,
r= 9, r= 10.
10!
p( 6 ) 
0.256 ( 1  0.25 )4
10  6!6!
 0.0162
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10!
p( 7 ) 
0.257 ( 1  0.25 )3
10  7 !7!
 0.0031
10!
p( 8 ) 
0.258 ( 1  0.25 )2
10  8!8!
 0.000386
10!
p( 9 ) 
0.259 ( 1  0.25 )1
10  9!9!
 0.0000286
10!
p( 10 ) 
0.2510 ( 1  0.25 )0
10  10!10!
 9.53  10 7
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p( r  5 )  p( r  6 )  p( r  7 )
 p( r  8 )  p( r  9 )  p( r  10 )
 0.0162  0.0031 0.000386
 0.0000286 9.53  10  7
 0.0197
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Example: A telecommunication system sends a binary
sequence using two levels of ±1V. A sequence of
“A=101010” was sent through a channel which is disturbed
by random noise of mean squares voltage 0.1V.
(i) Find the error rate, assuming that an instantaneous decision
is made at the centre of each received pulse with decision
level zero.
(ii) What is the probability of correct transmission of the
sequence A?
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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,  
n( t )  0
z
0.3162
0.1  0.3162
n( t )  0
1
p( n( t )  1 )  p(

)
0.3162 0.3162
1
 p( z 
)
0.3162
 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)
n(t) is zero mean and mean square of 0.1
  0,  
n( t )  0
z
0.3162
0.1  0.3162
n( t )  0
1
p( n( t )  1 )  p(

)
0.3162 0.3162
1
 p( z 
)
0.3162
 0.0008
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(ii) A has 6 bits, each bit has error probability of p=
0.008. For all 6 bits to be correctly transmitted, using
binomial distribution to calculate p(0),
6!
6
0
p( 0 ) 
0.992 ( 0.008 )
( 6  0 )!0!
 0.9528
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