Download 8.4 Approximations

Appendix: Poisson and Normal Approximations
(I)
Poisson Approximation:
Example:
Let X be the binomial random variable over 250 trials with p  0.01 . Then, it might
not be easy to obtain
P( X  3) 
250!
0.013 0.99247
3!247!
directly. However, if we only want to obtain an approximation, Poisson
approximation is a good choice.
Poisson approximation:
Let X be a binomial random variable over n trials and let
p  0.05, n  20.
Let Y be a Poisson random variable with mean np.
Then, the probability of X taking value i can be approximated by the
probability of Y taking value i. That is,
i
 n i
e  np np 
n i
  p 1  p  
, i  1, 2, , n.
i!
i
Example:
In the above example, the Poisson random variable with mean np  250  0.01  2.5
can be used for approximation. Thus,
250!
e 2.5 2.5
3
247
0.01 0.99 
P( X  3) 
 0.2138
3!247!
3!
3
Note that the exact probability is
1
 250 
0.013 0.99247  0.2149 .
P( X  3)  
 3 
Therefore, the normal approximation is reasonably accurate.
(II)
Normal Approximation:
Normal approximation:
Let X be a binomial random variable over n trials and the probability
of success be p. Let Y be the normal random variable with mean np
and variance np(1-p). Then, the probability of X taking value i can be

1
1
approximated by the probability of Y taking values in i  , i   .
2
 2
That is,
 n i
1
1
  p 1  p ni  P(i   Y  i  )
2
2
i
1
i
2
1
i
2

Note: the probability
P (i 
1
e
2 np 1  p 
1
1
Y i )
2
2
 x np 2
2 np (1 p )
dx .
can be obtained by
transforming Y to the standard normal random variable Z.
Example:
Let X be the binomial random variable over 100 trials and let the probability of a
success be 0.1. What is the probability of 12 successes by normal approximation?
[solutions:]
The normal random variable with mean
2
np  100  0.1  10
and variance
np(1  p)  100  0.1  (1  0.1)  9 can be used for approximation. Thus,
1
1
11.5  10 Y  10 12.5  10
P( X  12)  P(12   Y  12  )  P(


)
2
2
3
3
3
 P(0.5  Z  0.83)  P(0  Z  0.83)  P(0  Z  0.5)
 0.2967 - 0.1915  0.1052
Note that the exact probability is
100 
0.112 0.988  0.0988 .
P( X  12)  
 12 
Therefore, the normal approximation is reasonably accurate.
Note:
Let X be a binomial random variable over n trials and the probability
of success be p and let Y be the normal random variable with mean
np and variance np(1-p). Then,
1
Y  np
P( X  i)  P(Y  i  )  P(

2
np(1  p)
1
1
 np
i   np
2
2
)  P( Z 
),
np(1  p)
np(1  p)
i
where Z is the standard normal random variable. Similarly,
1
P( X  k )  P(Y  k  )  P( Z 
2
1
 np
2
)
np(1  p)
k
Example:
In the previous example, what is the probability of at most 13 successes by normal
approximation?
[solution:]
1
13   10
2
P( X  13)  P( Z 
)  P( Z  1.17)  P( Z  0)  P(0  Z  1.17)
3
 0.5  0.3790  0.8790
Note that the exact probability is
3
100 
0.1i 0.9100i  0.8761 .
P( X  13)   
i 0  i 
13
Therefore, the normal approximation is reasonably accurate.
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