# CONVERGENCE OF BINOMIAL AND NORMAL DISTRIBUTIONS FOR LARGE NUMBERS OF TRIALS

```10/20/04
MWG
22.38 PROBABILITY AND ITS APPLICATIONS TO
RELIABILITY, QUALITY CONTROL AND RISK ASSESSMENT
Fall 2004
CONVERGENCE OF BINOMIAL AND NORMAL
DISTRIBUTIONS FOR LARGE NUMBERS OF TRIALS
We wish to show that the binomial distribution for m successes observed out of n trials can be
approximated by the normal distribution when n and m are mapped into the form of the standard
normal variable, h.
(1)
P(m,n ) ≅ Prob. (h) , where
↑
↑
Binomial
Normal
Distribution Distribution
Binomial Distribution:
⎛ n⎞
P ( m,n ) = ⎜ m⎟ p m q (n −m ) ,
⎝ ⎠
(2)
p + q = 1 , where
(3)
p = probability of success in a single trial, and
q = probability of failure in a single trial.
− h2 2
e ( )
,
Prob. ( h) =
2π σ
Normal Distribution:
(4)
⎛ m − &micro;⎞
h ≡⎜
⎟ ,
⎝ σ ⎠
&micro; = np ,
(5)
(Binomial Distribution Mean)
σ = np q . (Binomial Distribution Standard
Deviation)
Recall Sterling Approximation:
m! ≅ 2 π m m m e − m
⎞
n
1 ⎛⎜
⎛ n⎞
n!
⎟
≅
⇒ ⎜⎝ m ⎟⎠ ≡
⎜
2 π ⎝ m(n − m )⎟⎠
m!(n − m )!
P(m,n ) ≅
⎞
1 ⎛⎜
1
⎟
2 π ⎜⎝ m(n − m)⎟⎠
1
1 of 2
1
m
m
n⎞ ⎛ n ⎞
⎟ .
⎜ ⎟ ⎜
⎝ m⎠ ⎝ n − m⎠
2⎛
m
n− m)
n p⎞ ⎛ n q ⎞ (
⎜ ⎟ ⎜
⎟
⎝ m ⎠ ⎝ n − m⎠
2⎛
(6)
(7)
≅
⎛ np ⎞ m ⎛ nq ⎞ ( n − m )
.
⎟
⎜ ⎟ ⎜
2 π (n pq ) ⎝ m ⎠ ⎝ n − m ⎠
1
The result above uses the relationships:
m = np+ h npq
(n − m ) = nq − h n p q
to obtain result
⎛
⎛ m (n − m )⎞
pq ⎞
pq⎞⎛
⎟⎟
⎟⎟ ⎜⎜ q − h
⎜
⎟ = n ⎜⎜ p + h
⎝
⎠
n ⎠
n ⎠⎝
n
⎝
≅ n pq .
x2
Then, use expansion of ln(1+ x ) ≅ x − , about x=0 to evaluate Eq. 1, using 2 π n pq Prob. (h )
2
as:
−ln ( 2 π n pq Prob. (h)) ≅ −ln ( 2 π n pq P(m, n))
⎡ ⎛ n p ⎞ m ⎛ n q ⎞ ( n − m) ⎤
⎥
= ln ⎢ ⎜ ⎟ ⎜
⎟
⎢⎣ ⎝ m ⎠ ⎝ n − m ⎠
⎥⎦
⎛
⎛
⎞
q ⎞⎟
⎜ 1− h p ⎟
= (n p + h n pq )ln ⎜⎜ 1+ h
ln
+
n
q
−
h
n
pq
(
)
⎜
n p ⎟⎠
nq ⎟⎠
⎝
⎝
⎛
⎛
q
h2 q ⎟⎞
p
h2 p ⎟⎞
⎜
≅ (n p + h n pq )⎜⎜ h
−
−h
+
nq
−
h
n
p
q
−
(
) ⎜⎝ n q 2n q ⎟⎠
n p 2n p ⎟⎠
⎝
⎛
⎞ ⎛
⎞
h2
h2
+ q h2 ⎟⎟ + ⎜⎜ −h n p q − p
= ⎜⎜ h n pq − q
+ ph 2 ⎟⎟
2
2
⎝
⎠ ⎝
⎠
h2 h2
= (p + q)
=
.
123 2
2
1
Thus, the result is obtained; verifying Eq. 4:
e ( )
e ( )
Prob. (h) =
=
. QED!
2π n p q σ 2π
− h2 2
2 of 2
− h2 2
```