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The Law of Large Numbers

Theorem 8.6. (de Moivre-Laplace theorem) Let p be a probability of some event A in n
repeated independent trials, m is a frequency of the event A, then the following approximate
formula is valid
( ) ( )
m 2 − np m 1 − np
P(m 1 ≤ m ≤ m 2 ) ≈ Φ √ − Φ √ . (8.8)
npq npq
⋆

∑ n
PROOF. A random variable m can be represented in the form m = X k , where X k = 1 if the event
k=1
A occurs in the k-th trial, and X k = 0 if it does not occur.
n n
∑ ∑
Since E(X k ) = p, Var(X k ) = pq, then E(m) = E( X k ) = np, Var(m) = Var( X k ) = npq.
k=1 k=1
On substituting these values in the formula (8.7) we obtain
( ) ( )
m 2 − np m 1 − np
P(m 1 ≤ m ≤ m 2 ) ≈ Φ √ − Φ √ .
npq npq
This theorem can be used for the determination of a probability of a deviation of the frequency m
from the mean np and also for the probability of the deviation of the relative frequency m from the
n
probability p. In the first case we have

( ) ( ) ( )
ε ε ε
P(|m − n| ≤ ε) = P(np − ε ≤ m ≤ np + ε) ≈ Φ √ − Φ −√ = Φ √ .
npq npq npq
Thus
( )
ε
P(|m − np| ≤ ε) ≈ 2Φ √ . (8.9)
npq
In the second case we get

( √ )
m n
( )
P − p ≤ ε = P(|m − np| ≤ nε) ≈ Φ ε . (8.10)
n pq
Example 8.1. The probability that an item will fail during reliability tests is
p = 0, 05. What is the probability that during tests with 100 items, the number
failing will be

1. at least five;

2. less than five;

3. between five and ten. ,

Solution. By the de Moivre-Laplace theorem,
( ) ( )
m 2 − np m 1 − np
P(m 1 ≤ n ≤ m 2 ) ≈ Φ √ − Φ √ .
npq npq

if n is sufficiently large. By assumption, n = 100, p = 0.05, q = 1 − p = 0.95.

1. The probability that at least five items fails is
( ) ( )
100 − 5 5 − 5
P(m ≥ 5) = P(5 ≤ m < 100) ≈ Φ √ − Φ √ =
4.75 4.75
= Φ(43.6) − Φ(0) = 0.5.

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