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Central limit theorem

2. The probability that less than five items fail is

( ( ) ( ))
5 − 5 0 − 5
P(m < 5) = P(0 ≤ m < 5) ≈ Φ √ − Φ √ =
4.75 4.75
= (Φ(0) − Φ(2.29)) = 0.489.

3. The probability that five to ten items fail is
( ( ) ( ))
10 − 5 5 − 5
P(5 ≤ m ≤ 10) ≈ Φ √ − Φ √ = (Φ(2.29) − Φ(0)) = 0.489.
4.75 4.75

Example 8.2. How many independent trials should be performed so that at least
five occurrences of an event A will be observed with probability 0.8, if the
probability of A in one trial is P(A) = 0.05. ,

Solution. From the Moivre-Laplace theorem, we see that

( ( ) ( )) ( )
n − 0.05 5 − 0, 05n √ 5 − 0, 05n
P(m ≥ 5) ≈ Φ √ − Φ √ = Φ(4.36 n) − Φ √ = 0.489.
0.0475n 0.0475n 0.0475n
√
For n = 1 we have Φ(4.36 n) = 1; therefore, substituting P(m ≥ 5) = 0.8 we obtain
( )
1 5 − 0.05n
− Φ √ ≈ 0.8,
2 0.0475n

or ( )
5 − 0.05n
Φ √ = −0.3.
0.0475n
From the table we find the argument x = −0.8416 corresponding to the value of the function
Φ(x) = −0.3.
5−0.05n
Solving the equation √ = −0.8416, we find the unique root n = 144. Thus, in order
0.0475n
that A occur at least five times with probability 0.8, 144 trials are necessary.

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