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Bernoulli trials and limit theorems

Using this theorems for large n we can write an approximate formula of probability P{k 1 ≤ µ n ≤
k 2 }. We denote
√ √
k 1 = a npq + np, k 2 = b npq + np.
Then
{ }
µ n − np √ √
P a ≤ √ ≤ b = P{a npq + np ≤ µ n ≤ b npq + np} = P{k 1 ≤ µ n ≤ k 2 } ≈
npq
b
∫ ∫ x 2
1 x 2 1 x 2
≈ e − 2 dx = √ e − 2 dx,
2π 2π
a x 1
where
k 1 − np k 2 − np
x 1 = √ , x 2 = √ .
npq npq
For simplicity we introduce an Laplace function
x
∫
1 t 2
Φ(x) = √ e − 2 dt
2π
0
But
x 2 x 2 x 1
∫ ∫ ∫
x 2 x 2 2
e − 2 dx = e − 2 dx − e −x 2dx,
x 1 0 0
then we obtain
( ) ( )
k 2 − np k 1 − np
P{k 1 ≤ µ n ≤ k 2 } ≈ Φ(x 2 ) − Φ(x 1 ) = Φ √ − Φ √ (4.9)
npq npq
1. Φ(x) is odd function: Φ(−x) = −Φ(x).
−x 2
1 ∫ − t
PROOF. In integral Φ(−x) = √ e 2 dt we substitute t = −u. Then dt = −du, Φ(−x) =
2π
0
x
∫ u 2
1 −
− √ e 2 du = −Φ(x). 2
2π
0
2. Φ(0) = 0; Φ(+∞) = 0, 5 i Φ(−∞) = −0, 5 ⇒ lines y = 0, 5 and y = −0, 5 are asymptotes
of function for x → +∞ and x → −∞ respectively.
∞ 2 √ √
∫ t
1 −
PROOF. InintegralΦ(+∞) = √ e 2 dtwesubstitutet = u 2.Thenwehavedt = 2du,
2π
0
√ ∫ √ √
2 2 2 π 1
∞ −u
Φ(+∞) = √ +0 e du = √ · = .
2π 2π 2 2
3. Φ(x) is an increasing function.
Φ(x) is tabulated for 0 ≤ x ≤ 5. For negative values −5 ≤ x ≤ 0 we can use that Φ(x) is an odd
function.
Example 4.6. If a single six-sided die is rolled 600 times, what is the
probability that a ’six’ is thrown exactly 100 times? Find probability that
’six’ is thrown from 90 to 110 times? ,
5
Solution. We have n = 600; k = 100; p = 1/6; np = 100; npq = 100 · 6 = 83, (3);
√ k−np f(0) 0,3989
npq = 9, 1287; x 0 = √ = 0; P 600 (100) ≈ √ = = 0, 0437; P{90 ≤ µ 600 ≤ 110} ≈
npq npq 9,1287
( ) ( )
Φ 110−np − Φ 90−np = Φ(1, 095) − Φ(−1, 095) = 2Φ(1, 095) = 2 · 0, 36324 = 0, 72648.
√
√
npq npq
We remark that (4.8) and (4.9) provide good approximation when npq ≥ 10. In other cases
approximate Poisson formulas are better.

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