Lois normales, cours, terminale S - MathsFG

N(µ, σ2)

XN(0; 1)
fR
f(x) = 1
√2πe−x2
2
x
P(X≤x) = Zx
−∞
f(x)dx
fR+∞
−∞ f(x)dx = 1
•fR
•f
•f
•P(X≤0) = 1
2[0; +∞[
1
2
•x P (X≤ −x) = P(X≥x) = 1 −P(X≤x)
•x P (−x≤X≤x)=2P(X≤x)−1
•x7→ 1
√2πe−x2
2R R
•
•f(−x) = f(x)x∈]− ∞;∞[
•
•P(−x≤X≤x) = P(X≤x)−P(X≤ −x) = P(X≤x)−P(X≥x)
P(−x≤X≤x) = P(X≤x)−(1 −P(X≤x)) = 2P(X≤x)−1

X∼ N(0; 1) E(X)=0 σ(X) = 1
P(a≤X≤b)
•a b µ σ
•a b µ σ
X∼ N(0; 1)
•P(X≥1,5) P(a≤X≤b)
P(X≥1,5) = 1 −P(X≤1,5)
P(X≥1,5) = 1 −(P(X≤0) + P(0 ≤X≤1,5))
P(X≥1,5) = 1 −0,5−P(0 ≤X≤1,5) ≈0,0668
b P (1,5≤X≤1000) ≈0,0668
•P(X < −1,5)
P(X < −1,5) = P(X < 0)−P(−1,5≤X≤0) = 0,5−P(−1,5≤X≤0)
XN(0; 1)
α∈]0; 1[ uα
P(−uα≤X≤uα) = 1 −α
Jx P (−x≤X≤x) = 2P(0 ≤X≤x) =
2Rx
0f(u)du
x H(x)=2Rx
0f(u)du
HRH]0; +∞[H0(x) = 2f(x)
xlimx→+∞H(x) = 1 H(0) = 0
α∈]0; 1[ 1 −α∈]0; 1[
uαH(uα) = 1 −α
u0,05 ≈1,96 u0,01 ≈2,58
P(−1,96 ≤X≤1,96) ≈0,95 P(−2,58 ≤X≤2,58) ≈0,99
[−1,96,1,96]
[−2,58; 2,58]

u P (X≤u)≈α α
•α µ σ
•α µ σ
X∼ N(0; 1) u P (−u≤X≤u) = 0,99
•P(−u≤X≤u)P(X≤u)
u
P(−u≤X≤u) = P(X≤u)−P(X < −u) = P(X≤u)−(1 −P(X≥ −u))
P(−u≤X≤u) = P(X≤u)−1 + P(X≥ −u) = P(X≤u)−1 + P(X≤u) = 2P(X≤
u)−1
•u2P(X≤u)−1=0,99 P(X≤
u) = 1,99
2u≈2,58

X E(X) = m
V(X)σ(X) = pV(X)
•X−m
•Z=X−m
σ
E(X−m) = E(X)−m
E(X−m) = E(X)−m=m−m= 0 E(X−m
σ) = 1 σE(X−m)=0
V(X−m) = V(X)V(X−m
σ) = 1 σ2V(X) = σ2 σ2= 1
n p ]0; 1[ Xn
B(n;p)Zn
Zn=Xn−np
√np(1−p)
Zn
Xn
Xn∼ B(n;p)E(Xn) = np σ(Xn) = pnp(1 −p)
E(Zn) = E(Xn−np
√np(1−p)) = E(Xn)−np √np(1−p)
E(Zn) = np−np
√np(1−p)= 0
V(Zn) = 1
np(1−p)V(Xn−np)V(aX) = a2V(X)
V(Zn) = 1
np(1−p)V(Xn)V(X+b) = V(X)
V(Zn) = 1
np(1−p)np(1 −p)=1
(Xn)
B(n;p)Zn=Xn−np
√np(1−p)a b
a<b
lim
n→+∞
P(a≤Zn≤b) = Zb
a
1
√2πe−x2
2dx
n Zn
1
√2πe−t2
2
n≥30 np ≥5n(1 −p)≥5