Cours de Statistique Licence 2-S4 SI-MASS

p
n p
p p
p(x)F(x)
X
F(x)X x
F(x) = P(X≤x).
a b b > a
P(a < X < b) = F(b)−F(a).

E(X)X
E(X) = X
i
xiP(xi).
µX
σ2
X
σ2
X=E[(X−µX)2] = X
i
(xi−µX)2P(xi) = X
i
x2
iP(xi)−µ2 X.
σX
X
µ σ σ2
X p(x)
x
p(x) = 1
σ√2πe−1
2(x−µ σ)2.
N(µ, σ)µ
σ X N(µ, σ)
E(X) = µ V (X) = σ2.
µN(0,1)
YN(µ, σ)Z=Y−µ
σN(0,1)
Φ
Φ(x) = P(Z < x)
ZN(0,1)
Φ(−x)=1−Φ(x)
Φ(0) = 0.5,Φ(1.645) ≈0.95,Φ(1.960) ≈0.9750
|x|<2 Φ
Φ(x)≈0.5 + 1
√2πx−x3
6+x5 40 .

zα/2
P(Z > zα/2) = α/2
α0.01 0.02 0.05 0.10
zα/22.58 2.33 1.96 1.645
c99% 98% 95% 90%
zα/2
P(Z < zα/2) =
P(Z < −zα/2) =
P(−zα/2< Z < zα/2) =
P(|Z|> zα/2) =
X Y N(µ1, σ1)
N(µ2, σ2)X+YN(µ1+µ2,pσ2
1+σ2 2)
χ2
k X1, . . . , Xk
X
X=
k
X
i=1
X2
i
χ2k χ2(k)X
α7→ χ2
α;ν(α;ν)),
χ2
α;νP(χ2(ν)> χ2 α;ν) = α
α= 0.990 ν= 5 χ2
α= 0.554 = χ2
0.99;5
χ2
N(0,1) Zzzα(1 −α)
χ2(ν)K2k;ν; 1 χ2
α;να;ν; 1
St(ν)Tt;ν; 1 tα;να;ν
F(ν1, ν2)Ff;ν1;ν2)fα;ν1,ν2α;ν1;ν2)

X µ σ2
α
P(|X−µ| ≥ α)≤σ2
α2.
p
p
n f F
n F
N(p;σ)σ=ppq/n n (n > 30)
σ0=rf(1 −f)
n
f n
σ p
σ=σ0rn
n−1=rf(1 −f) n·rn n−1=rf(1 −f) n−1.
Z
Z=F−p
σ
N(0; 1)
p p
α α ∈[0; 1] α
c= 1 −α p
α p α
zα/2
P(Z > zα/2) = α/2
ZN(0; 1) P(Z < −zα/2) = α/2
P(−zα/2< Z < zα/2)=1−α
2−α 2= 1 −α
P(−zα/2< Z < zα/2)=1−α⇔P−zα/2<F−p
σ< zα/2= 1 −α
⇔P(−zα/2·σ < F −p<zα/2·σ)=1−α
⇔P(F−zα/2·σ < p < F +zα/2·σ)=1−α
⇔P F−zα/2rf(1 −f)
n−1< p < F +zα/2rf(1 −f) n−1!= 1 −α

p1−α
#f−zα/2rf(1 −f)
n−1;f+zα/2rf(1 −f) n−1".
n n n −1
#f−zα/2rf(1 −f)
n;f+zα/2rf(1 −f) n".
α= 5% f≈0.5
f−1
√n;f+1 √n.
XN(µ, σ)X1, ..., Xnn
X
X=1
n
n
X
i=1
XiS0
n
2=1
n−1
n
X
i=1
(Xi−X)2.
zα/2P(−zα/2< Z < zα/2) = 1 −α
XN(µ;σ/√n)
1−α=P(−zα/2< Z < zα/2)
=P(−zα/2<X−µ
σ/√n< zα/2)
=PX−zα/2·σ/√n < µ < X +zα/2·σ/√n
σ2
x−zα/2
σ
√n< µ < x +zα/2
σ
√n
I=x−zα/2
σ
√n;x+zα/2
σ
√n.
n
µX−µ
S0
n/√n→
St(n−1) n−1
x−tα/2
s
√n< µ < x +tα/2
s
√n
tα/2=tα/2;(n−1) α/2
ν=n−1