Échantillonnage et estimation, cours, terminale S - MathsFG

•p
n
p
n
•n
f p
n
XnB(n;p)α
0< α < 1X
N(0; 1) uα
P(−uα≤X≤uα) = 1 −α
lim
n→+∞
P(Xn
n∈[p−uα
pp(1 −p)
√n;p+uα
pp(1 −p)
√n]) = 1 −α
In= [p−uα
√p(1−p)
√n;p+uα
√p(1−p)
√n]
Xn
n1−α n
1−α
[p−1,96pp(1 −p)
√n;p+ 1,96pp(1 −p) √n]
Xn∼ B(n;p)E(Xn) = np σ(Xn) = np(1 −p)Zn=Xn−np
√np(1−p)
lim
n→+∞
P(−uα≤Zn≤uα) = P(−uα≤X≤uα) = 1 −α

lim
n→+∞
P(−uα≤Xn−np
pnp(1 −p)≤uα) = 1 −α
lim
n→+∞
P(−uαpnp(1 −p)≤Xn−np ≤uαpnp(1 −p)) = 1 −α
lim
n→+∞
P(np −uαpnp(1 −p)≤Xn≤np +uαpnp(1 −p)) = 1 −α
lim
n→+∞
P(p−uα
pp(1 −p)
√n≤Xn
n≤p+uα
pp(1 −p)
√n) = 1 −α
u0,05 = 1,56
n≥30 np ≥5n(1 −p)≥5
XnB(n;p)Fn=Xn
n
p
•[Fn−1
√n;Fn+1 √n]n
p
•f
n[f−1
√n;f+1 √n]
p
lim
n→+∞
P(p−uα
pp(1 −p)
√n≤Fn≤p+uα
pp(1 −p)
√n) = 1 −α
f(p) = 1,96pp(1 −p)p∈[0; 1] f]0; 1[ f0(p) = 1,96 −2p+1
2√p(1−p
−2p+ 1
p1
2
f0(p)
f(p)% &

p∈[0; 1] 0 ≤f(p)<1p−uα
√p(1−p)
√n≤Fn≤p+uα
√p(1−p)
√n
p−1
√n< Fn< p +1 √n
p>Fn−1
√np>Fn+1 √n
Fn−1
√n< p < Fn+1 √n
2
√n
n