Les enseignants, les moyennes, les écarts types

∼
ECardE=|E|
f:E→Re f(e)
M σ
M σ
g(x) = 1
σ√2πe−(x−M)2
2σ2,
M σ 1
σ√2π
1 = 100%
ZR
g(x)dx = 1 = 100%.
M'10.86
σ'3.32
h: [0,20] →N
h(n) = n=
P20
n=0 h(n) = 1 = 100%
g

h h
h(n) ]n−1
2, n+1 2[n∈[0,20]N
Z20+ 1
2
0−1
2
h(x)dx =
20
X
n=0 Zn+1
2
n−1
2
h(x)dx =
20
X
n=0
h(n) = 1 = 100%.
x x ∈[0,20]N
f(e)∈[0,1
2, ...] =
1
2[0,40]Nh
Z20+ 1
4
0−1
4
h(x)dx =
40
X
k=0 Zk
2+1 4
k
2−1 4
h(x)dx =
40
X
k=0
1
2h(k 2) = 1.
N
N N
MFσF
MFσF
M σ
1
σ√2πe−(x−M)2
2σ21
σF√2πe−(x−MF)2
2σ2
F.
M=Pe∈Ef(e)
|E|.
M=f(1)+f(2)+f(3)
3=15+10+12 3'
12.33
f X
e∈E x f(e) = X(e) = x
1
|E|=p e
E(X) = X
e∈E
X(e)1
|E|=X
x
x p =X
V=Pe∈E(f(e)−M)2
|E|.
V(X) = Px(x−X)2p
σ V
σ=√V .
σ= 0
M σ
= 0 = 20 M= 10
σ=q1
100 P100
e=1 102=√102= 10
σ= 0
M g σ →0δM
[M−σ, M +σ]'
68.26%
RM+σ
x=M−σ
1
σ√2πe−(x−M)2
2σ2dx =1
2πR1
z=−1e−z2
2dz '68.26%.