(d1, d2)
P[Y ∈ [a, a + ]] = lim →0
a+
a
a < b
=
d
db
b
a
0 . 0
0 . 4
0 . 8
0 . 0
0 . 2
0 . 4
0 . 0
d
= d
db
d t
(I z) =
x
p r o b
a b
i l i t y
0 1 2 3
x
p r o b
a b
i l i t y
0 1 2 3
x
p r o b
a b
i l i t y
0 1 2 3
x
p r o b
a b
i l i t y
0 10 30
x
p r o b
a b
i l i t y
0 10 30
x
p r o b
a b
i l i t y
0 10 30
x
p r o b
a b
i l i t y
0 200
0 . 0
x
p r o b
a b
i l i t y
0 200
0 . 0
x
p r o b
a b
i l i t y
0 200
k! .
3! ≈ 0.06
3! ≈ 0.18
3! ≈ 0.22
3! ≈ 0.14
0 . 0
0
2
4
6
8
0 . 0
0 . 2
0 . 4
0 . 6
0 . 8
1 . 0
1 . 2
mu = −0.5 ; sigma = 3
4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5
0 . 0
0 . 2
0 . 4
0 . 6
4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5
0 . 0
0 . 2
0 . 4
0 . 6
0 . 0
−2 −1 0 1 2
0 . 0
0 . 4
0 . 8
2π e−
1 2
2 0
3 0
4 0
5 0
6 0
X
= np n−1
= np
0
(yi−y)2
= E(Y 2) − (EY )2
= 30
= 15 29
v=0
0 . 0
0 . 0
0 . 1
0 . 2
0 . 3
0 . 4
h(y) p(y)
E[X ] =
= a
= E[b2(Y − µ)2]
X
Y
f X,Y (x, y) = f X (x) ·
f Y | X (y | x) = f Y (y) ·
f X |Y (x | y)
f X (x) =
x
0
2
4
6
N
X
(N = 3, X = 0)
(N = 3, X = 1)
(N = 3, X = 2)
(N = 3, X = 3)
6 (1− θ)3 f N,X (3, 1) =
e−λλ3
6 3θ2(1− θ) f N,X (3, 3) =
e−λλ3
6 θ3
f N,X (n, x)
= f N (n)f X |N (x | n) =
e−λλn
n!
n
x
(n− x)!
e−λθ(λθ)x
∞
∞
0
1
2
3
4
(a)
x
y
0 . 0
0 . 4
0 . 8
0 . 0
0 . 2
0 . 0
0 . 4
0 . 8
0
2
4
6
(f)
x
E(X ) = E (E(X | Y ))
= E(X | Y = 2)P[Y = 2] +
E(X | Y = 3) P[Y = 3]
+ E(X | Y = 4) P[Y = 4] + E(X |
Y = 5)P[Y = 5]
+ E(X | Y = 6) P[Y = 6] + E(X |
Y = 7)P[Y = 7]
+ E(X | Y = 8) P[Y = 8] + E(X |
Y = 9)P[Y = 9]
+ E(X | Y = 10) P[Y = 10] + E(X |
Y = 11) P[Y = 11]
+ E(X | Y = 12) P[Y = 12]
= 0 × 1
(10/36)w = 4/36
w = 4/10.
4 . 5
5 . 5
6 . 5
7 . 5
2 . 0
2 . 5
3 . 0
3 . 5
4 . 0
Cov(aX + b,cY + d) = E
((aX + b− (aµX + b))(cY + d−
(cµY + d)))
= E(ac(X − µX )(Y − µY )) =
ac Cov(X, Y )
d
e−kλλ
1960 1980
− 3
− 2
− 1
0
1
2
3
(c)
Index
(d)
Time
1 . 0
1 . 5
2 . 0
=
n
Var(X 1 + X 2) = E((X 1 +
X 2)2)− (µ1 + µ2)2
= E(X 21 ) + 2E(X 1X 2) + E(X 22 )− µ2 1
− 2µ1µ2 − µ2
2
= σ2 1 + σ2
= µ2
2
1− P[|yn − µ| ≥ ] ≥ lim n→∞
1− σ2/n2 = 1.
b
a
0.3 0.4 0.5 0.6 0.7
0
1
2
3
4
5
6
0.3 0.4 0.5 0.6 0.7
0
2
4
6
8
0.3 0.4 0.5 0.6 0.7
0
5
Y 2 = 2X 1
)
n−1)
a
y
0 . 0
0 . 4
0 . 8
0 . 0
0 . 0
0 . 4
0 . 8
0 . 0
1 . 0
2 . 0
3 . 0
VC, 1
0
1
2
3
4
0 . 0
0 . 5
1 . 0
1 . 5
2 . 0
OJ, 0.5
0
1
2
3
4
5
0 . 0
0 . 5
1 . 0
1 . 5
2 . 0
OJ, 2
0 . 0
1 . 0
2 . 0
3 . 0
100 140 180 0 . 0
0 0
0 . 0
1 5
0 . 0
3 0
100 140 180 0 . 0
0 0
0 . 0
1 0
0 . 0
2 0
0 . 5
0 . 5
O J
V C
0
2
4
6
8
Student averages
Admit
Admitted Rejected
Gender
Male Female
e e c e
(a)
0 . 0
0 . 1
0 . 2
0 . 3
0 . 4
0 . 5
40 60 80 100
5 0
6 0
7 0
8 0
9 0
duration of eruption
0 50 100 150 200 250
1 . 5
2 . 5
3 . 5
4 . 5
data number
w a
i t i n
g t i m
e
0 50 100 150 200 250
5 0
7 0
9 0
0 50 100 150
2 . 0
4 . 0
data number
w a
i t i n
g t i m
e
0 50 100 150
data number
w a
i t i n
g t i m
e
0 20 40 60 80 100 120 140
5 0
8 0
5
Given : lon
5
Given : lat
1
0
2
0
3
0
4
0
5
0
6
0
0 4 8 1 3 1 8 2 4 2 9 3 4 3 9 4 4 5 0 5 5 6 0 6 5 7 0 7 6 8 1
a
d
s
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•
•
•
•
f (X i | θ).
0 . 0
0 . 2
0 . 4
0 . 6
0 . 8
1 . 0
θ
= P[
&nb
