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Question #304838
X 3 6 9 12 15
P(X) 4/9 2/9 1/9 1/9 1/9
Expert's answer
m
e
a
n
=
E
(
X
)
=
∑
i
x
i
p
(
x
i
)
mean=E(X)=\sum_ix_ip(x_i)
m
e
an
=
E
(
X
)
=
i
∑
x
i
p
(
x
i
)
=
3
(
4
9
)
+
6
(
2
9
)
+
9
(
1
9
)
+
12
(
1
9
)
+
15
(
1
9
)
=3(\dfrac{4}{9})+6(\dfrac{2}{9})+9(\dfrac{1}{9})+12(\dfrac{1}{9})+15(\dfrac{1}{9})
=
3
(
9
4
)
+
6
(
9
2
)
+
9
(
9
1
)
+
12
(
9
1
)
+
15
(
9
1
)
=
20
3
=\dfrac{20}{3}
=
3
20
E
(
X
2
)
=
3
2
(
4
9
)
+
6
2
(
2
9
)
+
9
2
(
1
9
)
+
1
2
2
(
1
9
)
E(X^2)=3^2(\dfrac{4}{9})+6^2(\dfrac{2}{9})+9^2(\dfrac{1}{9})+12^2(\dfrac{1}{9})
E
(
X
2
)
=
3
2
(
9
4
)
+
6
2
(
9
2
)
+
9
2
(
9
1
)
+
1
2
2
(
9
1
)
+
1
5
2
(
1
9
)
=
62
+15^2(\dfrac{1}{9})=62
+
1
5
2
(
9
1
)
=
62
V
a
r
(
X
)
=
σ
2
=
E
(
X
2
)
−
(
E
(
X
)
)
2
Var(X)=\sigma^2=E(X^2)-(E(X))^2
Va
r
(
X
)
=
σ
2
=
E
(
X
2
)
−
(
E
(
X
)
)
2
=
62
−
(
20
3
)
2
=
158
9
=62-(\dfrac{20}{3})^2=\dfrac{158}{9}
=
62
−
(
3
20
)
2
=
9
158
σ
=
σ
2
=
158
9
=
158
3
≈
4.19
\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{158}{9}}=\dfrac{\sqrt{158}}{3}\approx4.19
σ
=
σ
2
=
9
158
=
3
158
≈
4.19
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#340153
on Dec 2023
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