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Question #347823
If F(x) 1/39(3x-2)2;0<_x_<3
O: elsewhere
1. Verify that F(x) is a PDF
2. find E(x) and Var(x)
Expert's answer
1.
∫
−
∞
∞
F
(
x
)
d
x
=
∫
0
3
(
3
x
−
2
)
2
39
d
x
\displaystyle\int_{-\infin}^{\infin}F(x)dx=\displaystyle\int_{0}^{3}\dfrac{(3x-2)^2}{39}dx
∫
−
∞
∞
F
(
x
)
d
x
=
∫
0
3
39
(
3
x
−
2
)
2
d
x
=
[
(
3
x
−
2
)
3
3
(
3
)
(
39
)
]
3
0
=
343
+
8
351
=
1
,
T
r
u
e
=[\dfrac{(3x-2)^3}{3(3)(39)}]\begin{matrix} 3\\ 0 \end{matrix}=\dfrac{343+8}{351}=1, True
=
[
3
(
3
)
(
39
)
(
3
x
−
2
)
3
]
3
0
=
351
343
+
8
=
1
,
T
r
u
e
2.
E
(
X
)
=
∫
−
∞
∞
F
(
x
)
x
d
x
=
∫
0
3
x
(
3
x
−
2
)
2
39
d
x
E(X)=\displaystyle\int_{-\infin}^{\infin}F(x)xdx=\displaystyle\int_{0}^{3}\dfrac{x(3x-2)^2}{39}dx
E
(
X
)
=
∫
−
∞
∞
F
(
x
)
x
d
x
=
∫
0
3
39
x
(
3
x
−
2
)
2
d
x
=
∫
0
3
9
x
3
−
12
x
2
+
4
x
39
d
x
=\displaystyle\int_{0}^{3}\dfrac{9x^3-12x^2+4x}{39}dx
=
∫
0
3
39
9
x
3
−
12
x
2
+
4
x
d
x
=
[
1
39
(
9
x
4
4
−
4
x
3
+
2
x
2
)
]
3
0
=[\dfrac{1}{39}(\dfrac{9x^4}{4}-4x^3+2x^2)]\begin{matrix} 3\\ 0 \end{matrix}
=
[
39
1
(
4
9
x
4
−
4
x
3
+
2
x
2
)]
3
0
=
1
39
(
729
4
−
108
+
18
−
0
)
=\dfrac{1}{39}(\dfrac{729}{4}-108+18-0)
=
39
1
(
4
729
−
108
+
18
−
0
)
=
123
52
=\dfrac{123}{52}
=
52
123
E
(
X
2
)
=
∫
−
∞
∞
F
(
x
)
x
2
d
x
=
∫
0
3
x
2
(
3
x
−
2
)
2
39
d
x
E(X^2)=\displaystyle\int_{-\infin}^{\infin}F(x)x^2dx=\displaystyle\int_{0}^{3}\dfrac{x^2(3x-2)^2}{39}dx
E
(
X
2
)
=
∫
−
∞
∞
F
(
x
)
x
2
d
x
=
∫
0
3
39
x
2
(
3
x
−
2
)
2
d
x
=
∫
0
3
9
x
4
−
12
x
3
+
4
x
2
39
d
x
=\displaystyle\int_{0}^{3}\dfrac{9x^4-12x^3+4x^2}{39}dx
=
∫
0
3
39
9
x
4
−
12
x
3
+
4
x
2
d
x
=
[
1
39
(
9
x
5
5
−
3
x
4
+
4
x
3
3
)
]
3
0
=[\dfrac{1}{39}(\dfrac{9x^5}{5}-3x^4+\dfrac{4x^3}{3})]\begin{matrix} 3\\ 0 \end{matrix}
=
[
39
1
(
5
9
x
5
−
3
x
4
+
3
4
x
3
)]
3
0
=
1
39
(
2187
5
−
243
+
36
−
0
)
=\dfrac{1}{39}(\dfrac{2187}{5}-243+36-0)
=
39
1
(
5
2187
−
243
+
36
−
0
)
=
384
65
=\dfrac{384}{65}
=
65
384
V
a
r
(
X
)
=
E
(
X
2
)
−
(
E
(
X
)
)
2
Var(X)=E(X^2)-(E(X))^2
Va
r
(
X
)
=
E
(
X
2
)
−
(
E
(
X
)
)
2
=
384
65
−
(
123
52
)
2
=
4227
13520
=\dfrac{384}{65}-(\dfrac{123}{52})^2=\dfrac{4227}{13520}
=
65
384
−
(
52
123
)
2
=
13520
4227
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