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Question 41 - Week 5 Practice | Prasnya
Q41
00:00
4 Aug 2024
Q41.
What is the distribution of
X
X
X
?
@passage
A
f
X
(
x
)
=
1
2
π
[
e
−
1
2
(
x
+
1
)
2
+
e
−
1
2
(
x
−
1
)
2
]
f_X(x)=\frac1{\sqrt{2\pi}}[e^{-\frac12(x+1)^2}+e^{-\frac12(x-1)^2}]
f
X
(
x
)
=
2
π
1
[
e
−
2
1
(
x
+
1
)
2
+
e
−
2
1
(
x
−
1
)
2
]
B
f
X
(
x
)
=
1
2
4
π
[
e
−
(
x
+
1
)
2
/
4
+
e
−
(
x
−
1
)
2
/
4
]
f_X(x)=\frac1{2\sqrt{4\pi}}[e^{-(x+1)^2/4}+e^{-(x-1)^2/4}]
f
X
(
x
)
=
2
4
π
1
[
e
−
(
x
+
1
)
2
/4
+
e
−
(
x
−
1
)
2
/4
]
C
f
X
(
x
)
=
1
2
2
π
[
e
−
1
2
(
x
+
1
)
2
+
e
−
1
2
(
x
−
1
)
2
]
f_X(x)=\frac1{2\sqrt{2\pi}}[e^{-\frac12(x+1)^2}+e^{-\frac12(x-1)^2}]
f
X
(
x
)
=
2
2
π
1
[
e
−
2
1
(
x
+
1
)
2
+
e
−
2
1
(
x
−
1
)
2
]
D
f
X
(
x
)
=
1
2
π
e
−
x
2
f_X(x)=\frac1{2\sqrt{\pi}}e^{-x^2}
f
X
(
x
)
=
2
π
1
e
−
x
2
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Details
Q41
00:00
4 Aug 2024
Q41.
What is the distribution of
X
X
X
?
@passage
A
f
X
(
x
)
=
1
2
π
[
e
−
1
2
(
x
+
1
)
2
+
e
−
1
2
(
x
−
1
)
2
]
f_X(x)=\frac1{\sqrt{2\pi}}[e^{-\frac12(x+1)^2}+e^{-\frac12(x-1)^2}]
f
X
(
x
)
=
2
π
1
[
e
−
2
1
(
x
+
1
)
2
+
e
−
2
1
(
x
−
1
)
2
]
B
f
X
(
x
)
=
1
2
4
π
[
e
−
(
x
+
1
)
2
/
4
+
e
−
(
x
−
1
)
2
/
4
]
f_X(x)=\frac1{2\sqrt{4\pi}}[e^{-(x+1)^2/4}+e^{-(x-1)^2/4}]
f
X
(
x
)
=
2
4
π
1
[
e
−
(
x
+
1
)
2
/4
+
e
−
(
x
−
1
)
2
/4
]
C
f
X
(
x
)
=
1
2
2
π
[
e
−
1
2
(
x
+
1
)
2
+
e
−
1
2
(
x
−
1
)
2
]
f_X(x)=\frac1{2\sqrt{2\pi}}[e^{-\frac12(x+1)^2}+e^{-\frac12(x-1)^2}]
f
X
(
x
)
=
2
2
π
1
[
e
−
2
1
(
x
+
1
)
2
+
e
−
2
1
(
x
−
1
)
2
]
D
f
X
(
x
)
=
1
2
π
e
−
x
2
f_X(x)=\frac1{2\sqrt{\pi}}e^{-x^2}
f
X
(
x
)
=
2
π
1
e
−
x
2
Save
Read
Check
Details
