Solution Given that, boto] (19 ·X43 1-9/3 › X73 F(x) = f(x) = 1/² F(x) = 20. XL3 cx @PIXEN)= P(OLX200) = 10 f(x) dx x=3 = [² 27 dx = 27 X 27X+ 81 27 X4 -3 1 f(x) = 81 x ²3 xt = 3 = 1/2 #1 Therefore, given put is not valid bat because I flxdx#1 Now. making valid pdf by multiplying 3, we get, * x=-1 Apien I 7 (6) P (52X²7) = √" f(x) dx x=5 =81 X73 = 81 X 81 dx X+ 0[*²] 5 1 + 1 1029 375 P(54X27) = 0.1373/ T Answer
Solution Given that, boto] (19 ·X43 1-9/3 › X73 F(x) = f(x) = 1/² F(x) = 20. XL3 cx @PIXEN)= P(OLX200) = 10 f(x) dx x=3 = [² 27 dx = 27 X 27X+ 81 27 X4 -3 1 f(x) = 81 x ²3 xt = 3 = 1/2 #1 Therefore, given put is not valid bat because I flxdx#1 Now. making valid pdf by multiplying 3, we get, * x=-1 Apien I 7 (6) P (52X²7) = √" f(x) dx x=5 =81 X73 = 81 X 81 dx X+ 0[*²] 5 1 + 1 1029 375 P(54X27) = 0.1373/ T Answer
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
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TRANSCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT
![To find P(X>E(X)) first we find the E(x)ie
CO
E(X)= √ x f(x) dx
x=-6
√x B1
3 X4
81 dx
= 81[= ²70
= 81 [ + ]
9 =4.5
2
E(X) =
Now. P(x > 4-5) = f(x) dx
✓
X=4,5
81 dx
=
4.5 x4
jos
81 [23³] 0
x
-3
14:5
nakuin stand bited Powe
=81/273.375
Ledam pd [4
P(x24.5) = 0.2963,
Answer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbca5d3c4-52d4-45d8-b039-8f47715b8c01%2F659758bb-ec92-470c-aa00-ede4748e8b28%2Fcpxkmgm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:To find P(X>E(X)) first we find the E(x)ie
CO
E(X)= √ x f(x) dx
x=-6
√x B1
3 X4
81 dx
= 81[= ²70
= 81 [ + ]
9 =4.5
2
E(X) =
Now. P(x > 4-5) = f(x) dx
✓
X=4,5
81 dx
=
4.5 x4
jos
81 [23³] 0
x
-3
14:5
nakuin stand bited Powe
=81/273.375
Ledam pd [4
P(x24.5) = 0.2963,
Answer
![Solution - Given that, fou (vt in T 2
20 .X43
F(x) =
1-
f(x) = a
dx
오
d. F(x)=
X3
=
› X7,3
@PIXEN)= P(OLX200)
= 100 f(x) dx
x=3
SO 27
=
Į
dx
-3 B
- 27 [1²³] 30
0. XL3
27
x4
27x11
81
= 1/32 #1
ㅎ
=
Therefore given pdf is not valid bat because of(x) dx #1
Now. making valid pdf by multiplying 3, we get, x=
X=-1
√
I
f(x) = 81₁ x ²3
7
P ( 5 ≤ x² 7) = √ + f(x) dx
x=5
X73
=81
781 dx
X+
= 81 / x²-377
5
+1
1029 375
P(52X47) = 0.1373)
Answer
KA-MANTA](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbca5d3c4-52d4-45d8-b039-8f47715b8c01%2F659758bb-ec92-470c-aa00-ede4748e8b28%2F2mt8f1s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Solution - Given that, fou (vt in T 2
20 .X43
F(x) =
1-
f(x) = a
dx
오
d. F(x)=
X3
=
› X7,3
@PIXEN)= P(OLX200)
= 100 f(x) dx
x=3
SO 27
=
Į
dx
-3 B
- 27 [1²³] 30
0. XL3
27
x4
27x11
81
= 1/32 #1
ㅎ
=
Therefore given pdf is not valid bat because of(x) dx #1
Now. making valid pdf by multiplying 3, we get, x=
X=-1
√
I
f(x) = 81₁ x ²3
7
P ( 5 ≤ x² 7) = √ + f(x) dx
x=5
X73
=81
781 dx
X+
= 81 / x²-377
5
+1
1029 375
P(52X47) = 0.1373)
Answer
KA-MANTA
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