Suppose· X · has probability-density-function (p.d.f) fx(x) = 7(1+x²)³ ° which one is the p.d.f.. for Y=2X. A 2 fr(y)= π(4+ y²) B f(y)=- 1 π(1+4y²) fx(y)= 1 л(1+y²) 1 fr(y) == arctan y π C D
Suppose· X · has probability-density-function (p.d.f) fx(x) = 7(1+x²)³ ° which one is the p.d.f.. for Y=2X. A 2 fr(y)= π(4+ y²) B f(y)=- 1 π(1+4y²) fx(y)= 1 л(1+y²) 1 fr(y) == arctan y π C D
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![3. (单选题 3.0分)
1
Suppose X has probability density function (p.d.f) f (x)=.
π(1+x²)'
for Y=2X.+
A
2
fx(y) =
x(4+ y²)
B
fy(y)=-
1
π(1+4y²)
fr(y) =
1
π(1+ y²)
1
fr(y) == arctan y
π
C
D
which one is the p.d.f..](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbc553f1-6131-4f5e-b5e9-0aa3b55fe7e0%2F807bc06d-4e93-4845-ad6e-ca1bb6336f24%2Fizrfknk_processed.png&w=3840&q=75)
Transcribed Image Text:3. (单选题 3.0分)
1
Suppose X has probability density function (p.d.f) f (x)=.
π(1+x²)'
for Y=2X.+
A
2
fx(y) =
x(4+ y²)
B
fy(y)=-
1
π(1+4y²)
fr(y) =
1
π(1+ y²)
1
fr(y) == arctan y
π
C
D
which one is the p.d.f..
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