Question 2 The transformation of velocity vector, v m/s of a moving-body of mass m kg into corresponding kinetic energy (KE) is given by: KE=y= [(1/2)x mxv²] joules. Suppose v denotes a RV with its pdf specified as follows: fv(v) = av =0 in the range in the range (0≤v≤+1) otherwise And, a is a constant. Hence, determine the following: (i) Value of α; (ii) pdf of y (= KE); (iii) range of the transformed random variable, y; and, (iv) sketch the pdf of y over its range determined.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.6: Variation
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Question 2
The transformation of velocity vector, v m/s of a moving-body of mass m kg into
corresponding kinetic energy (KE) is given by: KE=y= [(1/2)x mxv²] joules. Suppose
v denotes a RV with its pdf specified as follows:
fv(v) = av
=0
in the range in the range (0≤v≤+1)
otherwise
And, a is a constant. Hence, determine the following: (i) Value of α; (ii) pdf of y (=
KE); (iii) range of the transformed random variable, y; and, (iv) sketch the pdf of y
over its range determined.
Transcribed Image Text:Question 2 The transformation of velocity vector, v m/s of a moving-body of mass m kg into corresponding kinetic energy (KE) is given by: KE=y= [(1/2)x mxv²] joules. Suppose v denotes a RV with its pdf specified as follows: fv(v) = av =0 in the range in the range (0≤v≤+1) otherwise And, a is a constant. Hence, determine the following: (i) Value of α; (ii) pdf of y (= KE); (iii) range of the transformed random variable, y; and, (iv) sketch the pdf of y over its range determined.
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