Answered: PROBABILITY AND STATISTICS (COLLEGE…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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PROBABILITY AND STATISTICS (COLLEGE LVL)

I HAVE SOME CONFUSION WITH THE QUESTION: Transformation: Given the f(x) below determine the pdf of U where U = Y + 2, f(y) = 2y^2, for 0 < y ≤ 2 and zero elsewhere.

I DON’T UNDERSTAND WHICH IS THE CORRECT FINAL ANSWER FOR THE LIMIT OF f(u), BY CHEGG (ANS: LIMIT 2 < U ≤ 4) OR MY FINAL ANSWER (ANS: LIMIT 2 < U ≤ 3.144) I WROTE THIS ANSWER BCZ PROFESSOR SAID “SINCE f(u) is the derivative of F(U), THEN THE U-LIMITS WILL DIFFER SLIGHTLY”

SO PLEASE CHECK EACH AND EVERY STEP OF MY ANSWER AND CORRECT ANYTHING THAT IS A MISTAKE ( I ALSO MEAN THE SYMBOL OF INEQUALITIES WHETHER “ <, >, ≤, ≥” )AND ALSO PLEASE STRICTLY FOLLOW/STICK WITH MY METHOD.

PLEASE HELP ME CLEAR THIS DOUBT.

f(u) is pdf

F(U) is cumulative

My SOLUTION

Set = 1
O set =0
-2)
(U-2)°= 0
(u-2)²= }
we (aun
skip
this
Simple
Algebra
U-2 =0
Cube rout of both sides
U-2=332
U =2
So: F(U)=0,
%3D
U= 3,144
2 and all valees to
the left
So: F(U) = 1,>B.144
To the right of 3.144
F(U): 2UE.144
. The pof qf U is between 2 and 3.144
feo)-dFu)
2
2LUS 3.14
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Q) Sdution
fy)=2y
given,
Fp =2y² for OLy2
U = Y+2
ニ
OTHERWISE
Formula
F(UE
F(U) Si IF U93
Y=U-2-
F(u)<0 = 0
F(u) asusb
F(U)@>I = 1
IF 0く)
U-2
asb
U-2
(U=
().
dy
U-2
3
FU) =
25U<3
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2/2

Definition Definition Probability of occurrence of a continuous random variable within a specified range. When the value of a random variable, Y, is evaluated at a point Y=y, then the probability distribution function gives the probability that Y will take a value less than or equal to y. The probability distribution function formula for random Variable Y following the normal distribution is: F(y) = P (Y ≤ y) The value of probability distribution function for random variable lies between 0 and 1.

Expert Solution

Step 1

Given

$f (y) = 2 y^{2}, 0 < y \leq 2$

Consider,

$F (y) = \int_{0}^{y} f (t) d t = \int_{0}^{y} 2 t^{2} d t = 2 {[\frac{t^{3}}{3}]}_{0}^{y} = 2 [\frac{y^{3}}{3}] . . . . . . . . . . . . . . . . . . . . . (1)$

Given U=Y+2

Step 2

Consider,

$F (U) = P (Y + 2 \leq u) = P (Y \leq u - 2) = F_{y} (u - 2) = \frac{2}{3} {(u - 2)}^{3}$

Therefore,

$f (U) = \frac{d}{d u} F (U) = \frac{2}{3} * 3 {(u - 2)}^{2} = 2 {(u - 2)}^{2}$

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