Question 252363: Please help me with this problem. I'm not really good at probability and statistics. A quick response would be nice. Thank you.
Given a PDF of variable x:
0 if x<-1
f(x)= ax+a if -1=
-ax+a if 0=
0 if x>1
a)Find a?
b)Find the probability that the random variable X has values within the interval (1/2,1) and within the interval (-1/3,1/3)?
c)Find P(X=1/2)?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Given a PDF of variable x:
0 if x<-1
f(x)= ax+a if -1<= x <0
-ax+a if 0<= x <=1
0 if x>1
a)Find a?
Solvd:
Integral from -1 to 0 of (ax+a) + Integral from 0 to 1 of (-ax+a) = 1
[(a/2)x^2+ax evaluated -1 to 0 + (-a/2)x^2+ax evaluated 0 to 1] = 1
[(a/2)0 + a*0]-[(a/2)(-1)^2+a(-1)] + [(-a/2)(1^2)+a(1)]-[(-a/2)0+a*0]= 1
-(a/2)+a + (-a/2)+a = 1
a = 1
========================================================================
So now f(x) is defined as
Given a PDF of variable x:
0 if x<-1
f(x)= x+1 if -1<= x <0
-x+1 if 0<= x <=1
0 if x>1
==========================================
b)Find the probability that the random variable X has values within the interval (1/2,1) and within the interval (-1/3,1/3)?
P(1/2
= -(x^2/2)+x eval btwn 1/2 and 1
= [-(1/2)+1]-[-(1/4)/2 + (1/2)]
= [(1/2)]-[(-1/8)+(1/2)]
= (1/2) - (3/8)
= 1/8
============================================
P(-1/3 < x < 1/3) =
Int(x+1)eval -1/3 to 0 + Int(-x+1) eval 0 to 1/3
Then same procedure as above.
Comment: Since these two areas are symmetric you
could just evaluate one side, then double it.
============================================
c)Find P(X=1/2)?
The probability of any particular value in a continuous
distribution is always zero.
Your problem would read Int(-x+1) eval from 1/2 to 1/2
That is zero.
===================
Cheers,
Stan H.
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