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ruth_lou
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#1
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A continuous random variable X has p.d.f defined by:

f(x) = cx/3 0<x<3
C 3<x<4
0 otherwise
a) find the value of c
b) the mean value of x
c) the value ''a'' for there to be a probability of 0.15 that a randomnly chosen value of X will exceed ''a''

im bit confused esp on part C!!! :eek:
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amo1
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(Original post by ruth_lou)
A continuous random variable X has p.d.f defined by:

f(x) = cx/3 0<x<3
C 3<x<4
0 otherwise
a) find the value of c
b) the mean value of x
c) the value ''a'' for there to be a probability of 0.15 that a randomnly chosen value of X will exceed ''a''

im bit confused esp on part C!!! :eek:
a) intergrate between 0 and 3 take{ to be intergral sign on with those limits
{f(x)dx=1
{cx/3 dx=1
c[(x^2)/6]=1
c=2/3
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amo1
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b) mean in same as above but E(X)= { xf(x) dx
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amo1
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c/ so here u need F(x) u intergrate f(x) between 0 and x

get a term in x of x^2/4 and u then make P(x<a)=0.15 and work it out from there
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ruth_lou
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(Original post by amo1)
c/ so here u need F(x) u intergrate f(x) between 0 and x

get a term in x of x^2/4 and u then make P(x<a)=0.15 and work it out from there
well ive integrated but i get like x^2/9
where have i gone wrong?
and where does x^2/4 come from!!!
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amo1
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(Original post by ruth_lou)
well ive integrated but i get like x^2/9
where have i gone wrong?
and where does x^2/4 come from!!!
sorry i just checked it i times by 1.5 not 2/3 but nyways ur right.... i am not soing maths at uni... goes into corner to reaccess life
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ruth_lou
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#7
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phew aww thanx!
what do i do then with the a to find 0.15?
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ruth_lou
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#8
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is the answer 1.16??
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