Determine the following probabilities for the probability density function 
f(x)=kx^2 for 0<x<4 (a)P(x>2) (b)P(1<x<3) (c)P(x<1) (d)P(x<1)+P(1<x<3)

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Clarification of Question by wjs-ga on 27 Jan 2003 20:41 PST

f(x)=3/64x^2 for 0<x<4

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Request for Question Clarification by gwagner-ga on 28 Jan 2003 00:44 PST

Hi wjs,
 
do you mean f(x) = 3/(64x^2) or f(x) = (3/64)*x^2 ?
 
Thanks,
gwagner

Hi wsj!!
First of all I must clarify the problem:
In order to satisfy the conditions to be a probability density
function f(x) must be:
f(x)=(3/64)*x^2 0 < x < 4 and f(x)=0 otherwise.
This function satisfies 
1)f(x)>=0 
and 
2)Integral between -oo and +oo is equal to 1. (The "symbol" oo means
infinitus).

(I will note Integral of f(x) between a and b by Int(f(x),a,b), we are
very limited to use the correct notation in this form).

Int(k*x^2) = k*(x^3)/3
By definition P(a<x<b) = Int(f(x),a,b), then 
P(a<x<b) = (3/64)*(b^3)/3 - (3/64)*(a^3)/3 = (b^3 - a^3)/64 for a>=0
and b<=4
and
P(a<x<b) = (b^3 - 0^3)/64 = (b^3)/64 for a<0 and b<=4
and 
P(a<x<b) = (4^3 - a^3)/64 = (64-a^3)/64 for a>=0 and b>4
and
P(a<x<b) = 1 for a<0 and b>4;


(a)P(x>2)

P(x>2) = P(2 < x < +oo) = (64 - 2^3)/64 =(64-8)/64 = 7/8;
-----------------------------------------

(b)P(1<x<3)

P(1<x<3) = (3^3 - 1^3)/64 = 26/64 = 13/32;
----------------------------------------

(c)P(x<1)

P(x<1) = P(-oo < x < 1) = 1^3/64 = 1/64;
-----------------------------------------

(d)P(x<1)+P(1<x<3)

P(x<1)+P(1<x<3) = 1/64 + 13/32 = 1/64 + 26/64 = 27/64

As you can see P(x<1)+P(1<x<3) is the probability for (x<1) OR (1<x<3)
such is the probability for (x<3) but x different to 1 then:
P(x<1)+P(1<x<3) = P(x<3) - P(x=1);
but P(x=1)=0 then:
P(x<1)+P(1<x<3) = P(x<3) = P(-oo < x < 3) = 3^3/64 = 27/64; that
confirms the previous result.
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I did this exercise based in my own knowledge, if you need a
clarification please post a request for it.
I hope this helps you.

Best Regards.
livioflores-ga