continous random variable questionsWatch

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Thread starter 14 years ago
#1
1. f(x)= 3/16 (4-x^2) 0<= x <=2
0 otherwise

Use this model to find the probability that x>100. ans=5/16

my other question is, when you find the c.d.f or p.d.f, how do you define it or how'd you know what he range is?
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14 years ago
#2
(Original post by hotboi01)
1. f(x)= 3/16 (4-x^2) 0<= x <=2
0 otherwise

Use this model to find the probability that x>100. ans=5/16

my other question is, when you find the c.d.f or p.d.f, how do you define it or how'd you know what he range is?
If that is a pdf and is zero for x>2 then the probability that x>100 will be 0 as it's the integral from 100->infinity of this pdf. Have you copied this down right?
0
14 years ago
#3
(Original post by hotboi01)
1. f(x)= 3/16 (4-x^2) 0<= x <=2
0 otherwise

Use this model to find the probability that x>100. ans=5/16

my other question is, when you find the c.d.f or p.d.f, how do you define it or how'd you know what he range is?
I guess, looking at it now that you mean prob that x>1 as I get

Int[1->2] 3/16 (4-x^2) =

3/16 [4x - x^3/3] [1->2] =

5/16.
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14 years ago
#4
(Original post by hotboi01)
my other question is, when you find the c.d.f or p.d.f, how do you define it or how'd you know what he range is?
The pdf is the derivative of the cdf.

if f(x) is a pdf its cdf F(x) is given by

F(x) = Int[-infinity->x] f(x) dx
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