Asked by assma
Let x1,x2,x3,x4,x5 be random sample of f(x)=(x+1)/2,-1<x<1,zero elsewhere, find P(four of{x1,x2,x3,x4,x5}>0)
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Answered by
MathMate
Given f(x)=(x+1)/2, -1<x<1 and zero elsewhere.
So for each individual sample,
P(X>0)=∫f(x)dx from 0 to 1
=∫(1/2)(x+1)dx
=(1/2)[x²/2+x] 0 to 1
=3/4 = p
Assume x1...x5 are independent and random samples, binomial theorem is appropriate:
P(four of 5 X >0)
=(5,4)p^4(1-p)^1
where (5,4) is combination 5 choose 4, or
(5,4)=5!/((5-4)!4!)=5
and p=3/4, (1-p)=1/4.
The answer should be around 0.4.
So for each individual sample,
P(X>0)=∫f(x)dx from 0 to 1
=∫(1/2)(x+1)dx
=(1/2)[x²/2+x] 0 to 1
=3/4 = p
Assume x1...x5 are independent and random samples, binomial theorem is appropriate:
P(four of 5 X >0)
=(5,4)p^4(1-p)^1
where (5,4) is combination 5 choose 4, or
(5,4)=5!/((5-4)!4!)=5
and p=3/4, (1-p)=1/4.
The answer should be around 0.4.
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