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Question

If the p.d.f. of a continuous r.v. X is f(x)=320x(x−2),−1<x<4=0,otherwise then the probability that X is negative, is?

A
1/5
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B
2/5
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C
3/5
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D
4/5
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Solution

The correct option is A 1/5
Given, p.d.f is given by f(x)={320x(x2),1<x<4}
{0,otherwise}
for c.d.f.
(i) If x(,1)
f(x)=xf(x)dx
=0

(ii) If x(1,4)
f(x)=xf(x)dx
=10dx+x1320(x22x)dx
=320[x33x2]σ1
=320[x33x2][320(131)]
=320[x33x2]+15

(iii) If x(4,),f(x)=1
p(x<0)=f(0)f(0)
=150
=15
Hence, the answer is 15.

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