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The pdf of a curve X is `f(x)={{:(k/sqrtx","0ltxlt4),(0","xle0" or "xge4):}` Then , `P(Xge1)` is equal toA. 0.2B. 0.3C. 0.4D. 0.5 |
Answer» Correct Answer - D `therefore" "oversetoounderset(-oo)intf(x)dx=1` `therefore" "overset4underset0intk/sqrtxdx=1` `rArr" "k[2sqrtx]_0^4=1` `rArr" "2k(sqrt4-sqrt0)=1` `rArr" "4k=1` `therefore" "k=1/4` Now, `P(Xge1)=P(1leXlt4)=overset4underset1intf(x)dx` `=overset4underset1intk/sqrtxdx=2k[sqrtx]_1^4` `=2(1/4)(2-1)=1/2=0.5` |
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