1 votes 1 votes If X is a continuous random variable whose probability density function is given by f(x) ={k(5x-2x^2) , 0<= x <= 2<br /> 0 ,otherwise} then p(x>1) is Probability random-variable + – admin asked Sep 9, 2015 admin 3.4k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes P(x>1)=1-p(x<1) integrating the given function net 0 and 2 gives k(5x^2/2-2x^3/3) Equate this to 1 to get value ok k value of k as 3/14 P(1)=3/14*11/6 =11/28 P(x>1)=1-11/28=17/28 Pooja Palod answered Sep 9, 2015 Pooja Palod comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes integrate the fun wrt x nd take the limit from 1 to 2 as the value of function is 0 for larger x. Saurav answered Sep 9, 2015 Saurav comment Share Follow See all 2 Comments See all 2 2 Comments reply admin commented Sep 9, 2015 reply Follow Share I tried that answer given as 17/28 but iam getting different answer 0 votes 0 votes Saurav commented Sep 9, 2015 reply Follow Share when u will integrete from 1 to 2 u will get a result which also contain k .The result u will get is 17/6 k. now since we know that integration of this density fun from -infinte to +infinite should be 1. integrate f(x) from -infinite to +infinite which should be 1. so,after integration the eqn will be 14k/3 = 1 ,so k=3/14 now put this k value into our required result = 17k/6 = (17/6 ) * (3/14) = 17/28 1 votes 1 votes Please log in or register to add a comment.