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(i) Let x be a random variable which takes the value  x 1 , x 2 , x 3 ,   x 4 ,   p ( x = x 4 )   = 0 . 5 ,   p x > x 2 = 0 . 8   a n d   i f     p x = x 1 ,   p ( x = x 3 )    are A.P. Find the probability distribution. 

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Given, px=x4=0.5=12Also, p(x>x2)=0.8px=x3+px=x4=0.8px=x3+0.5=0.8px=x3=0.3=310    ...(1)Also,px=x1+px=x2+px=x3+px=x4=1px=x1+px=x2+0.8=1   px=x3+px=x4=0.8px=x1+px=x2=0.2px=x1=0.2-px=x2     ...(2)Also, px=x1,px=x2,px=x3 are in A.PSo, 2px=x2=px=x1+px=x32px=x2=0.2-px=x2+0.3      [using 1 and 2]3px=x2=0.5px=x2=16Put the value of px=x2 in (2), we getpx=x1=0.2-16px=x1=130
Probability distribution is 
xi x1 x2 x3 x4
p(xi) 130 16 310 12

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