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Let X be a continuous random variable with p.d.f. f(x) = kx(1-x), 0b).. find k and determine anumber b such that `P(X le b) ` = `P (X ge b)` |
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Answer» `kint_0^1(x(1-x))dx=1` `k[int_0^1xdx-int_0^1x^2dx]=1` `k[x^2/2--x^3/3]_0^1` `k[1/2-1/3]=1` `k=6` `P(0,b)=P(b,1)` `6int_0^b(x-x^2)dx=1/2` `6[x^2/2-x^3/3]_0^b=1/2` `b^2/2-b^3/3=1/12` `4b^3-6b^2+1=0` `When b=1/2` `4/8-6/4+1=0` `4b^3-6b^2+1=(2b-1)(2b^2-2b-1)` `2b-1=0` `b=1/2` or `2b^2-2b-1=0` `b=(2pmsqrt12)/4` `=1/2pmsqrt3/2` Which is not possible so, `1/2+sqrt3/2>1,1/2-sqrt3/2<0`. |
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