A continuous random variable Y has a probability density function given by?

f(y) = { c(y - y^2) 0%26lt;=y%26lt;=1


{ 0 otherwise





(a) Determine the value of c.


(b) Calculate (i) P( .6 ≤Y ≤ .9 ), (ii) P( Y ≥ .3 ).


(c) Calculate the mean and variance of Y.


(d) Determine the cumulative distribution function for the random variable.








no idea what to do

A continuous random variable Y has a probability density function given by?
a) value of c = Integral from 0 - to c of f(y) dy





b) use th4e information you have from a)





c) mean : integral from 0 - 1 of f(y) dy / lentgh of interval which is 1





d) pffff
Reply:Integrate it...
Reply:Let Int mean the integral of a function, then:





a) We need total probability = 1, so Int[f(y) from y=0 to y=1] = 1.





b) Again, using integration.





c) E[Y] = Int[ y * f(y)]





VAR[Y] = Int[ y^2 * f(y)] - ( E[Y] ) ^ 2





d) the CDF si given by g(x) = Int[ f(y) from y=-infinity to y = x ]





Steve





Answers:





a) c=6





c) expected value = 1/2, variance = 1/20





d) CDF(x) = 0, x%26lt;0


= 3x^2 - 2x^3, 0%26lt;x%26lt;1


=1, 1%26lt;x

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