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

Question

f(y) = { c(y - y^2) 0<=y<=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

Answer 1

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

Answer 2

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<0
= 3x^2 - 2x^3, 0 =1, 1

Answer 3

Integrate it...