The cdf of checkout duration X for a book on a 2hour reserve

The cdf of checkout duration X for a book on a 2-hour reserve at a college library is given by: F(x) = {0 x lt 0 x^2/4 0 lessthanorequalto 2 1 2 lessthanorequalto x Use this cdf to compute the following: P(X lessthanorequalto 1) P(.5 lessthanorequalto X lessthanorequalto 1) P(X gt.5) The median checkout duration mu Fprime(x) to obmin the density function f(x)

Solution

(a)

P(X 1) = F(1)

= (1/4) x² dx --- integrate from 0 to 1 [formula :- xⁿ = x^{n+1/ n+1}]

= x³/12

= 1 /12 - 0 /12 = 1/12

b)
P(.5 X 1) = F(1) - F(.5)

we already know that F(1) = 1/12

Now F(.5) = (1/4) x² dx --- integrate from 0 to .5

= (.5)³ /12 = 0.125/12

Now substitute in the given relation then

F(3) - F(.5) = 1/12 - 0.125/12 = 0.0729

c)
P( x > .5) = 1 - P(x .5) = F(.5)

= (1/4) x² dx --- integrate from 0 to .5
= (.5)³ / 12 = 0.125/12

d)
(1/4) x² dx from 0 to M is 0.5

M³/12 = 0.5 ,

where M =median

M³ = (12)(0.5) = 6

M =1.817

e)

F(x) = x²/4 , 0 x 2

F\'(x) = 2x/4 , 0 x 2

f(x) = (2/4) x , 0 x 2