The time (in minutes) for a lab assistant to prepare the equipment for a certain experiment is a random variable taking values between 25 and 35 minutes with p.d.f `f(x) = {{:((1)/(10), 25 le x le 35),(0, "otherwise"):}` What is the probability that preparation time exceeds 33 minutes ? . Also find the c.d.f of X.

Explanation

Required probability `= P(X gt 33)`
`= underset(33)overset(oo)intf(x)dx`
`=underset(33)overset(35)intf(x) dx + 0`
`=underset(33)overset(35)int f(x)dx + 0" "......[because f(x) = 0, "where" x gt 35]`
` = underset(33)overset(35)int(1)/(10)dx = (1)/(10)underset(33)overset(35)int 1 dx`
`=(1)/(10)[x]_(33)^(35) = (1)/(10)[35-33] = (2)/(10) =(1)/(5)`
Let F(x) be the c.d.f of X
`therefore F(x) = P[X le x]`
`=underset(-oo)overset(oo) intf(x)dx`
`= underset(-oo)overset(oo)intf(x) dx +underset(25)overset(x) f(x)dx`
`=0+underset(25)overset(x)intf(x) dx " "......[becausef(x) =0 , "Where"f(x) lt 25]`
` = underset(25)overset(x)int(1)/(10)dx = (1)/(10)underset(25)overset(x)int 1dx`
` = (1)/(10)[x]_(25)^(x)= (1)/(10)(x - 25)`
`therefore F(x) = (x-25)/(10)`