Banner Ad 1

Collapse

Announcement

Collapse
No announcement yet.

please help, stuck on random variable problem

Collapse
This topic is closed.
X
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • please help, stuck on random variable problem

    Can anyone explain this to me?

    Let X be a random variable with distribution function
    0 for x < 0
    x/8 for 0< or = x < 1
    1/4 + x/8 for 1 < or = x <2
    3/4 + x/12 for 2 < or = x < 3
    1 for x> or = 3

    Calculate P[1 < or = X < or = 2]

    I know the answer is 19/24, but the Broverman study guide does not explain how it came up with the answer. Should I pick up a better study guide or can someone help me out?

  • #2
    Originally posted by rletson
    Can anyone explain this to me?

    Let X be a random variable with distribution function
    0 for x < 0
    x/8 for 0< or = x < 1
    1/4 + x/8 for 1 < or = x <2
    3/4 + x/12 for 2 < or = x < 3
    1 for x> or = 3

    Calculate P[1 < or = X < or = 2]

    I know the answer is 19/24, but the Broverman study guide does not explain how it came up with the answer. Should I pick up a better study guide or can someone help me out?
    Sets: {x<=2} = {x<1}U{1<=x<=2}

    It follows that

    P{x<=2} = P{1<=x<=2} + P{x<1}, and

    P{1<=x<=2} = P{x<=2} - P{x<1}.

    Using definition of cumulative distribution function:

    F(t) = P{x<=t)
    and
    F-(t) = P{x<t}.

    P{1<=x<=2} = F(2) - F-(1).

    F-(t) is equal to F(t) in points where F(t) is continous, and is equal to limit of F(t) from the left in the points of discontinuity. Using the above formula, F(2) = 3/4+t/12 = 3/4 + 2/12, and F-(1) = t/8 = 1/8.

    P{1<=x<=2} = F(2) - F-(1) = 3/4+2/12 - 1/8 = (18+4-3)/24 = 19/24.

    Comment

    Working...
    X