Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JerJason

  • 2 years ago

If the probability density of a random variable is given by: f(x)= Kx^2 for 0<x<1 0 elsewhere find the value k and the probability that the random variable takes on a value between (a) 1/4 and 3/4 (b) greater than 2/3

  • This Question is Closed
  1. JerJason
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So that I'm understanding this correctly. I'm finding the value of "k" in order to verify that the function f(x)=kx^2 is a probability density function based on these conditions: f(x)>=0 and f(x) dx = 1 when x is between -infinity and infinity.

  2. JerJason
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Therefore, \[f(x)=\int\limits_{0}^{1}kx^2dx=k \frac{ x^3 }{ 3 }_{0}^{1}=k/3=1\] so k=3

  3. JerJason
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    From here I'm plugging in the limits between 1/4 (.25) and 3/4 (.75). \[\int\limits_{.25}^{.75}kx^2dx=k \frac{ x^3 }{ 3 }_{.25}^{.75}=k(\frac{ .4218-.0156 }{ 3 })=.4062/3=.1354k\]

  4. JerJason
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I have a filling I did that last step wrong.

  5. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    thats correct, 0.1354*3=0.4062 is the correct probability for a)

  6. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you know how to do part b), right ?

  7. JerJason
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ^ alright good. For (b) do I convert f(x) into F(x), which is the distribution function then plug in the limit .67 (or 2/3) for F(x) and then minus that from 1 like so? Find: \[P(x>.67)\] \[F(x)=\int\limits_{0}^{x}kx^2dx=kx^3/3\] \[F(.67)=3*.67^3/3=.902289/3=.3008\] \[1-.3008=.70\]

  8. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that is correct, but there is also another way. \(\huge \int \limits_{(2/3)}^1 kx^2dx=....\) and u get same result 0.703

  9. JerJason
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes that makes sense as well. Thanks.

  10. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    welcome ^_^

  11. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.