Results 1 to 4 of 4
Like Tree2Thanks
  • 1 Post By amul28
  • 1 Post By SlipEternal

Math Help - Phi of normal distribution

  1. #1
    Member
    Joined
    Nov 2011
    Posts
    87

    Phi of normal distribution

    Hi everyone!

    So, my problem is:

    Let X be birth weight of a randomly chosen child in Norway, given in gram, g. Assume that
    X is normally distributed, where E(X) = 3315 and Var(X) = 5752
    .
    a) Calculate the following probabilities,


    1) P(X > 3000) 2) P(3000 < X < 3500) 3) P(X > 3500 | X > 3000)

    What I have done to solve it:

    So, I managed to get the right answer on 1, and I thought I would get the right answer on number 2.

    I thought: P(3000 < X < 3500) = P(X > 3000) ∩ P(X < 3500). In 1) I calculated that P(X > 3000) is 0,709, this was correct. I calculated that P(X < 3500) = 0,6235. So 0,709 x 0,6235 = 0,4421 , but the right answer is: 0,335. What am I doing wrong?

    Thanks for taking a look!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2010
    From
    Mumbai
    Posts
    91
    Thanks
    2

    Re: Phi of normal distribution

    hello,

    P(a < X < b), equations like this generally written as
    P(X < b) - P(X < a).

    hope this helps
    Thanks from Nora314
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2011
    Posts
    87

    Re: Phi of normal distribution

    Let me see, I tried taking P(x < b) - P(x < a), which in this case would be 0,6235 - 0, 709, but then the answer is negative and way off. Not sure if this is what you meant I should do
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,937
    Thanks
    785

    Re: Phi of normal distribution

    Ok, P(x<b) = P(x<a \mbox{ AND } x<b) + P(x \ge a \mbox{ AND } x < b). This is the sum principle. If a<b and x<a, then it is guaranteed that x<b. So, P(x<a \mbox{ AND } x<b) = P(x<a). Now, P(x<b) = P(x<a) + P(x \ge a \mbox{ AND }x < b) \ge P(x<a) + 0 = P(x<a). If P(x<b) \ge P(x<a), then your data is wrong. If P(x<b) = 0.6235, then P(x<a) \le 0.6235.

    Oh, I see. For your answer in part 1, you have P(x>3000) = 0.709. What is P(x<3000)? It is the compliment of P(x>3000). So, P(x<3000) = 1 - 0.709 = 0.291. Now, 0.6235 - 0.291 = 0.3325 as you wanted.
    Last edited by SlipEternal; October 8th 2013 at 10:42 AM.
    Thanks from Nora314
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 23rd 2013, 05:35 PM
  2. Replies: 1
    Last Post: March 21st 2013, 03:59 AM
  3. Replies: 2
    Last Post: March 15th 2013, 08:13 PM
  4. Replies: 2
    Last Post: March 29th 2010, 03:05 PM
  5. Replies: 1
    Last Post: April 20th 2008, 07:35 PM

Search Tags


/mathhelpforum @mathhelpforum