Results 1 to 5 of 5

Math Help - Statistics Problem. Please Help.

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    4

    Question Statistics Problem. Please Help.

    I can't seem to get the right answer for this problem...

    Given that the population of scores is normally distributed with (mean=110) and (standard deviation=8), determine the following:

    What is the percentage of scores that are between a score of 101 and 122?

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kelly3223 View Post
    I can't seem to get the right answer for this problem...

    Given that the population of scores is normally distributed with (mean=110) and (standard deviation=8), determine the following:

    What is the percentage of scores that are between a score of 101 and 122?

    If X = 101 then Z = \frac{X - \mu}{\sigma} = \frac{101 - 110}{8} = z_1.


    If X = 122 then Z = \frac{X - \mu}{\sigma} = \frac{122 - 110}{8} = z_2.

    (Your job by the way is to get the z-values ...)

    Then \Pr(101 < X < 122) = \Pr(z_1 < Z < z_2) = \Pr(Z < z_1) - \Pr(Z < z_2) = p.

    (Your job by the way is to get the value of p by calculating \Pr(Z < z_1) and \Pr(Z < z_2). I assume you've been taught how to do this using tables etc. ....)

    The percentage of scores is therefore 100p.

    If you need more help please show your working and say where you get stuck.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    4
    I'm still confused. I did the equation to find z1 and z2 and got -1.125 for z1 and 1.5 fo z2, but what do I do next to get the answer?


    Thanks for your help!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kelly3223 View Post
    I'm still confused. I did the equation to find z1 and z2 and got -1.125 for z1 and 1.5 fo z2, but what do I do next to get the answer?


    Thanks for your help!
    Pr(z < -1.125) = Pr(z > 1.125) = 1 - Pr(z < 1.125).

    Now calculate Pr(z < 1.125) (and hence Pr(z < -1.125)) and Pr(z < 1.5) and go back and read my first post for what to do.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2009
    Posts
    4

    I think I got it?

    So, 1.13 = .3708 (according to our table)
    1.50= .4332 ('' ")

    Percentage between 101 and 122: (0.3708+0.4332)x100=80.4%


    Is this correct?!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Statistics problem, can any one help?
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: December 26th 2009, 03:10 PM
  2. statistics problem
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: May 1st 2009, 04:04 PM
  3. Statistics problem
    Posted in the Statistics Forum
    Replies: 1
    Last Post: May 17th 2008, 05:38 AM
  4. Statistics Problem Please Help!
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 26th 2007, 05:32 AM
  5. Statistics problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: December 9th 2006, 10:46 PM

Search Tags


/mathhelpforum @mathhelpforum