+ Reply to Thread
Results 1 to 2 of 2

Thread: Please help! ***

  1. #1
    Points: 5,339, Level: 46
    Level completed: 95%, Points required for next Level: 11

    Posts
    16
    Thanks
    0
    Thanked 0 Times in 0 Posts

    Exclamation Please help! ***




    I'm having a difficult time figuring this out... confused on how to go about solving, & which formula/steps to use....


    A population has a mean of 180 and a standard deviation of 24. A sample of 64 observations will be taken. The probability that the mean from that sample will be between 183 and 186 is?
    Last edited by ecohelp; 10-06-2005 at 12:14 AM.

  2. #2
    Admin
    Points: 20,637, Level: 90
    Level completed: 58%, Points required for next Level: 213
    quark's Avatar
    Location
    Canada
    Posts
    479
    Thanks
    25
    Thanked 196 Times in 73 Posts

    Hi ecohelp,

    First you want to find se_Xbar, the standard error of sample mean.

    se_Xbar
    =sd/sqrt(n)
    =24/sqrt(64)
    =3

    Then you can standardize the sample mean and find the probability.

    P(183 < Xbar < 186)
    =P[(183-mu)/se_Xbar < (Xbar-mu)/se_Xbar < (186-mu)/se_Xbar]
    =P[(183-180)/3 < Z < (186-180)/3]
    =...

    Let me know if you have further questions.

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats