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Solving a normal distribution problem

  1. Mar 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Assume X is normally distributed with a mean of 10 and a standard deviation of 2. Determine the value for x that solves P(-x < X - 10 < x) = 0.95


    2. Relevant equations

    P(X < x) = P(Z < z) = P(Z < (x - mean)/stdev)

    3. The attempt at a solution

    P(10 -x < X < 10 + x) = 0.95
    P(X < 10 + x) - P(X < 10 - x) = 0.95
    P(Z < (10+x-10)/2) - P(Z < (10-x-10)/2) = 0.95
    P(Z < x/2) - P(Z < -x/2) = 0.95

    That's as far as I get, and then I don't know what to do next, or if what I did was correct.
     
  2. jcsd
  3. Mar 9, 2009 #2
    I hope you have some normal tables for this. You are making some mistakes in that P(X < 10 + x) - P(X < 10 - x) = 0.95 isn't true. First of all, what is the distribution of X-10?
    Can you use symmetry in any way to simplify P(-x < X - 10 < x)? After that can you make your distribution N(0,1)?

    If you have any problems on these then give me a shout :)
     
  4. Mar 10, 2009 #3
    Is P(-x < X - 10 < x) the same as P(10 -x < X < 10 + x)?
     
  5. Mar 10, 2009 #4

    Mark44

    Staff: Mentor

    Yes, sure.
     
  6. Mar 10, 2009 #5
    Hi Mark. Is P(10 -x < X < 10 + x), the same as P(X < 10 + x) - P(X < 10 - x)?
     
  7. Mar 10, 2009 #6

    Mark44

    Staff: Mentor

    Yes.
     
  8. Mar 10, 2009 #7
    So if P(10 -x < X < 10 + x) is the same as P(X < 10 + x) - P(X < 10 - x), then if P(-x < X - 10 < x) = 0.95 then does that mean that P(X < 10 + x) - P(X < 10 - x) = 0.95?
     
  9. Mar 10, 2009 #8

    Mark44

    Staff: Mentor

    That works for me. I got into this late, but I can't see how Focus's comment in post 2 could be correct.
     
  10. Mar 10, 2009 #9
    So I guess that what I have down so far is correct, right? It appears that there is symmetry where I left off at P(Z < x/2) - P(Z < -x/2) = 0.95. Could I then do the following?

    2[P(Z < x/2) - P(Z < 0)] = 0.95
     
  11. Mar 10, 2009 #10

    Mark44

    Staff: Mentor

    Absolutely. Did you make this realization after making a sketch of the bell curve?
     
  12. Mar 10, 2009 #11
    Yeah, -x/2 and x/2 are like -a and a. I drew those points on a bell curve and saw the symmetry.

    So would the following be correct?

    2[P(Z < x/2) - P(Z < 0)] = 0.95
    P(Z < x/2) - P(Z < 0) = 0.475
    P(Z < x/2) - 0.5 = 0.475
    P(Z < x/2) = 0.975
    x/2 = 1.96
    x = 3.92
     
  13. Mar 10, 2009 #12

    Mark44

    Staff: Mentor

    Looks good. There are some rules of thumb you can use for N(0, 1) distributions:
    Probability within 1 sigma of the mean (i.e., -1 < z < 1), [itex]\approx[/itex] .68
    Probability within 2 sigmas of the mean, [itex]\approx[/itex] .95
    Probability within 3 sigmas of the mean, [itex]\approx[/itex] .99+

    Your x values are a little under 2 s.d. away, so the probability/area should be around .95, which is right on the money for your problem.
     
  14. Mar 10, 2009 #13
    What's an N(0, 1) distribution? Does that just mean normal distribution? What does the 0 and the 1 mean?
     
  15. Mar 10, 2009 #14

    Mark44

    Staff: Mentor

    Sorry. That is just notation for the standard normal distribution: mean = 0, and standard deviation = 1. Your original distribution was N(10, 2).
     
  16. Mar 10, 2009 #15
    So can I use that checking probability trick for N(0,1) distributions for the N(10,2) distribution? Here the z value is 1.96, which is just under 1 s.d. away.
     
  17. Mar 10, 2009 #16

    Mark44

    Staff: Mentor

    The idea is that you can transform a random variable X that is N(mean, sd) to a z distribution (i.e., N(0, 1)) in this way:

    z = (x - mean)/s.d.

    You can go the other way, too:
    x = z*s.d. + mean

    Any probability written in terms of X can be rewritten in terms of z, which is preferable, since there are tables of z values.
     
  18. Mar 10, 2009 #17
    You are making this much too complicated. You can't justify the step for P(X < 10 + x) - P(X < 10 - x) = 0.95. X-10 is normal with mean zero and standard deviation 2. You can use the symmetry about zero to say thats the same as 2P(X-10<x). I'll let you finish it off :)
     
  19. Mar 10, 2009 #18

    Mark44

    Staff: Mentor

    Sure he can. The expression above represents the probability that X lies between 10 - x and 10 + x.
    We're way past that. You must not have read the other posts in this thread.
    If you think about it at all, 2P(X - 10 < x) is larger than 1 if x < 10, so this is not a helpful start.
     
  20. Mar 10, 2009 #19
    Sorry you are right I'm way off the ball, I meant to say 2P(X - 10 < x)-1.
     
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