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Normal distribution

  1. Mar 8, 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) = 0.2


    2. Relevant equations

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

    3. The attempt at a solution

    P(X < 10) - P(X < x) = 0.2
    P(Z < (10-10)/2) - P(Z < (x-10)/2) = 0.2
    P(Z < 0) - P(Z < (x-10)/2) = 0.2
    0.5 - P(Z < (x-10)/2) = 0.2
    P(Z < (x-10)/2) = 0.3
    (x-10)/2 = 0.617911
    x - 10 = 1.235822
    x = 11.235822

    The answer doesn't make sense. x is supposed to be smaller than 10.
     
    Last edited: Mar 8, 2009
  2. jcsd
  3. Mar 8, 2009 #2

    Mark44

    Staff: Mentor

    P(x < X < 10) = 0.2 <==> P(z < Z < 0) = 0.2, where Z is the usual standard normal distribution.

    From a table of areas under the standard normal distribution, I find a z value of about -.525.

    z = (x - 10)/2 ==> 2z = x - 10 ==> x = 2z + 10

    Substituting the value of z = -.525 that I found earlier, I get an x value of 8.95.

    So, P(8.95 < X < 10) = 0.2, approximately
     
  4. Mar 8, 2009 #3
    So when I tried to solve the problem, what did I do wrong?
     
  5. Mar 8, 2009 #4

    Mark44

    Staff: Mentor

    Your error is in the next line. Your z-value (which is what you're getting from the table) should be negative. Apparently you picked the wrong value. If you recall, my value was -.525.
    In working these kinds of problems, I find that it is much easier to switch right away to a probability involving z (or t, or whatever), do my calculation and look up the value, and then change back to the original random variable X.

    Mark
     
  6. Mar 8, 2009 #5
    I checked my book again, and it shows that the z value for 0.3 is 0.511967. Is that incorrect?
     
  7. Mar 8, 2009 #6

    Mark44

    Staff: Mentor

    I think you might be using your table incorrectly. The table I used has probability values (areas under the bell curve) only to 4 decimal places, but that's just a detail.

    For z = 0.3, my table shows a probability of 0.6179.
    For z = 1.0, it shows 0.8413.
     
  8. Mar 8, 2009 #7
    Yes, I was looking at the table incorrectly. For z = 0.3, the probability is 0.617911. This is similar to what you got and it's what I originally found.

    So for z = 0.3, if the probability is 0.617911, then shouldn't the following steps be correct?

    P(Z < (x-10)/2) = 0.3
    (x-10)/2 = 0.617911
     
  9. Mar 8, 2009 #8

    Mark44

    Staff: Mentor

    No. Since the probability P(Z < zp) = 0.3 (zp is the particular z value you're looking for), you have to be looking for a z-value in the left half of the bell curve. IOW, for negative values of z. Keep in mind that for z = 0, half of the area is to the left, and half to the right.

    If you're not working with a sketch of the bell curve, with the area you want shaded in, you should be.
     
  10. Mar 8, 2009 #9
    Why does it have to be in the left half of the curve? 0.3 is still a positive number. What part in my answer should I change, and how should I change it?
     
  11. Mar 8, 2009 #10

    Mark44

    Staff: Mentor

    Because you want P(Z < zp) = 0.3. This probability represents the area under the bell curve from z = [itex]-\infty[/itex] to some z value to the left of zero. If the inequality went the other way, as in P(Z > zp) = 0.3, you would be looking for a positive z-value.

    If you had to solve P(Z < zp) = .5, what would you get for what I'm calling zp?
     
  12. Mar 8, 2009 #11
    For P(Z < zp) = .5
    zp = 0.691462
     
  13. Mar 8, 2009 #12

    Mark44

    Staff: Mentor

    No. zp = 0
    I don't think you get the connection between probability and area under the bell curve. For example, P(-1 < Z < 0) represents the area under the curve between z = -1 and z = 0. The area/probability is about .34.
     
  14. Mar 8, 2009 #13
    I thought that the probability is the area under the curve.

    Although now I can see how you're using the table. So I guess for P(Z < zp) = .3, then zp = -0.53 or -0.52?
     
  15. Mar 8, 2009 #14

    Mark44

    Staff: Mentor

    Yes, probability is the area under the curve, but if you have a probability like P(Z< a), the probability is the area under the curve between z = [itex]-\infty[/itex] and z = a. If it's a probability like P(a < Z < b), it's the area under the curve between z = a and z = b. Finally, for a probability like P(Z > b), it's the area under the curve from z = a to z = [itex]\infty[/itex].

    As for the values, if you recall, I first said that it was about -.525.
     
  16. Mar 9, 2009 #15
    What I was asking is how do you know if the answer is -0.53 or -0.52? The z value doesn't correspond to one of those numbers but rather a number in between. Which value do I use, or does it not matter?
     
  17. Mar 9, 2009 #16

    Mark44

    Staff: Mentor

    Since the probability I was looking for was about midway between those numbers, I interpolated to get -.525, which is a better choice than either -.52 or -.53.

    If you want to find out more about this, do a search on "linear interpolation."
     
  18. Mar 9, 2009 #17
    I think I've pretty much got this problem figured out now. Thanks!
     
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