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Phi- normal distribution (how to look normal tables )

  1. Dec 11, 2004 #1
    Phi- normal distribution (how to look normal tables!!!)

    hello, can anyone please tell me how to look up values for the following from the "normal table" distribution.

    [tex] \phi^-1(0.25) [/tex]


    ans. is -0.68 but i can't figure out how the **** it is so!!!

    so please someone reply fast 'cause this simple thing is unnecessarily wasting my time!!!
     
  2. jcsd
  3. Dec 11, 2004 #2

    HallsofIvy

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    A nice online table for the Normal distribution is at
    http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/normaltable.html

    To find φ-1(0.25), I would first note that the table,as shown, only gives positive values for z. That is, F(0)= 0 and F(infinity)= 0.5. The standard φ (normal function) has domain -infinity to infinity and φ(0)= 0.5.

    To find φ-1(0.25) then, I note that I want the area under the normal curve from -infinity to z (a negative number since 0.25< 0.5) to be 0.25. Because of symmetry, the area under the normal curve from -z (a positive number) to +infinity is also 0.25. That means that the area from 0 to -z (which is what the table gives) must be 0.5- 0.25= 0.25 also. Since I am looking for φ-1, I look in the body of the table to find the number closest to 0.25 and find that it is in the row with 0.6 on the left in the column with 0.08 at the top (actually 0.25 would be between 0.07 and 0.08) that tells me that -z= 0.68 and so z= -0.68.

    That can be a little misleading because of the coincidence that 0.5- 0.25= 0.25 Here's a little different problem: what is φ-1(0.3)?

    That is: we want to find z such that the area under the normal curve from -infinity to z is 0.3. Because of symmetry, the area from -z to infinity must also be 0.3.
    Since the table tells us the area under the curve from 0 to -z, we must look up the z for 0.5- 0.3= 0.2.

    Looking in the table, I see that closest number to 0.2, 0.1985 is in the row with 0.5 on the left and in the column headed by 0.02. -z= 0.52 so z is -0.52.

    φ-1(0.3)= -0.52.
     
  4. Dec 11, 2004 #3
    hi thanks for the reply,

    but the table in our course is a "normal table" not "standard normal" ,

    so in my table it starts with 0.0 | 0.5.

    so when i look value for 0.0 i get 0.5 in the body.

    can you please explain in terms of the table that i am having where we don't have to add or subtract 0.5 to get the answer.

    thanks!

    instead of explain if you can use LaTex and find ans. like solving a sum would be better.

    once again thanks!
     
  5. Dec 12, 2004 #4

    HallsofIvy

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    "but the table in our course is a "normal table" not "standard normal" ,

    so in my table it starts with 0.0 | 0.5.

    so when i look value for 0.0 i get 0.5 in the body."

    Okay, I assume, then, that the only problem is that your table does not give negative z values.

    To find z such that &phi;(z)= 0.25, you are looking for a z such that the area under the Normal curve to the left of z is 0.25. By symmetry, this is the same as area to the right of -z. Since your table starts with &phi;(0)= 0.5 up to &phi;(infinity)= 1, it is already including the left half. You need to find z such that &Phi;(z)= 1- 0.25= 0.75.
    Look in the body of the table for 0.75 (or the closest number to that) and read off the corresponding z value. If &Phi;(z)= 0.75, &Phi;(-z)= 0.25.
     
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