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#1 2012-12-02 07:50:36
- genericname
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- Registered: 2012-05-16
- Posts: 52
What is the formula for finding Z in this case?
Hi, the problem is:
Find Z.025; that is find Z which cuts off a right-hand tail probability 0.025
The "Z.025" part looks like this:
http://i.imgur.com/CnZdc.png (didn't know how to type it out)
The answer is .70, but I don't know how it was calculated.
Last edited by genericname (2012-12-02 07:59:54)
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#2 2012-12-02 10:21:42
- Bob
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- Registered: 2010-06-20
- Posts: 10,626
Re: What is the formula for finding Z in this case?
hi genericname,
I'm assuming this is a normal distribution problem.
I cannot make those figures (P = 0.025 and Z = 0.7) go together. Is there more to this problem or have I got the wrong model?
Have a look at the interactive normal curve / table at
http://www.mathsisfun.com/data/standard … table.html
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob 
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#3 2012-12-02 10:41:20
- genericname
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- Registered: 2012-05-16
- Posts: 52
Re: What is the formula for finding Z in this case?
Ah, the answers were mixed up. The answer is actually 1.96. How did they get that?
Last edited by genericname (2012-12-02 10:49:03)
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#4 2012-12-02 11:29:30
- Bob
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- Registered: 2010-06-20
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Re: What is the formula for finding Z in this case?
Arrrhhh! That's more like it. See picture of normal distribution graph.
And further down that page you'll find that 1.96 gives the right probability.
Values for the table cannot be easily calculated which is why people use the table.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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#5 2012-12-02 13:58:03
- genericname
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- Registered: 2012-05-16
- Posts: 52
Re: What is the formula for finding Z in this case?
Oh! I see. As always, thanks!
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