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Old December 1st, 2008, 12:46 PM
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Default Z problem

I calculated my z value to be z=(23-18)/3 = 5/3 = 1.66

I am trying to find P(Z>1.6666) = ?

can someone please explain how to get this probability using the Z table?

thanks in advance

i got 0.0485

0.5 - 0.4515
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Old December 1st, 2008, 01:01 PM
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Hello,
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Originally Posted by ndcruz View Post
I calculated my z value to be z=(23-18)/3 = 5/3 = 1.66

I am trying to find P(Z>1.6666) = ?

can someone please explain how to get this probability using the Z table?

thanks in advance

i got 0.0485

0.5 - 0.4515
Well, if this is what you got by using the Z table, then you know how to use it, because your answer is correct...

The z table represents the area bounded by the Gaussian curve, the x axis, the y axis and the line x=a, where a is the z value of the table.

So in fact, tables give you this probability : P(0<z<a)

Let a>0 and b<0. Let z_a and z_b be the z scores of a and b.
P(z<a)=0.5+z_a
P(z>a)=0.5-z_a
P(z>b)=0.5+z_b
P(z<b)=0.5-z_b
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