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Zero's question at Yahoo! Answers regarding interpolating to find a z-score

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MarkFL

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Feb 24, 2012
13,775
Here is the question:

Z score approximation ?

How do you approximate a z score when you have something like 2.3756


Additional Details: how do you interlope ?
Here is a link to the question:

Z score approximation ? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

Administrator
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Feb 24, 2012
13,775
Re: Zero's quastion at Yahoo! answers regarding interpolating to find a z-score

Hello Zero,

Consulting a table, we find that the $z$-score associated with an area of 2.37 is 0.4911 and for an area of 2.38 is 0.4913.

To interpolate, we may simply use the ratio:

\(\displaystyle \frac{2.38-2.37}{0.4913-0.4911}=\frac{2.3756-2.37}{\Delta z}\)

\(\displaystyle \frac{0.01}{0.0002}=\frac{0.0056}{\Delta z}\)

\(\displaystyle 50=\frac{7}{1250\Delta z}\)

\(\displaystyle \Delta z=\frac{7}{62500}=0.000112\)

And so, we may state:

\(\displaystyle z\approx0.4911+\Delta z=0.491212\)

Using numeric integration, we find:

\(\displaystyle z=\frac{1}{\sqrt{2\pi}}\int_0^{2.3756}e^{-\frac{x^2}{2}}\,dx\approx0.49124\)

To Zero and any other guests viewing this topic, I invite and encourage you to register and post any other normal distribution question in our Basic Probability and Statistics forum.

Best Regards,

Mark.
 

zzephod

Well-known member
Feb 3, 2013
134
Re: Zero's quastion at Yahoo! answers regarding interpolating to find a z-score

Hello Zero,

Consulting a table, we find that the $z$-score associated with an area of 2.37 is 0.4911 and for an area of 2.38 is 0.4913.

To interpolate, we may simply use the ratio:

\(\displaystyle \frac{2.38-2.37}{0.4913-0.4911}=\frac{2.3756-2.37}{\Delta z}\)

\(\displaystyle \frac{0.01}{0.0002}=\frac{0.0056}{\Delta z}\)

\(\displaystyle 50=\frac{7}{1250\Delta z}\)

\(\displaystyle \Delta z=\frac{7}{62500}=0.000112\)

And so, we may state:

\(\displaystyle z\approx0.4911+\Delta z=0.491212\)

Using numeric integration, we find:

\(\displaystyle z=\frac{1}{\sqrt{2\pi}}\int_0^{2.3756}e^{-\frac{x^2}{2}}\,dx\approx0.49124\)

To Zero and any other guests viewing this topic, I invite and encourage you to register and post any other normal distribution question in our Basic Probability and Statistics forum.

Best Regards,

Mark.
There is something wrong here, probabilities (areas) are in the range 0 to 1 and z scores can be any real number, here they look the other way around. (Which is what the integral at the end would be saying if the z on the left hand side were replaced with p and the variable of integration were z.)

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