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

[MarkFL]

Here is the question:

Z score approximation ?

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

 

[MarkFL]

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 http://www.mathhelpboards.com/f23/ forum.

Best Regards,

Mark.

 

[zzephod]

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 http://www.mathhelpboards.com/f23/ 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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