Here is a link to 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 ?
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.)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.