# Instantaneous Z-Score Value

$$P(z=2)$$

I think that this is 0, as z-score is the area under the curve and could be found by integrating the PDF over the range. As the range is 0, the area is also zero. Is that correct?

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Since $P(z=2)$ is the same as

$$\frac1{\sqrt{2\pi}}\int_2^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$$

then yes, the result is zero; this can be interpreted also as saying

$$\frac1{\sqrt{2\pi}}\int_{-\infty}^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u-\frac1{\sqrt{2\pi}}\int_{-\infty}^2 \exp\left(-\frac{u^2}{2}\right)\mathrm{d}u$$

via the Fundamental Theorem of Calculus.

Geometrically, the "integral" covers no area on the Gaussian curve, which agrees with the result of $0$.

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It isn't P(0<z<2) but rather z at exactly 2 so it would be the integral from 2 to 2 rather than from 0 to 2. –  stimms Aug 29 '10 at 1:52
Thanks for your help –  stimms Aug 29 '10 at 2:11
No problem, stimms, thank you for clarifying what you wanted to see. :) –  Ｊ. Ｍ. Aug 29 '10 at 2:15