Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I had a question about an instantaneous z score. The question reads as

$$P(z=2)$$

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?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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 $0$.

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.