Find the following probabilities for the standard normal distribution Pr(z>2.5)

$z=\frac{x-\mu }{\sigma }$
The normal distribution is symmetric about mean 0.5.
$P(z>2.5)=0.5-Pr(z\le 2.5)$
The probability that the z lies between 0 and 2.5 is equal to the area that lies under the curve from 0 and 2.5.
In the column headed by A across 2.5 the corresponding value is 0.4938.
The probability $P(0\le z\le 2.5)$ is $A_{2.5}$.
So the value of $A_{2.5}$ from the appendix C is 0.4938.

The value of $P(z>2.5)$ 
$P(z>2.5)=0.5—0.4938=0.0062$

Aswer is 0.0062.